Why does the dissolution of calcium hydroxide have a negative entropy?

When ionic compounds dissolve in water, water molecules form a complex ion around the calcium cation.  The presence of the hydroxide anion has a similar effect, though the polarities of the water ligands is reversed.
Thus, dissolution of Ca(OH)2 in water is actually a chemical reaction.  In some cases, complex ions may be more structured than the crystals of undissolved ions.       In such a case, the entropy change of solution will be negative.


What does higher electron gain enthalpy mean here?

I take for granted that what you call "electron gain enthalpy" is generally described as "electron affinity". This is the energy gained (or lost) when an electron is added to an atom, ion or molecule. In general the electron affinity is a measure for the tendency of elements to form anions. The trouble is that you may find tables giving these numerical values as positive or negative values. For example, for fluorine $\ce{F}$, the electron affinity is $\pu{338 kJ/mol}$. But you may find books where the same affinity is given as $\pu{-338 kJ/mol}$. It depends on the sense you consider the reaction $\ce{A + e- -> A-}$.


How are $\sigma$ and $\sigma^\pm$ determined in Hammett plots?

Hammett wanted to find a way to quantify the effects electron-withdrawing (EWG) and - donating (EDG) groups have on the transition state or intermediate during the course of a reaction. Initially he took the $pK_{\mathrm{a}}$-values of benzoic acids carrying the respective functional group in para or meta position (ortho acids and aliphatic acids weren't used because steric effects would overlay with the electronic effects) as a guide and plotted them against the logarithm of the reaction rates.
So, he based his scale on the following reaction reaction:

Later Hammett decided not to use the $pK_{\mathrm{a}}$s themselves for his correlation but defined a new parameter, which he called $\sigma$.
This $\sigma$ shows how electron-donating or -withdrawing a group is relative to $\ce{H}$ as a ratio of
the $\log K_a$s or the difference of the $pK_{\mathrm{a}}$
between the substituted benzoate and benzoic acid itself.
If the acid required to determine $\sigma$ for a new substituent was not available, $\sigma$ could be determined by correlation with other reactions.
A different value of $\sigma$ for any given substituent was needed for the meta and para positions (sometimes called $\sigma_{\mathrm{m}}$ and $\sigma_{\mathrm{p}}$, respectively).
The equation for $\sigma$ is

\begin{equation}
\sigma_{\ce{X}} = \log \left( \frac{ K_{\mathrm{a}}(\ce{X-C6H4COOH}) }{ K_{\mathrm{a}}(\ce{H-C6H4COOH}) } \right) = pK_{\mathrm{a}}(\ce{X-C6H4COOH}) - pK_{\mathrm{a}}(\ce{H-C6H4COOH})
\end{equation}

or in short notation

\begin{equation}
\sigma_{\ce{X}} = \log \left( \frac{ K_{a} (\ce{X}) }{ K_{a} (\ce{H}) } \right) = pK_{\mathrm{a}}(\ce{X}) - pK_{\mathrm{a}}(\ce{H})
\end{equation}

where $\ce{X}$ is the functional group, i.e. the EWG or EDG, whose effect on the reaction rate shall be evaluated.
It is also possible to determine the $\sigma$ values via the logarithm of the ratio of the rate constants for the aforementioned reaction with substituent $\ce{X}$ (in short notation: $k_{\ce{X}}$) and with $\ce{X}=\ce{H}$ (in short notation: $k_{\ce{H}}$)

\begin{equation}
\sigma_{\ce{X}} = \log \left( \frac{ k_{\ce{X}} }{ k_{\ce{H}} } \right)
\end{equation}

Update

AcidFlask's comment reminded me that I forgot to say something about $\sigma^{+}$ and $\sigma^{-}$ values.
As mentioned above the "normal" $\sigma$ values are based upon the difference of the $pK_{\mathrm{a}}$ between a substituted benzoate and benzoic acid itself.
But for benzoates one cannot draw resonance structures that delocalize the negative charge onto the benzene ring via the $\pi$ electron system.
Yet, many reations of interest create negative or positive charges that can be stabilized by delocalization via resonance with the substituent.
For these reactions, one finds that Hammett plots using $\sigma$ values have considerable scatter.
Therefore, two new substituent effect scales were produced, one for groups that stabilize negative charges via resonance ($\sigma^{-}$), and one for groups that stabilized positive charges via resonance ($\sigma^{+}$).
The $\sigma^{-}$ scale is based upon the ionization of para-substituted phenols (I've found different accounts for this. According to the German Wikipedia the $\sigma^{-}$ scale is based upon the ionization of para-substituted anilins.),

for which groups like nitro can stabilize the negative charge via resonance.
The $\sigma^{+}$ scale is based upon the heterolysis ($\mathrm{S}_{\mathrm{N}}1$) reaction of para-substituted cumyl chlorides (phenyldimethyl chloromethanes),

for which groups like amino can stabilize the positive charge via resonance.
The $\sigma^{\pm}$ values can then be evaluated by the aforementioned equation

\begin{equation}
\sigma_{\ce{X}}^{\pm} = \log \left( \frac{ K_{a}^{\pm} (\ce{X}) }{ K_{a}^{\pm} (\ce{H}) } \right) = \log \left( \frac{ k_{\ce{X}}^{\pm} }{ k_{\ce{H}}^{\pm} } \right)
\end{equation}

but with the rate constants $k_{\ce{X}}^{\pm}$ and $k_{\ce{H}}^{\pm}$ (or the equilibrium constants $K_{a}^{\pm} (\ce{X})$ and $K_{a}^{\pm} (\ce{H})$) measured for their respective defining reactions.


It helps to write down the formulas, before you plug in the numbers.

Heat you need to extract to cool the water:

$$Q=m_1 c_p\Delta T$$

Heat needed to evaporate water:

$$Q=m_2L_v$$

So:

$$m_1 c_p\Delta T=m_2L_v$$

So:

$$m_1=\frac{m_2L_v}{c_p \Delta T}$$

Now plug in the numbers.


When does a reaction equation require the reversible sign and when does it not?

You just need to learn and know. Luckily, there is a certain logic to the question which reactions can be reversible and which cannot. However, a full list is impossible and would exceed the space of this margin.

For most mechanisms, there is (at least) one point-of-no-return step. Learn to identify this step. All other arrows should probably be $\ce{<=>}$, if you’re strict. These non-reversible steps are often (but not always!) carbon-carbon bond forming reactions, as a very rough guideline on which is which.


Can fluorine ever have a positive oxidation state?

As noted in the referenced question, fluorine is not in the +1 oxidation state in hypofluorous acid (which, incidentally, is the only hypohalous acid that has been isolated) nor in any other compound where its bonds give it a complete octet.  Even where the fluorine has a positive formal charge, attaching it to less electronegative atoms will lead to an oxidation state of -1.
However, it is possible for fluorine to show a zero oxidation state and for a fluorine atom to apparently show a positive charge.  As one might expect, this occurs in one of the most unusual compounds in all of chemistry.
Striking gold -- and a lucky 7
That compound is "gold heptafluoride", $\ce{AuF7}$, whose unusual structure is reflected in the IUPAC name: instead of gold(VII) fluoride, it is officially called difluorinegold(V) fluoride.  It is more accurately formulated $\ce{AuF5\cdot F2}$.
Himmel and Riedel [1] give calculations showing that unlike other heptafluorides, not all seven fluorine atoms are bonded to the central (gold) atom.  Rather, the gold is bonded octahedrally to five individual fluorine atoms and a difluorine ($\ce{F2}$) ligand, the latter attached end on and rotated so that in the most stable conformation (according to most of the several optimization schemes used), the remote fluorine atom is eclipsed with one of the other ligands as if attracted to it (a, below, picture from [1]).  The staggered arrangement one might expect (b) is instead a transition state in this fluxional molecule.  If we assume that the individual fluorine ligands are negatively charged as usual, this interaction indicates that at least the remote fluorine atom, labeled $\text{F}_b$, is positively charged.

And indeed, Table 2 of the paper shows the following charge distribution:

Thus the relatively electronegative gold atom, it's effective elecyltronegativity further enhanced by the other fluorine ligands, actually imparts positive charge to both fluorine atoms in the molecular fluorine ligand, as if there were a more electronegative moiety even than atomic fluorine.  Note the significantly greater portion of positive charge on the remote fluorine, the one which wheels around to attract one of the fluoride ligands in the eclipsed conformation.
To sum up, difluorine gold (V) fluoride has:

A difluorine ligand, apparently the first known case of thus

A gold-fluorine moiety, either $\ce{AuF5}$ or $\ce{AuF6}$ depending on how one views the structure, that is effectively more electronegative than fluorine itself

A resulting positive charge on some fluorine atoms.

Reference
1.
Daniel Himmel and Sebastian Riedel, "After 20 Years, Theoretical Evidence That 'AuF7' Is Actually AuF5‚F2", Inorganic Chemistry 46 (13), 5338-5342 (2007).


Yes, a common synthesis of benzotriazole starts with o-phenylenediamine as shown in the following figure.

Image Source

The first step in the reaction involves straightforward diazotization of one of the amino groups (exactly like what's done in the first part of the Sandmeyer reaction) to produce the ortho-amino diazonium ion. Instead of eliminating nitrogen as in the Sandmeyer case, the molecule loses a proton and then the diazomium ion internally captures the ortho-imino group to yield the triazole product.

The reaction is not reversible, the benzotriazole is very stable and the intermediate diazonium ion is relatively high energy making the back reaction unlikely under normal conditions.  The reaction will also work with substituted o-phenylenediamines as long as the substituent doesn't interfere with the diazotization step.  So for example, alkyl- or halo-substituted o-phenylenediamines will produce the corresponding alkyl- or halo-benzotriazoles.

Edit: response to comment

Intramolecular cyclization to produce a triazole will only occur if the amino groups in the starting diamine are ortho to one another.  If m- or p-phenylenediamines or non-ortho naphthyl diamines are used, then products resulting from intermolecular azo coupling will occur. See here for some examples.


Dissolution of lead chloride because of the formation of- Na2PbO2 or [Pb(OH)4]-2?

Actually there are two reactions:

$$\ce{Pb(OH)2 + NaOH → Na[Pb(OH)3]}$$

  Lead(II) hydroxide react with sodium hydroxide to produce sodium
  trihydroxoplumbate(II). Sodium hydroxide - concentrated solution. (chemiday.com)

$$\ce{2NaOH + Pb(OH)2 → Na2[Pb(OH)4]}$$

  Sodium hydroxide react with lead(II) hydroxide to produce
  tetrahydroxoplumbate(II) sodium. (chemiday.com)

In the Wikipedia article of lead hydroxide, it is written that:

  In solution, lead(II) hydroxide is a somewhat weak base, forming
  lead(II) ion, $\ce{Pb^2+}$ under weakly acidic conditions. This cation
  hydrolyzes and under progressively increasing alkaline conditions,
  forms $\ce{Pb(OH)+}$, $\ce{Pb(OH)2(aq)}$, $\ce{Pb(OH)3−}$, and other
  polynuclear species like $\ce{Pb4(OH)4^4+}$, $\ce{Pb3(OH)4^2+}$,
  $\ce{Pb6O(OH)6^4+}$

Sodium plumbite ($\ce{Na2PbO2}$) is formed by the reaction of lead(II) oxide and sodium hydroxide.

$$\ce{PbO + 2NaOH -> Na2PbO2 + H2O}$$

Some extra info.

The oxoions and hydroxoions of lead are called plumbates like  hydrated plumbate anion, $\ce{[Pb(OH)6]^2−}$ or the anhydrous anions $\ce{[PbO3]^2−}$
(meta-plumbate) or $\ce{[PbO4]^4-}$ (ortho-plumbate). The hydrated plumbates are formed by dissolving lead(IV) oxide in alkali and the anhydrous ones are formed by heating alkali or metal oxides with $\ce{PbO2}$. All plumbate(IV) salts are very strong oxidising agents and the hydrated plumbate(IV) salts decompose upon dehydration.


Formation of ferrous chloride or ferric chloride

Acids can oxidize iron since the redox potentials $E$ for $\mathrm{pH} = 0$ show that $\ce{H+}$ can oxidize $\ce{Fe}$ to $\ce{Fe^2+}$; but non-oxidizing acids cannot further oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$:

$$\begin{alignat}{2}
\ce{2H+ + 2e- \;&<=> H2}\quad &&E^\circ = +0.000\ \mathrm{V}\\
\ce{[Fe(H2O)6]^2+ + 2e- \;&<=> Fe}\quad &&E^\circ = -0.440\ \mathrm{V}\\
\ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}\\
\end{alignat}$$

However, oxygen can oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$:

$$\begin{alignat}{2}
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad &&E^\circ = +1.229\ \mathrm{V}\\
\ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}
\end{alignat}$$

Therefore, iron is oxidized to $\ce{Fe^3+}$ in dilute hydrochloric acid when in contact with air.


Can I react a Grignard reagent (ethylmagnesium bromide) with water and then ammonia?

I came across this question where I had to convert
  $\ce{CH3-CH2-Mg-Br}$ to $\ce{CH3-CH2-NH2}$. If I need to do so can I
  just react the given compound with water [...]

Always be cautious about unwanted, side reactions that may occur. Proton exchanges are one huge class of possible side reactions. Protons transfers occur readily in many circumstances because the protons is small and can be shuffled around without much to get in its way, physically.

So in the case of your hypothetical reaction between a Grignard reagent and water - realize that Grignards are strongly basic and therefore will deprotonate water. Unfortunately protonated Grignards are no good in terms of reactivity. They just fall back apart into hydrocarbons and a basic halide salt.

Further reading regarding Grignards - It is tempting to use Grignards to create carbon-carbon bonds, but anything that is acidic as water or more acidic will be deprotonated by the Grignard before any carbon-carbon bonds are created. When preparing Grignards, glassware must be perfectly dry. Not just wiped dry with a paper towel but heated dry. Grignards also live in ether solvents since ethers are only very weakly acidic.

Grignards may be used to attack aldehydes and ketones in creating carbon-carbon bonds, but realize that if the aldehyde or ketone contains a functional group that is about as acidic as water or more so - then the first equivalent of Grignard you add isn't going to the aldehyde or ketone carbon - it's going to perform a proton-transfer.

  Is there any set of principles in organic chemistry which govern as to
  which substances can be replaced with what?

In general, it is good to keep in mind the motif nucleophile attacks electrophile in introductory organic chemistry.

Exceptions to motif: nucleophile attacks electrophile

Other heuristics you can use in predicting what reactions might commence ...

Bond energies. This is mainly useful for studying radical reactions - to which the motif nucleophile attacks electrophile doesn't apply as well. Are we going from weaker to stronger bonds? That's a favorable move, enthalpically.
Polarities. This falls under nucleophile attacks electrophile.
Kinetics. It's good to learn the difference between thermodynamics and kinetics early in organic chemistry. Sometimes chemical entities may be thermodynamically unstable but kinetically stable. Maybe a carbon is particularly electrophilic, but it's too well "hidden" by neighboring groups (steric hindrance). So will a nucleophile attack this carbon? Probably not to a great extent.


When there's a photoelectric effect, do the electrons leave altogether or leave one by one?

In 1905, Einstein demonstrated that the photoelectric effect was a quantum phenomenon. That is, for each photon of light of a minimal specific wavenumber, one electron in a specific orbital is released.

For example, for cesium in its base electron configuration, $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 6s^1}$, about $\pu{3.43E-19 J}$ is needed to raise that electron to, effectively, an "infinite" orbital (where it is unlikely to be found near the atom). This corresponds to a wavelength of $\pu{\approx 580 nm}$ or wavenumber of $\approx \pu{17200 cm-1}$.

That said, very energetic photons may knock out more than one electron apiece, with some of the photon's surplus energy applied to the kinetic energy of the first electron, and some applied to ionizing the atom further. However, at the minimum photon level, for every photon absorbed, one electron is freed. A brighter light has more photons, so a brighter light releases more electrons (with limitations due to the build-up of free electrons near the $\ce{Cs}$ cathode, the space charge).


Thermal equilibrium in reversible processes

In a cases where where the temperature is changing during the reversible process, instead of just using a single reservoir at a constant temperature, what you do is use a continuous sequence of reservoirs, all at slightly different constant temperatures (over the range from the beginning to the end of the process), with a tiny (differential) amount of heat exchanged with each.  So, for each reservoir, you have $$dS_{surr}=\frac{dq_{syst}}{T_{surr}}$$ and, if you add all these differential changes in entropy up, you end up with $$\Delta S_{surr}=-\int{\frac{dq_{syst}}{T_{surr}}}$$


Why are isotopes an issue in reading mass spectra?

Here is the "periodic table for biomolecules" (leaving out hydrogen, please ignore silicon):

The pie chart shows the isotope distribution. Of the elements C, N O, P, S and H, sulfur has the highest abundance minor isotope (around 4%). On the other end of the spectrum,phosphorous is a mono-isotopic element, with just one stable isotope. There is an interactive version at https://www.isotopesmatter.com/applets/IPTEI/IPTEI.html.

So for small metabolites, there will be little signal due to minor isotopes. However, in a macromolecule with hundred carbons, the 13-carbon isotope (around 1% abundance) does become relevant. The signal for the mass ion will approximately follow a Poisson distribution, with zero or one 13-carbon isotope in the molecule as most probable:

Source: https://homepage.divms.uiowa.edu/~mbognar/applets/pois.html

For smaller numbers of atoms in a molecule, the multinomial distribution is appropriate, as explained here: https://chemistry.stackexchange.com/a/87669/72973


Distinguishing between potassium chloride and sodium chloride

Per Wikipedia:

  ..., MgSO4·H2O, can be combined with a solution of potassium chloride to produce potassium sulfate.

So, cooling an aqueous solution KCl in the presence of Epsom's Salt:

$\ce{2 KCl + MgSO4 <=> MgCl2 + K2SO4}$

creates potassium sulfate, which falls out of the solution forming distinct orthorhombic crystals.

If Na2SO4.7H2O is created instead from aqueous NaCl in place of KCl, you will get something that resembles a gluey paste.


There is an iron hexacarbonyl, not as a neutral compound but as a dication which satisfies the 18-electron rule. This along with its heavier Group 8 congeners is reported by Finze et al. [1]. The authors who report the tetrafluoroborate salts, also reference earlier work with the corresponding fluoroantimonates.  The reference was found in Wikipedia.

Reference

Finze, M.; Bernhardt, E.; Willner, H.; Lehmann, C. W.; Aubke, F. (2005). "Homoleptic, σ-Bonded Octahedral Superelectrophilic Metal Carbonyl Cations of Iron(II), Ruthenium(II), and Osmium(II). Part 2: Syntheses and Characterizations of [M(CO)6][BF4]2 (M = Fe, Ru, Os)". Inorganic Chemistry. 44 (12): 4206–4214. https://pubs.acs.org/doi/10.1021/ic0482483


Calculating the charges of anions and cations

Some basic knowledge of the atoms in the periodic table is necessary to extract the ionic charges. In the case of $\ce{K2MnO4}$, for example, we know that the potassium ion is positive, and that the manganate ion is negative. We see that the compound is neutral: has zero net charge. So how can we place charges to get the molecule to be neutral, given the subscripts?

Potassium is a Group 1 element, meaning it only has 1 valence electron, placed in the  4s orbital. Potassium is eager to lose this electron to fulfil the octet rule. Therefore potassium ions have one less electron than protons, and the charge is +1. Since we have two $\ce{K^+}$ ions, the manganate ion must have a charge of -2 to make sure the molecule has zero net charge.

$$\ce{K2MnO4 <=>[\ce{H2O}] 2K+ + MnO4^2-}$$

Zinc is one of two transition metals that only have one (natural) oxidation state: +2. The electron configuration is [Ar]3d$^{10}$4s$^2$. When zinc becomes a cation, it loses the two 4s electrons. Knowing this, we see that the hypochlorite must have a charge of -1: the two hypochlorite -1 charges counter the 2+ charge of zinc.

$$\ce{Zn(ClO)2 <=>[\ce{H2O}] Zn^2+ + 2ClO^-}$$

I recommend not relying on simple rules, such as the one you mentioned. Rather, try to understand why the rule came about, and understand when the rule fails, and why. It really helps to be able to "read" the periodic table, as a lot of basic information is incorporated into it.


Is the smallest or largest diameter important for the color of a molecule?

The size of a molecule does not correlate with the color of a molecule.

Color depends upon which wavelengths of light are absorbed and which are transmitted by a molecule.  Visible light typically ranges from around 390 to 700 nm.  Light with a wavelength between approximately 635 to 700 nm appears red to the human eye.  If we have a molecule that absorbs all visible light below 635 nm, the molecule will appear red; it absorbs all visible light except for the red wavelengths, hence only the red wavelengths will be reflected or transmitted and be available for the eye to detect.

What determines which wavelengths of light are absorbed by a molecule?
This is controlled by the energy separation (energy gap) between the highest energy orbital that contains an electron (the HOMO or highest occupied molecular orbital) and the lowest energy orbital that an electron can be promoted to (the LUMO or lowest unoccupied molecular orbital). If this energy gap falls in the visible range, then the molecule will appear colored to the eye when the molecule absorbs light at the wavelength that corresponds to the energy gap (e.g. when a photon is absorbed that promotes an electron from the HOMO to the LUMO) .

Molecules like 2,2-dimethylpropane contain only sigma bonds.  The energy gap in typical sigma bonds is very large meaning that it takes a lot of energy to promote an electron from the HOMO to the LUMO.  The amount of energy required typically falls in the ultra-violet region, outside the visible range; therefore these molecules are usually colorless since they transmit all visible wavelengths.

Molecules containing pi bonds have smaller HOMO-LUMO energy gaps than molecules with just sigma bonds.  If there are enough pi bonds and they are conjugated then the gap can be small enough that light in the visible range can be absorbed and make the molecule appear colored.  Here is a diagram showing how the energy gap decreases as we increase the length of our conjugated pi system.

Even with 1,3,5-hexatriene the light absorption at 258 nm is still outside the visible  range, so the molecule will not be colored.  But as we continue to extend the conjugation, eventually the molecule will absorb light around 400-500 nm, the blue light range.  As the molecule absorbs blue light it will appear yellow. In addition to pi conjugation, certain substituents such as the nitro group and the carbonyl, to just name two, can also serve to extend the conjugation path in a molecule.

The molecule you asked about, penta-1,2,3,4-tetraene, looks roughly like this.

If we examine it closely, we'll notice that the orbitals really look like those in two distinct (orthogonal) 1,3-butadiene molecules.  As the chart above shows, 1,3-butadiene does not absorb in the visible range.  Nonetheless, because penta-1,2,3,4-tetraene does have an extended pi system it will have a smaller HOMO-LUMO gap than 2,2-dimethylpropane and will absorb light at longer ultra-violet wavelengths (lower energy) than 2,2-dimethylpropane.


Why do experiments not verify the law of mass action?

The rate of mass action is true: the rate of reactions is proportional to the activity of its reactants.

On the other hand, from a macroscopic point of view, it is also true that reaction rates may not correspond to apparent stoichiometric proportions.

To understand this disparity, we need to refer to the concept of mechanisms.

The rate of mass action applies to elementary reactions, that is, to reactions that happen the way we write them at the microscopic level.

If the molecular, physical reaction consists on two molecules colliding and reacting, and we write it as

$$\ce{A + B -> C}$$

The rate of that reaction will, per the law of mass action, be

$$r = k · a_A · a_B$$

However, most macroscopic chemical processes aren't that simple, and in fact are the product of multiple steps; for instance, in the previous example, perhaps A undergoes a transformation into an intermediate which then goes on to collide with B to form C:

$$\ce{A -> D}$$

$$\ce{D + B -> C}$$

Macroscopically, we still have $\ce{A + B -> C}$, but the kinetics of the reaction will be very different.

For instance, imagine that the second step ($\ce{D + B -> C}$) is much faster than the first, so that any amount of D created reacts with B almost immediately to produce C. Then the first reaction will act as a bottleneck, and the overall reaction rate will be roughly the rate of the first step:

$$r = k_1 · a_A$$

If we do not know the underlying mechanism, this will be an apparent contradiction between the law of mass action (from which we could naively conclude that $r = k · a_A · a_B$) and the actual rate law ($r = k_1 · a_A$).

Much of the work on chemical kinetics consists in proposing mechanisms, testing these proposals (for instance, looking for experimental evidence that D exists as an intermediate) and deriving rate laws that are consistent with the law of mass action and experimental observation (which applies to elementary steps).


how to calculate mols Ag from AgNO3 solution

If I titrate something with 1 molar AgNO3, adding 50 ml of AgNO3 solution... How would I calculate how many mols of Ag that equates to?

A 1 molar solution of $\ce{AgNO3}$ is 1 mole per liter. 50 ml is 0.050 liters

$ 1 \dfrac{\text{mole }\ce{AgNO3}}{\text{liter}}\times0.050\text{ liters} = 0.050 \text{ moles }\ce{AgNO3}$

  So then how many mols of Ag are there in 0.05 mol of AgNO3? Is there just 0.05 mols of Ag in AGNO3 or do I need to somehow take into consideration the proportion of the rest of the elements?

$0.050 \text{ moles }\ce{AgNO3}$ is correct since there is only one atom of silver per "molecule" of $\ce{AgNO3}$. There would also only be 0.050 moles of N atoms, but 0.150 moles of O atoms in 0.050 moles of $\ce{AgNO3}$.


The figure below illustrates these two ionization reactions.  The key difference between the two is that when the carboxylic ionizes the carboxylate anion is formed.  The carboxylate anion is much more stable than the alkoxide anion formed when the alcohol ionizes.  The carboxylate anion is much more stable because we can draw resonance structures to describe it.  This resonance stabilization of the carboxylate anion explains why carboxylic acids are so much more acidic than an alcohol.


How do I comment on the Gibbs energy values standard and reaction in different situations?

Here is a labeled and annotated version of your graph (from OpenStax Chemistry).


Are all amino acids except glycine chiral compounds?

In general, there are a variety of amino acids that are not chiral (achiral).  In physiological proteins, the amino acids are generally alpha-amino acids (one carbon between the amino and carboxylic acid groups), but there are also, beta-amino acids, gamma-amino acids, etc.

Just limiting ourselves to alpha-amino acids, other than glycine, the most interesting example of an achiral amino acid is amino malonic acid, which has been given its own three-letter abbreviation (Ama) along with the more-common alpha amino acids.  Ama has being found in the hydrolysates of E. Coli and human proteins.  See Aminomalonic acid: identification in Escherichia coli and atherosclerotic plaque PNAS vol. 81 page 722.

Also, the amino acids of protein orgin generally have an alpha H, but if this H is substituted the compound is still an alpha amino acid.  If the two side chains on the alpha carbon are the same, the compound is achiral. For example alpha-methyl alanine.  An example that occurs naturally in cranberries is 1-aminocyclopropane 1-carboxylic acid. The alpha amino acids with two side chains are sometimes referred to as alpha branched amino acids, and have been found in meteorites.

Another class of achiral alpha amino acids are those with a side chain double-bonded to the alpha carbon such as dehydroalanine.


Free automatic organic synthesis Linux command line program

The SAscore is now implemented in RDkit and available for download.


How to name this cyclic compound with an O in it?

Well, the molecule you draw is tetrahydrofuran, aka THF, but of course the only way you can guess that is if you already know the molecule. It's a very common organic solvent.

If you didn't know its common name, you could try to name it according to IUPAC nomenclature. It's an ether, and it's cyclic, so it can be named as oxacyclopentane. But IUPAC has specific names for cyclic ethers, which in this case is oxolane:

(from this organic chemistry book).


Hybridization - Mixing one atomic s orbital with 2p orbitals

When some number of atomic orbitals are mixed together to produce hybrid atomic  orbitals, the same number of hybrid atomic orbitals must result.  So, when N atomic orbitals are mixed together, N hybrid atomic orbitals are produced.  In the case you describe, mixing one s atomic orbital with two p atomic orbitals means that three hybrid atomic orbitals must result.

These three hybrid orbitals will each contain 33.333% s character and 66.666% p character because that is the ratio of the one s orbital to two p orbitals that we are mixing together.  We can describe such an orbital as $\ce{s^{33.333}p^{66.666}}$, but if we reduce that notation to the lowest common denominator we get $\ce{s^1p^2}$, or the commonly used description $\ce{sp^2}$.  This simply means that this hybrid orbital is composed of one part s character to two parts p character.

To summarize if we mix 3 atomic orbitals, one s and two p orbitals, then 3 $\ce{sp^2}$ atomic hybrid orbitals will result.


Nomenclature for two cyclohexane rings linked by a single bond

The relevant sections on ring assemblies of two identical cyclic systems in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) actually read as follows:

P-28.2.1 Ring assemblies with a single bond junction
Assemblies of two identical cyclic systems joined by a single bond are named by one of two methods:
(1) by placing the prefix ‘bi’ (see P-14.2.3) before the name of the corresponding parent hydride enclosed in parentheses, if necessary. Parentheses are used to avoid confusion with von Baeyer names;
(2) by placing the prefix ‘bi’ (see P-14.2.3) before the name of the corresponding substituent group (for names of substituent groups, see P-29), enclosed in parentheses, if necessary.
Each cyclic system is numbered in the traditional way, one with unprimed locants, the other with primed locants. Lowest possible locants must be used to denote the positions of attachment. These locants must be cited in preferred IUPAC names; they can be omitted in general nomenclature when no ambiguity would result.
(…)

P-52.2.7.1 Preferred IUPAC names for assemblies of two or more identical cyclic systems joined by a single bond are formed using the names of parent hydrides rather than the names of substituent groups, (…). For two-component assemblies, locants of one ring are unprimed; locants of the second ring are primed. Locants, including the locants 1 and 1', are necessary in preferred IUPAC names to indicate points of attachment of rings or ring systems (see P-28.2.1 ).

Therefore, the preferred IUPAC name for the compound that is given in the question is formed according to Rule (1) as 4,4'-dimethyl-1,1'-bi(cyclohexane).
The corresponding name according to Rule (2) is 4,4'-dimethyl-1,1'-bi(cyclohexyl). This name may be used in general nomenclature.


Significant figures in conversion factors?

I'm not sure where your conversion factor came from as it is incorrect but regarding the correct factor:
Because the inch is defined using the metre, and the gallon defined using the inch, there is an exact conversion between them—1 US gallon is exactly 3.785411784 L.
Thus, like any exact quantity or conversion factor, we treat the gallon to litre conversion factor as having infinite significant figures.
Because we have defined it exactly, there is no implicit uncertainty in the number.
There are other conversion factors that are not exact and do actually have uncertainties, but for most practical purposes, these have far more significant figures than a normal measured quantity.
See this question for more details.


Harmonic frequencies (IR) and basis set size

No, you cannot say that larger energies will necessarily give larger frequencies.

Consider, for instance, a hypothetical (and purely fictional) method of determining the electronic energy of a system for a given geometry. Call this set of nuclear coordinates (geometry) $\mathbf{G}$, and our hypothetical method is a function which acts on these coordinates to produce an energy, i.e. $$f(\mathbf{G})=E$$

This method, however, is a very special kind of method. It is one which gives an absolutely unreliable energy. It gives you a number that has absolutely no basis in reality and can't be trusted at all. But, one thing we know about this method is that the error, which is truly horrendous and unpredictable, is constant! From this fact, it follows that this method predicts the correct frequencies, the experimental ones. That is, at the minimum of the potential predicted by our hypothetical method and the actual minimum, the difference between our energy, $E_0$, and the true energy, $\mathcal{E_0}$, is,
$$E_0-\mathcal{E_0}=\epsilon$$
or,
$$f(\mathbf{G})-\mathcal{E}_{\mathbf{G}}=\epsilon$$
For all real methods, however, we would need to write the above equation as,
$$h(\mathbf{G})-\mathcal{E}_{\mathbf{G}}=\epsilon({\mathbf{G}})$$

The point of all this being that the hypothetical method I described was not variational or anything else you might desire out of a quantum chemical method. It only predicted energies for which the error term was constant. Whereas the real method, $h(\mathbf{G})$, predicts energies for which the error term is a function of the geometry $\mathbf{G}$.

The concise way of saying this is that frequencies depend only on the curvature of the potential well and not on the depth of the potential well.

With that established, we have answered the question that smaller basis sets, with their larger energies, do not necessarily give higher frequencies.

For instance, if you look at Sinha and Wilson's paper,[1] you will learn of something very empirical which computational chemists do sometimes that bothers me a lot but is illustrative for our purposes. In that paper, a big set of 42 organic molecules is taken, and the harmonic frequencies are calculated at a bunch of different methods and basis sets, then these frequencies are compared with the experimental values. A least squares fit is then performed to come up with scaling values to map the harmonic frequency onto the experimental frequency. As you say, these constants are numbers less than one, say 0.95. The reason these constants are less than one is what bothers me about doing this. Experimental values are anharmonic frequencies. We don't live in a harmonic world. Anharmonicity shifts the frequency down, by lowering one of the potential walls to accommodate bond-breaking, hence we need to scale our harmonic frequencies downwards. So we're basically comparing apples with oranges and there's no way to be sure that this will work in all cases.

More to your question, in this paper, the correction factors do not increase monotonically with basis set size as you suggested (this is the same as frequencies decreasing with basis set size). I won't reproduce their numbers here, because I don't know if that's frowned upon, but for both MP2 and B3LYP, the cc-pVnZ and aug-cc-pVnZ (n=D,T,Q) basis sets have T < D < Q for the scaling factor. So, the triple-zeta basis set invariably predicts a lower energy, but predicts a higher frequency.

Even more interestingly, if one only considers low-frequency modes, which are particularly poorly described by normal modes (these are torsions usually), the ordering becomes Q < T < D for the scaling factors. That's really a failure of the harmonic model more than anything, but it illustrates the point well that lowering the energy really doesn't guarantee you anything in terms of frequencies.

To address your comment, the reason there is any dependence on basis set size is because the error away from the equilibrium structure will change with basis set size, and hence the curvature of the potential changes which affects the frequency.

[1] Pankaj Sinha, Angela K. Wilson, J. Phys. Chem. A, 2004, 108 (42), 9213-9217. DOI: 10.1021/jp048233q


How to know when dissolution is complete?

You've described the general tendency for any equilibrium process, including solubility. So, the short answer is yes, you will dissolve more until the reaction quotient is equal to the equilibrium constant.


Is there a name for the intersection of the graphs of first and second order reactions?

As far as I know there isn't a specific term for that point.

If you have a one-step first order reaction

$$\rm A \xrightarrow{k_1} B + C, \quad \text{rate} = k_1[A]$$

that can run backwards as a second order reaction

$$\rm B + C \xrightarrow{k_{-1}} A, \quad \text{rate} = k_{-1}[B][C]$$

and the rates of those reactions are equal, you can call that point "equilibrium".

You can't really plot the rates separately the way you have unless you have a common concentration, e. g. plot a change in concentration $x$ where $\rm [A] = [A]_0 - x$, $\rm [B] = [B]_0 + x$, and $\rm [C] = [C]_0 + x$.


Why can't sodium borohydride reduce carboxylic acid and ester groups?

The reason for this is because OH and OR groups are much more electron-donating. Typically, when you have an aldehyde or ketone, the electronegative electron is slightly more negative and thus the carbonyl carbon is more positive. However, with an OH or OR group attached (i.e in carboxylic acid and esters), the group can push electrons towards the carbon since they are electron-donating groups. This makes the carbonyl carbon less positive and less open to nucleophilic attack.


Do dendrites effect lithium-ion polymer batteries?

According to this Berkley Labs publication, "Roots of the Lithium Battery Problem: Berkeley Lab Researchers Find Dendrites Start Below the Surface":

  Using X-ray microtomography at Berkeley Lab’s Advanced Light Source (ALS), a team led by Nitash Balsara, a faculty scientist with Berkeley Lab’s Materials Sciences Division, observed the seeds of dendrites forming in lithium anodes and growing out into a polymer electrolyte during cycling. It was not until the advanced stages of development that the bulk of dendrite material was in the electrolyte. Balsara and his colleagues suspect that non-conductive contaminants in the lithium anode trigger dendrite nucleation.
  “Contrary to conventional wisdom, it seems that preventing dendrite formation in polymer electrolytes depends on inhibiting the formation of subsurface dendritic structures in the lithium electrode,” Balsara says.

So, it would seem that dendrite formation through polymer electrodes is indeed a problem, though it may require a "non-conductive contaminant" to initiate nucleation if these authors are correct.


At what temperature do complex hydrocarbons become ashes?

What happens depends whether it is being heated in the presence or absence of oxygen.  For example, in the manufacture of coke (mostly carbon), coal is heated without oxygen and from Coke Production For Blast Furnace Ironmaking:

  The coal-to-coke transformation takes place as follows: The heat is transferred from the heated brick walls into the coal charge. From about 375°C to 475°C, the coal decomposes to form plastic layers near each wall. At about 475°C to 600°C, there is a marked evolution of tar, and aromatic hydrocarbon compounds, followed by resolidification of the plastic mass into semi-coke. At 600°C to 1100°C, the coke stabilization phase begins.

Without oxygen, the carbon will remain upto its boiling point of 3825 °C.

With oxygen, carbon as coke with start combustion at its autoignition temperature of about 700°C


Ionisation Energy vs First Ionisation Energy

There is no difference between "ionization energy" and "first ionization energy". Another definition from Wikipedia:

  The ionization energy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation.

Here it's more clear that they are the same, since "the most loosely bound electron" is a single electron, and each successive ionization only removes a single electron at a time.

Additionally, if one were to remove all electrons from an atom, it would be explicitly stated as "complete ionization".


Why is it ill advised to eat the solid aspirine formed from salicylic acid + acetic anhydride

Unless you worked to Pharma industry Good Manufacturing Practice you shouldn't trust your product

The simple reason why you should not trust the product is that that you don't know it is safe unless you have followed rigorous quality control processes. The drug industry follows carefully regulated quality-control processes known, in Europe, as Good Manufacturing Practice. This is far from easy to meet and even major drug firms sometimes have their factories closed because they fail to ensure it.

For your particular reaction there are many possible things you would need to check before knowing the product was safe to consume. You really don't want to guess whether all the acetic anhydride or acetic acid has been removed: even a small amount will be bad for you (you can't rely on intuition that it should all have been removed). And are you sure there are no nasty side products? Have you checked? Have you checked the purity to make sure? Have you had a check by an independent party?

If you can't say yes to all of these (and more: go read the GMP standards for aspirin manufacture), don't take the product.


How is combustion speed regulated in solid propellants?

(I apologize for being too lazy to find all of the citations for my answer.  However, I have a fair amount of experience as an organic chemist, at one point worked in an underground mine with explosives and have built (for entertainment) a solid fuel composite propellant)

Why does propellant "burn" instead of detonating like similar explosives?

A typical solid composite propellant looks like this:  (1) mainly ammonium nitrate (oxidizer) + "fuel" (typically aluminum flake or magnesium) + binder (polymer + perhaps plasticizer) + (maybe) catalysts OR substitute ammonium perchlorate for ammonium nitrate.

The most similar mining explosive is ANFO or variants thereof:  ammonium nitrate + diesel fuel (possible sensitizing substitutions are... some ammonium perchlorate for the ammonium nitrate, some aluminum in addition to fuel, partially oxidized "fuels" such as hexamine nitrate, nitromethane, possibly a small percentage of a secondary explosive such as TNT or RDX, or addition of plastic "bubbles" to simulate the bubbles that sensitize pure nitroglycerine.

The key point is that the latter mining explosive is called a blasting agent.  The taxonomy of high explosives looks like this:  primary explosives can be easily DETONATED (as opposed to deflagration, which is a type of "burning", detonation involves a supersonic thin region of reaction that moves through the explosive, giving it brisance that can shatter rock).  Primary explosives are used in primers, and can be detonated with sparks, physical shocks, etc.  Secondary explosives (dynamites, TNT, PETN) need to be detonate with a primary explosive.  They are relatively insensitive to other means of initiation (one measure of this is the deflagration to detonation distance, which really is the amount of explosive that can burn before detonation takes place.  It is usually quite large).  You can often shoot or burn secondary explosives without them detonating.

Finally, blasting agents are quite hard to detonate.  They require a booster that is a secondary explosive.  Blasting agents are very insensitive and can be burned without detonating, especially if not completely confined.

This answers your first question...  we should not expect solid propellants to detonate without a booster (actually I'd guess that solid propellants are even harder to initiate, as ANFO variants are usually oxygen balanced so that all of the fuel is oxidized, whereas propellants appear to be a bit carbon rich).

Note that blasting agents and solid propellants share the property that the oxidizer and fuel form distinct particles, as opposed to explosives such as nitroglycerine, which are self-oxidizing within the molecule.  The low explosive gunpowder shares the property of having a mixture of discrete particles of oxidizer and fuel, and it merely deflagrates.

TLDR:  The explosives that are most similar to solid composite propellants are difficult to detonate, so it isn't a surprise that they burn instead.

==============

How is the burn rate controlled?  (note, I use burn as burn typically is decomposition at a rate of a few cm per second, while deflagration is at a rate of many meters per second.  Detonation is thousands of meters per second).

A:

All propellant mixtures have (approximately) a burn rate that is a function of pressure.  This approximation is called Vieille's Law.  The formula is

$rate = {rate}_0 + aP^\beta$

where $P$ is the pressure.  Thus, as pressure increases in the rocket motor, burn rate increases.  Ideal values of $\beta$ are between 0.5 and 0.8, otherwise initiation can be difficult or else a slight overpressure could be amplified to cause the pressure to increase dramatically and destroy the rocket.  What seems to affect the burn rate?  I'll list a few:

(1)  Ammonium perchlorate burns much faster than ammonium nitrate.

(2)  Generally, the rate limiting property of a propellant is the particle size of the oxidizer, so more finely divided oxidizer means faster burn rate.

(3)  The geometry of the propellant is important.  The burning tends to occur at the exposed surface, so wide and not very long propellant castings will produce more thrust.

(4)  Metallic catalysts increase burn rate, such as iron oxides and chromates.

(5)  I've heard that a small amount of carbon black increases burn rate by increasing infrared absorption in the casting.  It also makes it look cooler.

(6)  Not so good:  cracks and voids in the casting will increase burn speed, as they lead to more exposed area.  That's why typical binders are rubbers, often including plasticizers, as these are less likely to crack under pressure.

Here's a couple of citations:

A amateur rocketry site

metallic wires as a burn rate catalyst

Rocket engine mathematics


Why is the enthalpy of solution for NaCl 3.88 and not 4.1 kJ/mol?

Your approach is correct. The small discrepancy of the results when using parameter values from different sources could be caused by inconsistent standard states (temperature and pressure), or simply by measurement uncertainty (in this case, the resulting difference would not be considered significant).

By way of comparison, I found the following parameter values for a temperature of $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ and a pressure of $p=1\ \mathrm{bar}=100\ \mathrm{kPa}$.

Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{NaCl(s)})=-411.2\ \mathrm{kJ\ mol^{-1}}$ [1]
Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Na+(aq)})=-240.1\ \mathrm{kJ\ mol^{-1}}$ [2]
Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Cl-(aq)})=-167.2\ \mathrm{kJ\ mol^{-1}}$ [2]
Standard molar enthalpy of solution $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl(s)})=3.88\ \mathrm{kJ\ mol^{-1}}$ [3]

Calculating the standard molar enthalpy of solution from the given values for the standard molar enthalpy of formation yields

$$\begin{align}
\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl(s)})&=\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Na+(aq)})+\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Cl-(aq)})-\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{NaCl(s)})\\[6pt]
&=-240.1\ \mathrm{kJ\ mol^{-1}}-167.2\ \mathrm{kJ\ mol^{-1}}+411.2\ \mathrm{kJ\ mol^{-1}}\\[6pt]
&=3.9\ \mathrm{kJ\ mol^{-1}}
\end{align}$$

Considering the number of significant digits, there is no significant difference between the calculated result of $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl})=3.9\ \mathrm{kJ\ mol^{-1}}$ and the corresponding literature value of $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl})=3.88\ \mathrm{kJ\ mol^{-1}}$.

References

[1] “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

[2] “Thermodynamic Properties of Aqueous Ions”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

[3] “Enthalpy of Solution of Electrolytes”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


Order of stability of substituted benzyl carbanion

I'm not a chemist, but I doubt whether the mesomeric effect is "more in case I than in III" as you put it.

I'm under the impression that the mesomeric effects (alone) in both cases activate the benzene ring to the same extent (If you try drawing the resonance structures, you'd probably feel the same way).

However, I suppose in case I, the minor $-I$ effect of the $\ce{-OCH3}$ group makes I slightly more stable than III (because the negative charge is somewhat diminished).

So to me, $\mathrm{II > I > III}$ looks about right.


Effect of concentration on rate of reaction

The fast reaction should be written as reversible reaction if $\ce{NOBr2}$ is a high energy intermediate:
$$\ce{NO + Br2 <=> NOBr2}$$

Then, the concentration of $\ce{NOBr2}$ can be estimated as:

$$[\ce{NOBr2}] = K_{eq} [\ce{Br2}] [\ce{NO}]$$

If the second reaction is an elementary step, the rate law follows the stoichiometry:

$$\text{rate} = k [\ce{NOBr2}] [\ce{NO}]$$

Substituting the concentration of the intermediate $\ce{NOBr2}$ into that rate law gives the overall rate law:

$$\text{rate} = k K_{eq} [\ce{Br2}] [\ce{NO}] [\ce{NO}]$$

and the dependency of the rate on doubling all reactants.

  Now I know that overall order of a reaction is equal to the molecularity of the slowest step

This is only true if the slowest step is the first step. Otherwise, you have to consider all the reactants that make the intermediates necessary for the slow step.

What if the first reaction goes to completion?
In this case, the concentration of the intermediate would depend on the limiting reactant. Changing all reactant concentration by a factor of two, the concentration of intermediate would increase by a factor of two as well (not like the factor of 4 if the first step is a pre-equilibrium).

What else do we need to know to solve this problem?
The problem description does not tell us the physical state of the reactants, intermediates and products. If $\ce{Br2}$ is a pure liquid, we would not be able to change its concentration. I would guess the intention is to have all species in the gas phase, though.


Perhaps the simple way to look at this is think of the sphere structures as three layers. Think of the (x,y) coordinates of the spheres, neglecting the z coordinate. The first (white balls) and second layers (black balls) are the same in HCP and FCC. So let's label the first layer the A layer and the second layer the B layer.

Now we add the third layer.

In HCP, which is shown on the left in the above image, the center of spheres in the third layer would be placed directly over the center of a sphere in the A (first) layer. So the third layer (white ball)is another A layer.
In FCC, which is shown on the right in the above image, the spheres in the third layer (green ball) would be offset by 1/2 the radius of the sphere, above a sphere in the first layer. (Note that the dashed circle denotes a position directly above the A level circle.) So the third layer is neither an A layer nor a B layer. Let's call it a C layer. Now the fourth layer would again have spheres whose center are directly above the center of the spheres in the first layer. So it would be another A layer.

So HCP has (AB)(AB)(AB)... layers but FCC has (ABC)(ABC)(ABC)... layers. Where A, B and C denote different sets of (x,y) coordinates. Again the z coordinate being the layer of the ball coming out of the 2D image.


Can you determine when the orbtials of an atom will hybridize?

The molecule $O_2$ does not have a double bond O=O. It is a biradical which has a single bond O-O, and two independant electrons.

Anyway in diatomic molecules, it is not necessary to consider hybridization. It does not help. Hybridization is an approach that has been invented by Pauling to explain the geometry of polyatomic molecules. Hybridization is not really necessary. It allows to easily describe the shape of the molecules, but the same result may be obtained without hybridization, although it is much more complicated to get to the same result.


Why is hexafluoridonickelate(IV) diamagnetic?

This is a common mistake to make when analysing transition metal compounds. The undergraduate or the non-coordination chemist looks at a complex, identifies a ligand, puts it into their box of weak-field or strong-field ligands, deduces an electronic structure and is then completely perplexed if something does not work the way the simple picture would assume.

It is then helpful to remember what the molecular orbital scheme of am octahedral complex looks like; refer to scheme 1. Note that scheme 1 ignores π contributions of the ligands but fret not for I shall briefly explain where (and how) they come in.

Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

On the left-hand side of the scheme you can see the metal’s orbitals, from top to bottom they are (in nickel’s case) $\mathrm{4p\ (t_{1u}), 4s\ (a_{1g})}$ and $\mathrm{3d\ (e_g + t_{2g})}$. On the right-hand side, we have the six ligand orbitals which transform as $\mathrm{a_{1g} + e_g + t_{1u}}$. Only orbitals with identical irreducible representation interact constructively (or destructively) leading to the corresponding molecular orbitals shown in the middle.

What is commonly termed field split, $10~\mathrm{Dq}$ or $\Delta_\mathrm{o}$ is the difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ — the former being entirely metal-centred if π interactions are not considered, the latter being formed by both the metal and the ligand in a bonding-antibonding interaction.

Adding the ligands’ π orbitals, realise that they would transform as $\mathrm{t_{1u} + t_{1g} + t_{2u} + t_{2g}}$. Most of these interactions are either strictly non-bonding or contribute only weakly to bonding since the corresponding symmetry $\mathrm{t_{1u}}$ is already being used to make σ interactions. Only the $\mathrm{t_{2g}}$ p orbitals introduce a new interaction with the central metal by turning what are nonbonding orbitals in scheme 1 into a bonding/antibonding pair of orbitals. Since the interaction is π type, it will be a weak splitting, but it contributes to the decrease of $10~\mathrm{Dq}$ by raising the bottom layer.

I have talked a lot but so far I did not mention the point why hexafluoridonickelate(IV) is a low-spin complex. However, the basis has been laid. It is important to visualise that the field split is — because it derives from the strength (or weakness) of bonding/antibonding interactions in the molecular orbital scheme — dependent both on the metal’s and the ligands’ properties. This can be visualised in the very simple (and thus not satisfying; when approaching a higher level) equation $(1)$:

$$10~\mathrm{Dq} = g_\mathrm{M} \times f_\mathrm{L}\tag{1}$$

In this formula, $g_\mathrm{M}$ represents the contribution of the central metal while $f_\mathrm{L}$ that of the ligands. Obviously, if we modify the metal in a certain way, we can significantly influence the value of $10~\mathrm{Dq}$, i.e. the field split, i.e. whether the complex prefers a high or a low-spin state. Enough circling the issue, let’s get to the point.

The preference for high or low spin can be greatly influenced by the metal’s oxidation state. Remember that a positive charge attracts electrons, so that their respective orbitals will (all!) be lowered in energy by oxidation. Thus, the energies of the metal’s d orbitals and those of the ligands’ σ-symmetric orbitals will move closer together creating a larger stabilisation/destabilisation. At the same time, the π symmetric ligand orbitals will be more removed energetically and thus the weak π interaction will be further diminished. These effects in combination lead to a greater field split for high central metal oxidation states. In hexafluoridonickelate(IV), the oxidation state is $\mathrm{+IV}$ — pretty high for nickel which is probably most commonly found as nickel(II) in coordination complexes. Thus, it is not surprising that this higher oxidation state leads to a low-spin complex. Low-spin $\mathrm{d^6}$ is, of course, diamagnetic.

For further exemplification, I quote two values from a table of $g_\mathrm{M}$ values Professor Klüfers included in the linked internet scriptum (in German):

$$\begin{array}{cc}\hline
\ce{M^n+} & g_\mathrm{M}/ 1000~\mathrm{cm^{-1}}\\ \hline
\ce{Co^2+} & 9 \\
\ce{Co^3+} & 18.2 \\ \hline\end{array}$$

As can be seen from these values, increasing cobalt’s oxidation state from $\mathrm{+II}$ to $\mathrm{+III}$ doubles the metal’s contribution to $10~\mathrm{Dq}$. And that was only an increase of one.


Experimental difficulty with the iodine clock reaction

The theory is correct. Increasing concentrations increase the number of effective collisions per unit time and the rate increases. So accordingly even in the experiment it should happen like that.

So in visualizing your situation, it must have been an experimental error and  not the theory. (Say there must have been less amount of $\ce{H2O2}$ in the mixture).

In this reaction what happens first is the conversion of $\ce{I-}$ to $\ce{I2}$ (have to add $\ce{H2O2}$). This $\ce{I2}$ will then react with metabisulphite or (even sodium thiosulphate) to produce $\ce{I-}$. Since you have added a constant value of meta in trials 6-10 there is a definite amount of $\ce{I2}$ removed in each trial by meta then the remaining $\ce{I2}$ will react with starch to give the color change.


Distinguish between glucose and Fructose samples

One of the simplest ways would be bromine-water test. Bromine water oxidizes glucose to gluconic acid, hence decolorize the solution. Being a mild oxidizing agent, Bromine water is not capable of oxidizing fructose (ketone).


Is there an abundance threshold of a 'synthetic' element for it to be considered natural?

Is there a natural abundance threshold of a 'synthetic' element for it
  to be considered natural?

$\mathrm{> 0}$

In other words, any element that can be detected outside of the laboratory (e.g. in the natural environment) is a naturally occurring element.  This would include the 80 stable elements along with the 18 radioactive elements with very long half-lives, and the shorter-lived radioactive elements produced upon their decay.

Therefore there are 98 naturally occurring elements; all of the others are synthetic, artificial, man-made, whatever you choose to call them.

This Wikipedia article provides nice background, especially the following sections:

the introductory section, especially the last paragraph therein
Occurrence and origin on Earth
List of the 118 known chemical elements - Occurrence


The nearest to an ideal fluid is a hard sphere fluid, but this is removed from the ideal gas or even solution concept of ideality in a critical way.

Ideal (also called perfect) gases are ideal because they lack intermolecular interactions. A statistical description that ignores the intermolecular potential suffices to describe an ideal gas. A first extension toward nonideality, following the vdW equation, describes particles in a gas as hard spheres (having an excluded volume).

The property of reduced compressibility encoded by a hard sphere potential also provides the simplest model for a liquid. The hard sphere potential suffices to describe a simple liquid, and it is about as ideal an abstraction of a liquid as you can get. However, while it shows a liquid-solid phase transition, it does not display a liquid-gas transition. In that sense it is not a universal model, it does not capture some key properties of liquids, in the same sense that the ideal gas law is not universal because it too does not address some simpler key properties of real gases (such as excluded volume). However, introduction of an additional attractive term as in the full vdW model, bringing us further away from ideality, allows us to observe a critical point below which a phase transition clearly occurs between liquid and gas. I should add that the hard sphere potential can be expressed in reduced units, making it universal within the scope of that limited model. It is a suitable first approximation for understanding some properties of nonpolar liquids, for instance condensed noble gases.

I would add that Raoult's law is another ideal model for a liquid, and consistent with the ideal gas concept (or rather Dalton's law). It is also a limiting law (or model), like the ideal gas, and applies to dilute solutions.


General expression for entropy and free energy

Since G and S are taken to be zero at T = 0, H is also 0 at T = 0.  So you can calculate H by integrating the heat capacity from 0 to T:  $$H=\int_0^TC_pdT=\frac{a}{2}T^2+\frac{b}{4}T^4$$
Then you substitute H into the equation $$\frac{\partial (G/T)}{\partial T}=-\frac{H}{T^2}=-\frac{a}{2}-\frac{b}{4}T^2$$
and integrate between 0 and T.  This will give you G.  You can then solve for S from G = H - TS.  This gives you the same result as integrating directly to get S.


Why some materials are transparent, but other ones aren't?

A brief summary: It is caused by the electrons in the particles in the substance. When electrons are orbiting the nucleus of an atom they can move to higher and lower energy levels. They do, however, need energy in the form of photons to do this, this means that when a ray of light hits the atom, the electrons in it will absorb some of the energy from the light ray. It needs a specific amount of energy to move up to the next energy level so will only absorb light from a specific part of the spectrum. This means that in some materials most of the light can pass through it and it may only absorb light waves not of visible light.

So wood, to some frequencies of light, is see through but not to the light waves we can see. That is why we can have mobile and radio signal from inside houses but only see through glass.

(Electrons are actually much more complicated and theoretical than this suggests but this)

Try watching this:  https://www.youtube.com/watch?v=Omr0JNyDBI0


What is the origin of the naming of deoxycorticosterone acetate?

Naming complex chemical compounds in a straightforward systematic fashion is usually not very easy. And often there are trivial names, that existed way before this systematic naming scheme was introduced. Especially when it comes to naturally occurring compounds like steroids. The IUPAC defines steroids as the following:

  Naturally occurring compounds and synthetic analogues, based on the cyclopenta[a]phenanthrene carbon skeleton, partially or completely hydrogenated; there are usually methyl groups at C-10 and C-13, and often an alkyl group at C-17. By extension, one or more bond scissions, ring expansions and/or ring contractions of the skeleton may have occurred. Natural steroids are derived biogenetically from triterpenoids.

  (source: iupac.org)

There is even a quite lengthy article about the nomenclature of steroids: [1] G. P. Moss, Pure & App. Chem. 1989, 61 (10), 1783-1822.

The name Deoxycorticosterone acetate is itself derived from a trivial name. The underlying principle compound is corticosterone and its systematic steroid name is 11β,21-dihydroxypregn-4-ene-3,20-dione, see [1] Table 2. It can be derived from cortisol, which has an additional hydroxy group in position 17. The systematic name therefore is (11β),17,21-trihydroxypregn-4-ene-3,20-dione.

You assigned correctly, that the prefix de(s)oxy- refers to losing an oxygen containing functional group. In this case a hydroxy group at position 11. The systematic name for 11-deoxycorticosterone would therefore be 21-hydroxypregn-4-ene-3,20-dione.

Finally the acetate refers to an acetyl ester group at position 21 (in the sidechain). Interestingly the systematic name for cortisol acetate is 21-O-acetylcortisol. Since it has already a trivial name, see [1], 3S-4.3:

  When the steroid acid has a trivial name that already specifies the hydroxyl group that is esterified, the prefix must be O-acyl-.

Applying the same logic, the systematic name for Deoxycorticosterone acetate should be 21-O-acetyldeoxycorticosterone. Which makes little to no sense. A more systematic approach is described again in  [1], 3S-4.3

  Esters of steroid alcohols
  Esters of steroid alcohols are named by the appropriate steroid substituent group followed by that of the acyloxy group in its anionic form. The steroid name is formed by replacing the terminal -e of the hydrocarbon name by -yl, -diyl, etc. and inserting this before the locant with a Greek letter to designate position and configuration. When necessary locants should be used with the anionic part of the name. Examples: [...]
  The prefix form acyloxy-, with the appropriate locants, is used if there is a preferred functional group as the suffix (e.g. an acid or lactone). Examples: [...]

So in this case since you have a ketone, which is probably more important, I would choose the second part of this rule. Hence the systematic name becomes 21-acetoxypregn-4-ene-3,20-dione.

As you can see, naming is quite a big, often confusing subject and sometimes the origins are not clear at all. And your analysis was completely correct in every way. Trivial names are often like the vocabulary of a specific language - if you don't know it, you have to look it up.

Additional fun fact:
The name acetate is a retained name which can be used in a semisystematic way.


Is Delta H = Delta U for bomb calorimeter or Delta H = Delta U + Delta n(g)RT?

First and foremost your $-0.5$ should have units of moles (i.e. should be $-0.5~\mathrm{mol}$). Don't leave out units. They're important.

  $$\Delta H = \Delta U + p\Delta V$$

If you were not taught what conditions this equation holds true, shame on your teacher. If you were taught it, you should probably try to remember that this holds for constant pressure.

In general

$$\begin{align}
H &= U + pV \\
H_2 - H_1 &= (U_2 + p_2V_2) - (U_1 + p_1V_1) \\
&= (U_2 - U_1) + (p_2V_2 - p_1V_1) \\
\Delta H &= \Delta U + \Delta (pV)
\end{align}$$

Only under constant pressure $p_2 = p_1 = p$ does this simplify to

$$\begin{align}
\Delta H &= (U_2 - U_1) + (pV_2 - pV_1) \\
&= \Delta U + p(V_2 - V_1) \\
&= \Delta U + p\Delta V \\
\end{align}$$

If this is not true (and think about it, if your volume is constant, then what do you think is happening to the pressure inside the calorimeter?), then you may use the ideal gas approximation

$$p_2V_2 = n_2RT_2;\quad  p_1V_1 = n_1RT_1$$

and if you further assume constant temperature $T_2 = T_1 = T$ (which is probably not true, but it is close enough to the truth since the heat capacity of a calorimeter is usually large leading to a very small difference between $T_2$ and $T_1$), then

$$\begin{align}
\Delta H &= (U_2 - U_1) + (n_2RT - n_1RT) \\
&= \Delta U + (n_2 - n_1)RT \\
&= \Delta U + RT\Delta n
\end{align}$$

The further approximation is that the non-gaseous species do not contribute significantly to the $\Delta(pV)$ term, so $\Delta n$ here can be identified with $\Delta n_g$, the difference in the number of moles of gas.

TL;DR

$$\color{red}{\textbf{Learn when the equations you are using are appropriate to use.}} \\
\large\color{red}{\textbf{Don't just blindly use them!}}$$


What happens if don't use MgCl₂ in buffer solution in Ca²⁺/Mg²⁺ titration?

Summary The Magnesium is added because the color change is easier to spot with it, even though calcium also changes the color of the solution in presence of TBT.

Long explanation and experiment

Even though your question might looks silly at first, I know it makes sense because I work with Soil Analysis, and the titrimetric determination of $\ce{Ca^2+/Mg^2+}$ is performed using the same pH 10 $\ce{NH3/NH4^+}$ buffer, with the addition of a small amount $\ce{MgSO4}$ and $\ce{EDTA}$ to the buffer.

As far as I know, and according to a fast online search, the addition of magnesium is to ensure the visibility of the end point when using Eriochrome Black T (EBT) as indicator. Supposedly, the color change wouldn't be easily spotted without the magnesium addition. This is how it's supposed to happen:

Considering there will be no change of color in EBT only due to Calcium, a small amount of magnesium is necessary to interact with the indicator and dye the solution. The equivalent number of moles of EDTA is added to the buffer, otherwise the results would be overestimated. Upon completion of the titration, the remaming $\ce{Mg^2+}$ complexed with TBT would be chelated by EDTA and the endpoint detected.

This is important in basically two scenarios:

There is no magnesium in your sample.
The buffer used is not pH 10, but pH 12, exactly to remove magnesium and detect only calcium.

In both cases, you need some (known) amount of magnesium to see the color change.

The problem with this is that Calcium is supposed to color TBT. It's even the example given on TBT's Wikipedia page, and there's also a full article about the interactions of magnesium and TBT. So why would you add magnesium?

To answer your queston - which also became my own - I did a qualitative experiment, because I'm in the lab right now. And the result was interesting.

I added small amounts of Calcium Chloride and Magnesium Sulphate to two different beakers containing purified water, and then added a few drops of TBT. I also added TBT to the purified water as a blank. This is what I got:

As you can see, the $\ce{Ca^2+}$ solution did change the color of the indicator, but not nearly as much as $\ce{Mg^2+}$ did. If you think about real-sample analysis, where usually the water (or soil extract) contains other coloured materials, making sure the endpoint is easily visible is very important. As a colorblind, I already find this change in color a bit hard, and for $\ce{Ca^2+}$ alone I suppose it would be even harder without the magnesium.

So, finally: The Magnesium is added because the color change is easier to spot, even though calcium changes the color of the solution in presence of TBT.


How to find the alcohol content of a probiotic liquid?

Complex instrumental analytics (HPLC, photometry using a $\ce{NAD+/\,NADH}$ testing kit, etc.) is out of question.

In principle, the most simple approach would be the use of a hydrometer to measure the density before and after fermentation.


How does water defy gravity in paper towels?

This is simply an example of capillary action expanded to a sheet of many tiny capillaries. In a narrow glass tube immersed in water, water will climb the inside of the tube to a level higher than that of the water outside because it is energetically favourable for the water to wet the surface due to adhesive forces and the cohesive forces of the water pull the bulk of the water up with the contact line of the water and the glass. Glass and water have strong adhesive forces as glass has many silicate groups which can interact with water through hydrogen bonding and other polar interactions. As the liquid rises, the volume of water above the surrounding liquid increases, thus increasing the weight of the column. When the weight of the column equals the adhesive and cohesive forces drawing the water upwards, the system is at equilibrium and the water will rise no further.

This can be calculated from: $$h = \frac{\gamma\cos\theta}{\rho gr}$$ where h is the height of the column, $\gamma$ is the liquid-air surface tension, $\theta$ is the equilibrium contact angle between the glass and the water, $\rho$ is the density of water, g is the acceleration due to gravity, and r is the radius of the tube. As you can see, increasing the surface tension (which comes from water's cohesive forces) or decreasing the contact angle (the lower the contact angle, the stronger the adhesive forces between liquid and surface) will increase the height the column will reach, and conversely a denser liquid or larger tube will decrease the height (greater increase in weight for a given rise).

I won't do the whole calculation but for a 1 mm radius glass tube with water ($\theta \approx 50°$, $\gamma = 72\ \mathrm{mN/m}$), the equilibrium height is 4.7 mm. Now, if we shrink the tube much further to approximate something like a pore in a piece of paper towel, we can make the liquid go much further. (it's obviously not perfectly cylindrical, but close enough for illustration and the cellulose-water contact angle seems to be close enough to glass' as cellulose's abundance of hydroxyl groups is similarly adhesive to water) At a diameter of 10 µm, we get a height of 47 cm.

When we expand this concept to a large sheet of interconnected pores, we find that capillary action can draw a relatively large volume of water across considerable distances, which in addition to being great for cleaning up household spills, opens the possibility of using paper as a substrate for analytical devices that don't need pumps, including the lateral flow assay (most commonly seen as the pregnancy test stick) and paper microfluidic devices.


Lestoil, PineSol or other organic-solvent cleaner should get out the unset hardener. Once set, epoxy is hard to remove; you might try methylene chloride-based paint removers [toxic!]. Some hardeners are organic peroxides, which may bleach spots on the clothing. –


Sodium will not take one electron, as you propose. No, it is a redox reaction : the sodium atom will donate one electron to the Oxygen atom of the $\ce{ROH}$ alcohol, so as to repel one $\ce{H}$ atom and make an alkoxide ion. At the end, a $\ce{Na^+}$ cation and a $\ce{RO^-}$ alkoxide anion are produced in the solution. Of course the $\ce{H}$ atom will join the next $\ce{H}$ atom to produce $\ce{H2}$ bubbles that gets out of the solution. $$\ce{Na + ROH -> Na^+ + RO^- + \frac{1}{2}H_2}$$ The reaction is not dangerous. It will not catch fire if done in a sufficiently large amount of alcohol. I have done it plenty of times in the lab. No problem ! It is even a good way for getting rid of old residues of partially oxidized pieces of metallic sodium, in case you don't know what to do with them.


Where did the acac come from in Cr(acac)₃ synthesis by CrCl₃ reagent?

$\ce{CH3COCH2COCH3}$ is the condensed formula of pentane-2,4-dione, commonly known as acetylacetone. The $\ce{acac^-}$ is just a shorthand for the acetylacetonate ligand, i.e., $\ce{C5H7O2^-}$, which is just the deprotonated form of the acetylacetone used in reaction two.


How to recognise the significant figures in given input values for the calculated output value?

You have a reasonable question: if the data supplied specified 1.00 $\mathrm{kJ}$, and the temperature change was given as $10.00~^{\circ}\mathrm{C}$, then "23.90 g" would be a justifiable answer.

However, since the data precision was given to only one digit for energy, then the mass would be "20 g".

Personally, given enough space, I'd write an explanation, of the sort you've give, e.g. "Though I calculated 23.90 g, it could be rounded to 20 g due to lack of significant digits in the data given."

BTW, you've shown good effort in thinking through the homework assignment, in my opinion, which is as important as a numerically correct answer.


Electrodes transport electrons produced from one half-cell to another, thus producing an electric charge. You need to have a reference value to compare it to in order to obtain the relative potential. Relative to the Standard Hydrogen Electrode (SHE), that is.

The SHE has a potential that is set arbitrarily to zero giving you a basis with which to compare the potentials from the electrode you're testing.

For further reading, I feel this article on the Standard Hydrogen electrode is pretty descriptive.


Why does the solubility of some salts decrease with temperature?

A saturated solution is at equilibrium (rate of dissolution is equal to rate of crystallization) with some equilibrium constant $K_1$. If you change the temperature of the system at equilibrium, you will observe a different equilibrium constant $K_2$. Whether the equilibrium constant increase or decreases is described by the Van't Hoff equation:
$$\ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$

So it depends on the sign of the enthalpy of reaction, in this case the enthalpy of dissolution.

  It is well known to us that the Solubility of solute in a solution increases with the increase in the temperature because, when the temperature increases the molecules of the solvent gain more kinetic energy.

The statement about solubility is not always true, and the explanation leaves out something. A lot of things change when you increase the temperature. The solvent gains kinetic energy, the solute gains kinetic energy, and the solid gains kinetic energy. How this influences the solubility depends on the specific system, and is hard to predict.

  However, there a few salts like cerium sulphate , lithium carbonate sodium carbonate monohydrate, etc. whose solubility decreases with the increase in temperature. How it is so ?

Now that we introduced the Van't Hoff equation, we know it must have to do with the enthalpy of dissolution. Usually, you expect the interactions in the solid that need to be broken to be stronger than the gains from solvating the solute. In these cases, it is apparently the opposite.


Is the Arrhenius Equation only valid for reactants in the gas state?

As Ben Norris explained, $R$ can be just more convenient so that one can deal with quantities per mole.

I.e. one could write either $e^{(-E_a/RT)}$ or $e^{(-E_a/K_BT)}$, where in the first instance one would use $E_a$ in $\ce{J~mol^{-1}}$, in the second instance just in $\ce{J}$. The term in the exponent should of course be dimensionless.

Since the numbers in $\ce{kJ~mol^{-1}}$ are much more familiar for chemists, one would often use the first one to avoid having to use very small numbers.

The crux is that the Boltzmann factor in the Arrhenius expression is related to the chance for particles having enough energy at a given temperature, $T$, to surmount the activation barrier.


What will be the coordination number in Fe(0)?

This is another data point for my constant mantra that using hybridisation to explain coordination chemistry is not helpful in the slightest. And even ‘worse’: you are trying to invoke some strong meaning of the 18-electron guideline even if not explicitly. The 18-electron guideline is a guideline not a rule. Even the octet guideline (that I personally like to refer to as a rule) is not as strict a rule as proper physical rules.

As Ben points out, in diiron nonacarbonyl iron(0) has a coordination number of six, which immediately voids your ‘excitation hybridisation’ argument. I am unaware of any fully characterised tetracoordinate iron(0) species. However, I guess they exist due to bulky ligand design and that they will be either square planar or tetrahedral in nature depending on the type of ligand in question. The type of ligand and its interaction is much more important than a hypothetical ‘excitation hybridisation’ scheme.

Please forget any and all of the wrong things you learnt about hybridisation with respect to transition metal complexes. Please refer to molecular orbital theory to explain these instead.


Can't decide between E1 and E2 mechanisms for this reaction

You have a tertiary halide. These dissociate rather easily into the relatively stable tertiary carbocation and the free anion. That in and by itself should be indication enough that the reaction proceeds via E1. Alcoholates are only moderately strong bases so the mechanism cannot be $\mathrm{E1_{cb}}$.

You could consider E2 because the substrate is non-cyclic which means that it can always orient itself in such a way that a hydrogen atom is anti with respect to the halide. The only thing to take care of is steric hindrance; just draw the appropriate Newman projections to determine the likelyhood.

Once you have the tertiary carbocation, any of the β hydrogens could dissociate. You would likely initially get some kind of mixture based on the corresponding $\mathrm pK_\mathrm a$ values weighted by the number of hydrogens. Looking at the three different carbons, one is obviously much more acidic because it is benzylic. This explains the major product.

Even if initially another hydrogen is deprotonated, you can assume that equilibration takes place towards the conjugated styrene system given catalytic excess base.


Why does capping after each coupling in Solid phase peptide synthesis improve the results?

if you cap then you “block” all of the peptides that you are synthesizing.(?)

No, you don’t, the order of steps is important. First you try to couple your new amino acid to the solid phase. You will end up with uncoupled ends with free amino groups and coupled ends with a protected amino group. Then you cap, e.g. with phenoxyacetic acid anhydride. This will only cap the uncoupled ends with free amino groups. Then the Fmoc or whatever protective group is removed rendering free amino groups for new couplings.

Say you wanted to create the sequence $\ce{CHEMISTRY}$. And say at one specific spot on the resin, you failed to couple the serine residue; you have $\ce{H2N-TRY-CO-resin}$ at that one spot. If you now do not cap, you might generate the sequence $\ce{CHEMITRY}$ because isoleucine will couple to threonine’s free amino group etc. By capping, you essentially generate $\ce{Ac-NH-TRY-CO-resin}$ which won’t disturb you anymore.

And the benefits don’t stop there. Without capping, at the end you would end up with the sequences $\ce{CHEMISTRY}$ (the one you want), $\ce{HEMISTRY}$, $\ce{CEMISTRY}$, $\ce{CHMISTRY}$, $\ce{CHEISTRY}$, $\ce{CHEMSTRY}$, $\ce{CHEMITRY}$, $\ce{CHEMISRY}$, $\ce{CHEMISTY}$ and maybe even $\ce{CHEMISTR}$. These are all of similar length to your desired product and hard to purify away via HPLC. But if you cap, you can only get $\ce{HEMISTRY}$, $\ce{EMISTRY}$, $\ce{MISTRY}$, $\ce{ISTRY}$, $\ce{STRY}$, $\ce{TRY}$, $\ce{RY}$, and $\ce{Y}$ — most of these are significantly shorter and easier to purify away.

Also, assume a constant amount of amino acids always being added to the resin to couple. If only the so far correctly built sequences are available, these will ‘see’ a higher ‘concentration per site’ of their coupling partners maybe resulting in slightly increased yield, too. (Note that this paragraph used layman’s terminology.)


Are all organic molecules with odd-numbered electron count radicals?

By IUPAC definition of radical, the organic molecules with odd number of electrons must have at least one of them unpaired and are therefore radicals. The fundamental reason is the definition.

Bear in mind, that other spin states must be taken into account, as simplest example the oxygen. It exists as a triplet, so two unpaired electrons.

The situation gets different, when you do have a metal in your compound, but it is out of the scope of the question.


Both $\ce{SrTiO3}$ and $\ce{MgO}$ can safely be washed with methanol. These compounds are insoluble in most any solvent other than certain acids. I suspect that any online reference you've found saying methanol is corrosive is with respect to organic compounds, nothing like the composition of your substrates.

One (somewhat handwaving) way to convince yourself that this is the case is that we know the compounds are insoluble in water (see their Wikipedia and PubChem pages) and ethanol (from your experience, also confirmed for $\ce{MgO}$ in it's Wikipedia and PubChem pages). In terms of solvent polarity, methanol lies between water and ethanol. It would be extremely unlikely for an ionic solid to be completely insoluble in water and ethanol yet have some significant solubility in methanol.

Although we can be confident that the substrate materials will not be destroyed by methanol, only trial will let you know how effective it is for cleaning whatever contamination may be present in your situation. Methanol is fairly, though not completely, similar to ethanol in it's solvent properties, and there is a good chance that it will be as effective as ethanol. Again though, only trial will verify this in your situation.


The hydrogen atom wavefunctions can be useful for multielectron atoms as a means of looking up their size (by means of a parameterization). The hydrogen atom wavefunctions form one possible basis for the definition of effective nuclear charges, see for instance here, where  Hartree–Fock orbitals are used to compute effective charges:

$$ Z_{eff} = \frac{<r>_H}{<r>_Z}$$

Here $<r>_H$ and $<r>_Z$ are the mean hydrogenic and Hartree-Fock radii (for nuclear charge Z).

The mean hydrogenic radius for ground state hydrogen is related to the Bohr radius $a_0$ as
 $$<r>_H = \frac{3}{2}\frac{a_o}{Z}$$

Expressions for $<r>_H$ are available for other orbitals, allowing $<r>_Z$ to be computed from the tabulated values of $ Z_{eff}$ as

$$ <r>_Z = \frac{<r>_H}{Z_{eff}}$$

So using the hydrogen wavefunctions in this way is a little more complicated than using the equation for the Bohr model, but just the s-orbitals can already be useful to give you an idea about the size of atoms.


There was definitely a typo and the book meant to write $\ce{Ca}$ instead of $\ce{Co}$. In the magnetic properties of a coordination complex, both the counter-ions and the coordination sphere metal ion are contributors. Since $\ce{Hg^2+}$ is diamagnetic and the overall compoud is given to be diamagnetic, the counter-ion has to be diamagnetic as well. Hence, it cannot be $\ce{Co^2+}$, which happens to have unpaired electrons.

What the book might have meant to say is that calcium is the counter-ion here. When checking diamagnetism or paramagnetism, you have to instead check the metal ion inside the coordination sphere, which in this case is $\ce{Hg^2+}$. And note that $\ce{Hg^2+}$ will have a $\ce{5d^10}$ electronic configuration.


How do I convert zwitterions to neutral form in Avogadro?

I don't think the neutral protomer of an amino acid can be generated in Avogadro.  What can be done is adjusting the protonation state for different values of pH.  That option is in the Build menu - Build-> Add Hydrogens for pH.


Reduction of an Allylic Enoate to afford allylic alcohol in presence of Benzyl Acetal Groups

Your benzyl acetal groups should be stable to all varieties of hydride reducing agents.
For the enoate reduction I suggest Dibal-H or LiBH4/Et2O.

I don't have references for this as I no longer have Scifinder access, and Google is not finding me any references that are not behind paywalls.


Is it possible to obtain the amount of air required for a combustion analysis?

For the given problem there isn't much $\ce{CO}$ created. Given that the question is multiple choice an exact amount isn't needed. The problem only needs to be solve close enough to determine which choice is the answer. So assume complete combustion and then guess that the answer will be about 5% high. If octane, $\ce{C8H18}$, is completely burned with oxygen then the reaction is:

$$\ce{C8H18 + 12.5O2 -> 8CO2 + 9H2O}$$

since the problem states that 1.14 kg of octane was burned, the number of moles is:

$$\dfrac{1140}{114} = 10.0$$

Thus for complete oxidation:

125 moles of oxygen would be needed.
Air is only 21% oxygen so 125/0.21 = 595 "moles" of air are needed.
At 28.8 g/"mole" that would be 17.1 kg of air.

In reality there is about 5% less oxygen required since there is some $\ce{CO}$ produced, so $17.1*0.95 = 16.3\ \pu{ kg}$ is a good guess.

I'll leave finding the exact solution to the OP.


Why does reducing the pressure in an equilibrium reaction favour the side with more moles of gas?

For your reaction, $\Delta n_g = 3-(1+1)=1$ (assuming all A,B,C are gases). $K_p = \frac{[C]^3}{[A][B]} (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$, where $\Sigma n$ is the total number of moles of gases in the reaction mixture(even of those species which are not taking part in the reaction). Here, it is a constant.

When pressure is decreased($P_{total}$), $ (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$ decreases as $\Delta n_g >0$.

In order to keep $K_p$ constant (it is always constant at a given temperature), $[C]$ has to increase or $[A],[B]$ has to decrease. Therefore the reaction proceeds/shifts more in the forward direction, which has more number of gaseous moles.

Similarly, increase in pressure favours the side with lesser number of gaseous moles.


How can I figure out whether a reaction is endothermic or exothermic without its enthalpy value?

First of all, the equilibrium constant $K$ in general doesn't tell too much about the enthalpy change of the reaction as it is related to the (Gibbs) free energy change $\Delta G = \Delta H - T\Delta S$. It is true that in some simple cases (such as gas phase reactions) the  entropy change can be approximated by looking at the stoichiometry of the reaction but in general it's usually more complicated.

Regarding your distinction between one-way and reversible reactions I think you may as well assume all reactions are reversible, but some have incredibly huge $K$ so the equilibrium is shifted totally to one side of the reaction - this is the result of a very negative $\Delta G$.

In case of a "reversible" reaction you can always look at the change of the equilibrium constant with temperature. The constant is

$K = \exp\left(-\frac{\Delta G}{RT}\right)=\exp\left(-\frac{\Delta H}{RT}\right)\exp\left(\frac{\Delta S}{R}\right)$

so from the temperature dependence you can deduce the sign of $\Delta H$. Naturally, this is assuming that $\Delta H$ and $\Delta S$ doesn't depend on the temperature in the temperature range you're looking at.

The figure shows the change of equilibrium constant with temperature (arbitrary units) for an exothermic (yellow line) and an endothermic (blue line) reaction.

Just by looking at the reaction I don't think you can tell the enthalpy change. Normally you have bonds breaking on the reactant side and others forming on the product side and normally there is a delicate balance between the two. People have come up with bond strength values (which could be determined from known $\Delta H$ values) and you can use those in your reaction.


How many stereoisomers of 1-methylidene-2-(4,4,6-trimethylhepta-2,5-dien-1-yl)-1,2-dihydronaphthalene are there?

The text is correct. You are partly correct about the alkenes, but mistakenly assume 1) all are stereocenters. 2, it is incorrect to assume that even if stereogenic, alkenes have two stereocenters simply because they have 2 carbons. What makes them stereogenic is that there are geometric differences relative between the two carbons. Thus stereogenicity comes from that relationship, not the number of carbons. 3) While everyone know what you mean by cis and trans, those are not acceptable terms in organic chemistry unless you are referring to sides of rings or in relative terms only.

Looking at just olefins and going from left to right:

Next, the endocyclic alkene must be cis. You cannot place an E olefin in a ring smaller than 8 eight carbons but that would make the name change...this is asking for configurational (stereo) isomers...not constitutional variants.

Next, the exocyclic alkene is terminal. It has two of the same group (H) on the a carbon, it is therefore not a stereogenic center. How do you know? Switch the two hydrogens...does anything change? no, so it can't be.

Next, the E-olefin next to the quaternary carbon is the only stereogenic alkene...it could be Z. So that is a stereocenter (both carbons are)...though that would be incredibly high energy (the largest substituents (and neopentyl is enormous) are sterically congested).

Finally, the last olefin double bond is not a stereocenter. There are two of the same group on one of the carbons. Transposing them changes nothing.

But we're not finished. There is a chiral center. Do you see it? The maximum and actual number of stereoisomers is two. 2^2 is 4, the answer is 4. Here they all are.


The yellow colour in solution is likely from iron ions from the metal being oxidized and dissolved:
$$\ce{Fe_{(s)} + 2e- -> Fe^{2+}_{(aq)}}$$ (Probably $\ce{Fe^3+}$, too)

The gas bubbles/smoke is the product of the other half of the reaction, the reduction of something in your solution. Because you have chloride in solution, chlorine gas is a possibility:
$$\ce{2Cl^{-}_{(aq)} -> Cl_{2(g)} + 2e-}$$

Hydrogen may also be produced depending on what you're using as an electrode on the q-tip.


This is a typo. The prime 2' would refer to position 2 in a separate ring, but you are only numbering the central ring.

Note that you have drawn the structure of 1,4-bis(4-aminophenoxy)benzene.

You can test the names of chemicals and their structures using a tool such as OPSIN.


How is it possible to distinguish antimony from lead?

As mentioned in the comments, you could perform Archimedes' Experiment.

Firstly, take the mass ($m$) (in grams) of your sample.
Then, measure the volume $V$ displaced (according to the first link), in a method summarised in the image below:

You could use any kitchen graduated container. Even though these have graduations for mL (millilitres) and even though the water density changes with temperature (source:USGS), for the temperatures in a home experiment, the following conversion could be used 1 mL = 1 cm3 of water displaced.

Then determine the density by $\rho = \frac{m}{V}$ and compare it to the known values for lead at 11.3 g.cm-3 and antimony at 6.7 g.cm-3.

Another observed feature is that according to the LennTech site for lead is that as opposed to antimony, lead will tarnish on exposure to air.


Do we have some "rule of thumb" to estimate the miscibility of metals?

Bottom line: there are no simple rules of thumb.
To demonstrate this, I will show a round robin of phase diagrams for fcc metals ($\ce{Ag}$, $\ce{Cu}$, $\ce{Au}$, $\ce{Ni}$) with no intermetallic compounds - just fcc solid and liquid.

$\ce{Ag}$-$\ce{Au}$, full mutual solubility, liquid and solid enthalpies of mixing pretty close:

$\ce{Ag}$-$\ce{Cu}$, with limited mutual solid solubility and a eutectic:

$\ce{Ag}$-$\ce{Ni}$, no solid solubility, a huge liquid miscibility gap:

$\ce{Au}$-$\ce{Cu}$, complete mutual solid solubility plus a eutectic:

$\ce{Au}$-$\ce{Ni}$, solid miscibility gap at lower temperatures, but complete solid miscibility above that, a slightly irregular liquidus:

$\ce{Cu}$-$\ce{Ni}$, solid miscibility gap at lower temperatures, pretty ideal liquidus:

Bottom line: even for nominally 'simple' combinations of near-ideal fcc metals coming up with a workable rule of thumb just doesn't work.


Is this a possible explanation as to why sigma bonds are stronger than pi bonds?

This is a very short answer and it might as well have just been a comment. It is probably also tempting in it's simplicity. For starters, let's assume, that nuclei are point charges.

When two atoms form bonds, all the electrons experience the attractive force of the nuclei, of both of them. This is independent of the type of orbital the electrons occupy.

In a σ orbital, the path of maximum electron density is between the nuclei. It is also where the electronic field of the nuclei is strongest. For a π orbital the situation is different.

Let's take a step back and consider an atom. The s orbital is spherical symmetric. The maximum electron density is (formally) in the nucleus. For the p orbital, there is a node at the nucleus. The maximum electron density is further away from the opposite charge, causing attraction to be lower. The energy of the orbital will be higher.

Similarly we can see that for π orbitals. The node of these kind of orbitals is exactly where the field of the nuclei is strongest. Hence the attraction is lowered, the energy raised, because the maximum electronic density is further away from the nuclei.

In very short terms: In σ orbitals the maximum electron density is on the bonding axis, while in π (and δ) orbitals it is off the bonding axis.


1.) The covalent OH bond strength in water is slightly stronger than the NaCl ionic bond strength.

2.) Water might be neutral but each atom exhibits a partial charge.  Oxygen sucks up most of the electron density away from the hydrogen atoms leaving O with a partial negative charge and hydrogen with a partial positive charge.  This gives rise to the famous hydrogen bonding behavior of water.

3.) You would have to analyze the energies of each of the interacting components to know.

4.) Cl is more electronegative than Na due to the larger number of protons in the nucleus.  This means that Cl will suck electron density away from Na as opposed to the other way around.


How to activate polystyrene with carboxyl group?

It is certainly possible to nitrate polystyrene powder with nitric & sulfuric acid mixture and reduce the nitro groups to aniline with $\ce{SnCl2/HCl}$ as described in here1. How effective this would be for a microtitre plate I cannot say, but chemistry is an experimental subject.
Reference

Zhao, H. & Guo, J. & Bian, L. & Wei, Z. & Deng, K.. (2009). Synthesis of nitro and amino linear polystyrene. Hecheng Shuzhi Ji Suliao/China Synthetic Resin and Plastics. 26. 22-26. (link)