Will strong concentrated acid destroy litmus paper and pH paper if they are dipped in acid?

Liquid pH testing solutions or drops are also available. Other indicators like phenolphthalein, methyl orange , methyl red can be used if you expect your test solution to fall in the pH range of the respective indicators. Also it will be wise to take a small sample of the solution to test it rather than testing the whole thing.


As Ivan Neretin commented, parts of a compound can never dictate its own properties to the entire molecule's properties. Additionally benzene is not a very effective functional group by itself.

In short, a restricted use is only applied when that certain chemical is exposed by itself. In paracetamol or any other thing including benzene, you will not see any benzene vapor emanating from them or something. It would also be meaningful if benzene containing compounds degradation products included benzene, like in environmental degradation of chlorobenzenes, though this is not a usual case as I see.


$\ce{SrO}$ is an ionic compound comprising $\ce{Sr^2+}$ ion and $\ce{O^2-}$ ion.

The oxidation state of $\ce{Sr}$ changed from $0$ to $+2$, and $\ce{O}$ from $0$ to $-2$.

Therefore, $\ce{Sr}$ is oxidized while $\ce{O2}$ is reduced.

The ionic equation is just that, since there is no individual ion in the equation ($\ce{SrO}$ is a solid!).

Are you really told to find a half equation for a reaction not performed underwater?


At normal room temperature and pressure graphite is (slightly) more stable than diamond. But the melting point is not a good indicator of this.

Melting points are determined by the bonding structure of the solid and any potential liquid (indeed there may not be a liquid phase at some pressures). They don't necessarily tell you whether a different bonding structure is more thermodynamically stable.

The thermodynamic stability of different carbon structures depends on subtle tradeoffs among several factors in the bonding of the various solids (allotropes) than can be formed with different bonding arrangements. There is no easy way to tell which is the most thermodynamically stable other than by measurement or some very high-powered calculations about the quantum mechanics of the bonding.

The different allotropes have radically different structures (tetrahedral in diamond, a ball of hexagons and pentagons in buckminsterfullerene and flat plates of hexagons in graphite). Changing one into another requires a lot of bond breaking which isn't going to happen much at room temperature. Interconversion is possible at high pressures and temperatures when carbon is dissolved in liquid rock or metal and the carbon allotrope that forms is more likely to be whatever is stable at that whatever the pressure is (diamond is more stable at higher pressure which is why the earth or industry needs very high pressures to create diamond).

So, for most practical purposes, it doesn't matter than graphite is the more stable allotrope: your diamonds are not going to spontaneously turn to ash during your, or the universe's, lifetime.


Step 1: $\alpha$-bromination
$$\ce{CH3-COOH + Br2/P -> Br-CH2-COOH}$$

Step 2: Nucleophilic Substitution by Sodium Cyanide
$$\ce{Br-CH2-COOH + NaCN -> NaBr + NC-CH2-COOH}$$

Step 3: Acid-catalyzed Conversion of Nitrile to Amide or Acid

(1) First hydration... $$\ce{NC-CH2-COOH + H2O -> H2N-CO-CH2-COOH}$$
(2) Further heating... $$\ce{H2N-CO-CH2-COOH -> HOOC-CH2-COOH + NH4+}$$

Special Notes

Be VERY CAREFUL with Step 2 as you are potentially forming $\ce{HCN}$ which could get everyone in the lab killed if you're not careful.


What is the IUPAC Name of [Cr(NH3)5Cl]Cl2.NH3?

This answer is similar to this answer on $\ce{[Co(NH3)5(H2O)]Cl3}$, Thanks to Loong (Faded Giant) for the references.
According to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book), the systematic names of coordination entities are generated by using the following principles:

IR-9.2.2.1 Sequences of ligands and central atoms within names
The following general rules are used when naming coordination compounds:
(i) ligand names are listed before the name(s) of the central atom(s),
(ii) no spaces are left between parts of the name that refer to the same coordination entity,
(iii) ligand names are listed in alphabetical order (multiplicative prefixes indicating the number of ligands are not considered in determining that order),
(…)

Therefore the systematic name of the co-ordination entity $\ce{[Co(NH3)5Cl]}$ would be "pentaamminechlorido" since the ammine ligands are named before the chlorido ligands according to alphabetical order.
The ionic composition of the complex salt $\ce{[Co(NH3)5Cl]Cl2}$ may be indicated in several ways:

IR-9.2.2.4 Charge numbers, oxidation numbers and ionic proportions
The following methods can be used to assist in describing the composition of a compound:
(i) The oxidation number of the central atom in a coordination entity may be indicated by a Roman numeral appended in parentheses to the central atom name (including the ending ‘ate’, if applicable), but only if the oxidation state can be defined without ambiguity. When necessary a negative sign is placed before the number. Arabic zero indicates the oxidation number zero.
(ii) Alternatively, the charge on a coordination entity may be indicated. The net charge is written in arabic numbers, with the number preceding the charge sign, and enclosed in parentheses. It follows the name of the central atom (including the ending ‘ate’, if applicable) without the intervention of a space.
(iii) The proportions of ionic entities in a coordination compound may be given by using
multiplicative prefixes. (See Section IR-5.4.2.1.)

According to the above rules, the oxidation state of cobalt in the complex can be written in two ways - as (III) or (3+).
Similarly the chloride ions can be named in two different ways - chloride or dichloride.
Also, see:

IR-5.5 NAMES OF (FORMAL) ADDITION COMPOUNDS
[...]
The names of the individual components of such a generalized addition compound are each constructed by using an appropriate nomenclature system, whether compositional, substitutive or additive. The overall name of the compound is then formed by connecting the names of the components by ‘em’ dashes; the proportions of the components are indicated after the name by a stoichiometric descriptor consisting of arabic numerals separated by a solidus or solidi. The descriptor, in parentheses, is separated from the compound name by a space. The order of names of the individual components is, firstly, according to the increasing number of the components and, secondly, alphabetical. As the only exception, the component name ‘water’ is always cited last. (Note that this represents a change from the rule in Ref. 2 according to which the component names must follow the order given by the formula.) The numerals in the descriptor appear in the same order as the corresponding component names.
[...]

$\ce{CaCl2.8NH3} \equiv \quad$  calcium chloride—ammonia (1/8)

Therefore the systematic names for the compound $\ce{[Co(NH3)5Cl]Cl2.NH3}$ would be:

pentaamminechloridocobalt(III) chloride—ammonia (1/1)
pentaamminechloridocobalt(3+) chloride—ammonia (1/1)
pentaamminechloridocobalt(III) dichloride—ammonia (1/1)
pentaamminechloridocobalt(3+) dichloride—ammonia (1/1)


The borane reagent in question, disiamylborane, is one of the many bulky hydroborating agents which preferentially add to less hindered double bonds.

In this case, the 1,1-disubstituted double bond outside the ring is more prone to hydroboration than the trisubstituted double bond within the ring, simply due to sterics.

There are many more examples of hydroborations with this reagent which are dictated by sterics: for further examples see its corresponding entry in the Encyclopedia of Reagents for Organic Synthesis. In all cases, disiamylborane is selective for terminal alkenes (i.e. either monosubstituted alkenes, or 1,1-disubstituted alkenes).


Does Nitrogen dioxide change colour if we put it in a constant temperature of e.g. 100 kelvin?

The changing color of $\ce{NO_2}$ is because it exits in equilibrium with $\ce{N_2O_4}$ as shown in the following equation $$\ce{2 NO_2 <=> N_2O_4} + \mathrm{58~ kJ/m}$$ $\ce{NO_2}$ is a reddish-brown colored gas, while the $\ce{N_2O_4}$ is colorless.  Considering Le Chatelier's principle,

Taking heat out of the equilibrium will push it to the right
Putting heat in, will push it to the left
Increasing pressure will push it to the right

As we shift the equilibrium we change the concentration of the components, and consequently the color of the gaseous mixture.

Back to your question, in the solid state (which is where it would be at 100°K), the equilibrium is shifted completely to the right.  So if we had a tube of gaseous $\ce{NO_2}$ and $\ce{N_2O_4}$ at room temperature it would have a distinct red-brown color.  If we then plunged it into a bath at 100°K, it would solidify to $\ce{N_2O_4}$ and appear colorless.


Why is the pressure correction in the van der Waals equation proportional to (n/V)^2?

Summing up the argument from the Wikipedia derivation, the reason it is proportional to $(\frac{1}{V_\mathrm m})^2$ where $V_\mathrm m=\frac Vn$ (the molar volume) comes from how the molecules experience the attractive force. The only molecules that experience a net attractive force are those close to the edge of the container because as we move toward the center of the container, the attractive forces on either side of a molecule cancel out. The force toward the center experienced by a molecule at the edge should be proportional to $(\frac{1}{V_\mathrm m})$. But we can also say that the number of molecules at the edge of the container should be proportional to $(\frac{1}{V_\mathrm m})$. So, by combining the proportionality constants into a single constant $a$, we obtain that the correction to the pressure should be $(\frac{a}{V_\mathrm m^2})=\frac {an^2}{V^2}$ .


Bonding and anti-bonding orbitals in the light of time-dependent Schrödinger equation?

As F'x has indicated, the sum of atomic orbitals $\phi_k$ (AO) that form the molecular orbital $\psi_j$ (MO)
$$ \psi_j = \sum_k c_{jk} \phi_k $$
is just a solution (an approximate solution) to a self-consistent variational procedure to solve the time-independent Schrödinger equation, obtained after choosing a particularly well suited basis to represent the solution. They approximately solve this equation:
$${\sf F}\psi_j = \epsilon_j \psi_j$$
where ${\sf F}$ is the Fock operator or if you wish, an effective one-electron energy operator and $\epsilon_j$ is an effective "one-electron" energy that gives a good approximation to the ionization potential from that orbital.

Then you have to construct your overall n-electron wave function $\Psi({\bf x}_1,\ldots,{\bf x}_n)$, which in the simplest case (the one that follows from the minimal molecular orbital theory) is a Slater determinant. You can notice that for this wave function the coefficients $c_{jk}$ are not unique. Following the general properties of determinants, by a similarity transformation of the basis you can get a new set of coefficients $c'_{jk}$ that will not solve the Fock equation, but after which you obtain the same overall electronic wave function.

In everything that I have said so far, there are no time-dependent phases: we only have one "real" (observable) energy, an eigenvalue of the time-independent Schrödinger equations for the electronic Hamiltonian
$$ {\sf H}_{elec} \Psi({\bf x}_1,\ldots,{\bf x}_n) = E \Psi({\bf x}_1,\ldots,{\bf x}_n)$$
for a fixed geometry. Now is where you can put the time-dependent Schrödinger equation into work. In principle you could create electronic superposition states combining $\Psi({\bf x}_1,\ldots,{\bf x}_n)$ of different energy. These are the ones that would evolve in a manner similar to what you have assumed, with time-dependent evolving phases.


Why are powder XRD axes sometimes plotted in reverse?

Best reason I can think of is that you're being given data in momentum space rather than position space.

You can convert from "position-time" space, which is what we're comfortable with as humans, and describe the system in "momentum-energy" space, which is simpler from a physics point of view. Position is the reciprocal of momentum and energy is the reciprocal of time. If you convert the diffraction equations to momentum ($k$), which is proportional to $\frac{1}{sin(\theta)}$, then the diffraction equation $\lambda = 2dsin(\theta)$ is transposed to $k' = k + K$ where $K$ describes the peak in momentum space, $k$ describes the momentum of the incoming photon, and $k'$ describes the diffracted photon. This equation is linear and much easier to manipulate than the "real-space" form of $\lambda = 2dsin(\theta)$.

The equation in "momentum" space (k-space) is difficult to understand but is nice and linear from a physics standpoint.

https://en.wikipedia.org/wiki/Crystal_momentum


There is no way to answer your question accurately for this setup. It will depend on the temperature of the water in the bowl, the exposed surface area, the convective flow rate of air near the surface, and the relative humidity of the air - none of which are constant or even very predictable.

Your best bet is to set it up and measure the amount before and after distillation. The difference would be your loss.


Where can I get potential energy surface for simple molecules?

Extremely accurate state-of-the-art potentials for more than 20 different diatomic molecules (including excited states for many of them!) are given here.

Bob Le Roy has a reputation for building some of the best small molecule potentials in the world, and in recent years he has started to make them available online. But these are not just a Taylor series around equilibrium, they are analytic functions covering the entire landscape of possible geometries, and for diatomics the specific function is given as $V(r)$ here.

You can plug the parameters from the given publications into the $V$ function, and then Taylor expand the global analytic function into a Taylor series if you want, but the original (non-Taylor expanded) function will be more accurate globally.

If the molecule doesn't yet have a potential available, you can build one ab initio (which is just the Latin phrase for "from scratch") using a program such as the free version of MOLPRO which can be found in the row of the linked webpage saying 0.00 EUR.

The input for making potentials in MOLPRO is here.

If you have specific molecules you are interested in, and you can tell us what you wish to do with these potentials, I can try to direct you to the best potentials for the ground and excited states for the particular molecules you ask for. There is no comprehensive database that has all potentials in one place.


Is so called "Polymerized LiNb (Lithium Niobocene)" a possible compound?

Materials such as Terfenol-D and Galfenol are magnetostrictive ferrous alloys that expand and contract in the presence of magnetic fields, as well as creating magnetic fields when acted on by a physical force.  Both these materials were developed at Ames Lab (formerly the Naval Ordnance Laboratory, that's why the material names end with -NOL).  Terfenol-D has the highest magnetostriction of any Fe-alloy, but the high rare earth content or Terbium and Dysprosium limits its commercial applications.

The "polymer" moniker for Lithium polymer batteries does no refer to the lithium metal or oxides, it only speak to the polymer electrolyte or a polymer casing.  It just a marketing term, no a chemical designation

So on to your question... Polymer, no.  Monomer, no.  Niobium has an extremely high melting point (nearly 2500C/4500F).  It is a very difficult material to melt.  Lithium boils around 1300C.  From a metallurgical processing point of view, you would need a chamber pressurized to several MPa with an inert gas to prevent the lithium from volatilizing as you attempt to alloy it with niobium.  You would also need some powerful magnetic stirring to attempt to keep the mix homogeneous since the lithium would literally float out of the solution.


The author of the article has named the phenomenon incorrectly. Adsorption can indeed result in a lower volume than might be expected, for example when a gas interacts with a high surface area solid. In the case of hydrogen storage, for example, the adsorbed phase is dense enough at low temperatures that much more hydrogen can be stored in a container filled with a high surface area material, and at lower pressures, than an empty container of the same volume.

However, the situation that is described actually is a result of solvation, not adsorption, and the volume-reducing effect has been discussed in some. other. questions, have a look around!

In essence, adsorption is only discussed in the context of the interface between two phases, such as gas-liquid, gas-solid, or two immiscible liquids.

If someone were feeling really pedantic, they could make an argument that the solvated molecules in the liquid constitute a really small "adsorbent" particle, and the solvent molecules in the first solvation shell are "adsorbates", and more power to them. Really, the forces that are at work in both cases (solvation and adsorption) are very similar.

It might be an interesting study to investigate where the crossover between adsorption and solvation happens as a function of molecule size. After all, graphene is just a really big molecule and its adsorption properties are pretty great. A brief search on adsorption solvation crossover reveals that there is some investigation that has been done on the process of going from a 3D gas $\ce{->}$ 2D adsorbate $\ce{->}$ aqueous solvate here.

tl;dr adsorption does occur at the interface between immiscible liquids, but the OP's article is talking about solvation


Number of unpaired electron and total spin angular moment

The first choice lists $\ce{Fe^3+}$ as the central metal atom (CMA), so you have the configuration: $\mathrm{3d^5 4s^0}$. Remember that fluoride is a weak field ligand and would not cause pairing of electrons. This gives you 5 unpaired electrons.

The second choice has $\ce{V^2+}$ as the CMA, the configuration being $\mathrm{3d^3 4s^0}$. This gives you 3 unpaired electrons.

Finally, the third choice has $\ce{Fe^2+}$ as the CMA. Yet again water is a weak field ligand and does not cause pairing of electrons. The isolated configuration although is $\mathrm{3d^6 4s^0}$, but now you'd have 4 electrons in the $\mathrm{t_2g}$ set and 2 electrons in the $\mathrm{e_g}$ set. So now you've 4 unpaired electrons.

  more unpaired electron results in greater spin-only magnetic moment.

This is true and I can see it being followed as $\mathrm{A>C>B}$.

Also see my comment for the strength of ligands.


I am having problem with this question including topics of redox reactions and charges

As the formula of the substance is known, you should first state the formula of the ions produced when the substance is dissolved into water. Here, $\ce{K_4Fe(CN)_6}$ gets dissolved in water and produced $4$ ions $\ce{K^+}$ so that the $4$ corresponding negative charges  must be fixed on the remaining anion, which has the formula $\ce{[Fe(CN)_6]^{4-}}$ with $4$ negative charges. Now you have to think how to synthesize this complex anion from $\ce{1 Fe^x+}$ ion and $\ce{6 CN^-}$ ions. The equation is $$\ce{Fe^{x+} + 6 CN^-  -> [Fe(CN)_6]^{4-}}$$ Counting the charges gives x positive charges from $\ce{Fe^{x+}}$, 6 negative charges form the six cyanide ions, and this makes a total of $\ce{4-}$. So the initial $\ce{Fe}$ ion must have been charged $+2$
The nature of the ion $\ce{CN^-}$ remains to be explained : Well ! The carbon atom has 4 independent valence electrons. Nitrogen has 5 valence electrons, but 2 of them are taken in a doublet. So there is only 3 available electrons in N atoms for bonding. When 1 C is attached to 1 N, this makes 3 covalent bonds between C and N. As a consequence one electron is still available on the carbon atom. This is not a stable compound. The carbon atom needs one more electron to get 8 electrons in its outer layer. This electron can be offered by one H atom, which would make a covalent molecule HCN, with one covalence between H and C, and three covalences between C and N. But this last electron can also be given by a sodium Na or a potassium atom K. This would produce the compound NaCN or KCN, where  Na (or K) and the carbon atom are bound in a ionic bond between $\ce{Na^+}$ (or $\ce{K^+}$) and the ion $\ce{CN^-}$. This is the origin of the ion cyanide $\ce{CN^-}$.


Finding the sum of coefficients of the reaction

Linear equations aside, note that this is a redox reaction and you can use half-reactions method as well (BTW the reaction looks fine to me):

$$\ce{$x$ \overset{0}{Zn} + y H\overset{+5}{N}O3 -> $a$ \overset{+2}{Zn}(NO3)2 + b H2O + $c$ \overset{-3}{N}H4NO3}$$

$$
\begin{align}
\ce{\overset{+5}{N} + 8 e- &→ \overset{-3}{N}} \tag{red}\\
\ce{\overset{0}{Zn} &→ \overset{+2}{Zn} + 2 e-} &|\cdot 4 \tag{ox}\\
\hline
\ce{4 \overset{0}{Zn} + \overset{+5}{N} &→ 4 \overset{+2}{Zn} + \overset{-3}{N}} \tag{redox}
\end{align}
$$

So that the complete balanced equation is

$$\ce{4 Zn + 10 HNO3 -> 4 Zn(NO3)2 + 3 H2O + NH4NO3}$$

leaving $a = 4$, $b = 3$, $c = 1$ and the sum $a + b + c = 4 + 3 + 1 = 8$.


Mass fraction in a mixture of metal salts given the total mass of the metal

You need to do a bit more math than that.
In order to solve the majority of problems in chemistry you need to switch to the amounts  of substances.

So, let's use the following notations: $\omega$ – mass fraction; $m$ – mass; $n$ – amount; $M$ – molecular mass.
I refer to $\ce{NaNO3}$ as compound $1$ and to $\ce{Na2SO4}$ as compound $2$.
Let's solve this algebraically first.
from the definition:

$$\omega_1 = \frac{m_1}{m} = \frac{n_1M_1}{m}$$

In order to find $n_1$, we use the sodium content in both salts

$$n(\ce{Na}) = \frac{m(\ce{Na})}{M(\ce{Na})} = n_1 + 2n_2 \implies n_1 = \frac{m(\ce{Na})}{M(\ce{Na})} - 2n_2$$

and total mass:

$$m = n_1M_1 + n_2M_2$$

Combining both equations (I let you do the math), you find $n_1$:

$$n_1 = \frac{m(\ce{Na})}{M(\ce{Na})} - \frac{2\left(m - \frac{m(\ce{Na})}{M(\ce{Na})}M_1\right)}{M_2 - 2M_1}$$

Now you can find the mass fraction of sodium nitrate:

$$
\begin{align}
\omega_1 &= \frac{\left(\frac{m(\ce{Na})}{M(\ce{Na})} - \frac{2\left(m - \frac{m(\ce{Na})}{M(\ce{Na})}M_1\right)}{M_2 - 2M_1}\right)M_1}{m} \\
  &= \frac{\left(\frac{\pu{1.61 g}}{\pu{23.00 g mol-1}} - \frac{2\left(\pu{5.37 g} - \frac{\pu{1.61 g}}{\pu{23.00 g mol-1}}\cdot\pu{85.00 g mol-1}\right)}{\pu{142.04 g mol-1} - 2\cdot\pu{85.00 g mol-1}}\right)\cdot\pu{85.00 g mol-1}}{\pu{5.37 g}} \\
  &= 0.4513~\text{or}~45.13\%
\end{align}
$$


How hydrate isomerism results difference in colours?

Your idea appears to be correct.

Initially, the structure of $\ce{CrCl3.6H2O}$ corresponds to the dark green compound $\ce{[CrCl3(H2O)3].3H2O}$. This complex shows hydrate isomerism; i.e. the chlorido ligands are step by step replaced by water molecules:

$$\ce{$\underset{\text{dark green}}{\ce{[CrCl3(H2O)3].3H2O}}$ <=> $\underset{\text{dark green}}{\ce{[CrCl2(H2O)4]Cl.2H2O}}$ <=> $\underset{\text{blue green}}{\ce{[CrCl(H2O)5]Cl2.H2O}}$ <=> $\underset{\text{violet}}{\ce{[Cr(H2O)6]Cl3}}$}$$

Thus, the actual complexes are:

$\ce{[CrCl3(H2O)3]}$ (dark green)
$\ce{[CrCl2(H2O)4]+}$ (dark green)
$\ce{[CrCl(H2O)5]^2+}$ (blue green)
$\ce{[Cr(H2O)6]^3+}$ (violet)

Each of these complexes has octahedral geometry. Cr(III) has a $\mathrm{d^3}$ electronic configuration. In the ground state, three electrons occupy the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$). The two $\mathrm{e_g}$ orbitals ($\mathrm{d}_{x^2{-}y^2}$ and $\mathrm{d}_{z^2}$) are empty.

Three spin allowed d–d transitions are expected for Cr(III) complexes. The corresponding absorption maxima of $\ce{[Cr(H2O)6]^3+}$ are:

$$\begin{alignat}{2}
&\mathrm{^4A_{2g}\to{^4T_{2g}}}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\
&\mathrm{^4A_{2g}\to{^4T_{1g}(F)}}\quad&&\tilde\nu_2=24\,500\ \mathrm{cm^{-1}}\\
&\mathrm{^4A_{2g}\to{^4T_{1g}(P)}}\quad&&\tilde\nu_3=38\,600\ \mathrm{cm^{-1}}
\end{alignat}$$

The first transition $\mathrm{^4A_{2g}\to{^4T_{2g}}}$ corresponds to promoting one electron from the $\mathrm{t_{2g}}$ level to the $\mathrm{e_g}$ level. Thus, it directly gives the value of $10\ \mathrm{Dq}$ as $17\,400\ \mathrm{cm^{-1}}$.

The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the low-field ligand $\ce{Cl-}$, the value of $10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are lowered:

$$\begin{alignat}{2}
&\ce{[Cr(H2O)6]^3+}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\
&\ce{[CrCl(H2O)5]^2+}\quad&&\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\\
&\ce{[CrCl2(H2O)4]+}\quad&&\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\\
&\vdots \\
&\ce{[CrCl6]^3-}\quad&&\tilde\nu_1=13\,200\ \mathrm{cm^{-1}}
\end{alignat}$$

By way of comparison, if the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{CN-}$, the wavenumber of the corresponding absorption maximum is increased ($\tilde\nu_1=26\,700\ \mathrm{cm^{-1}}$ for $\ce{[CrCN6]^3-}$).

The wavenumbers can be converted to the wavelength of the absorbed light:

$$\begin{alignat}{2}
&\ce{[Cr(H2O)6]^3+}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\quad\lambda=575\ \mathrm{nm}\ \text{(yellow)}\\
&\ce{[CrCl(H2O)5]^2+}\quad&&\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\quad\lambda=610\ \mathrm{nm}\ \text{(orange)}\\
&\ce{[CrCl2(H2O)4]+}\quad&&\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\quad\lambda=640\ \mathrm{nm}\ \text{(red)}\\
\end{alignat}$$

Since the colour of the absorbed light in this series changes from yellow to orange to red, the corresponding colour of the remaining visible light changes from violet to blue green to green.


Which is more activating among Phenol, Sodium Phenoxide & Anisole? Also arrange them in increasing order of their activating effect

Sodium Phenoxide is much more activating than phenol itself. Because here directly one negative charge comes and bond with the C below and the pi bond shifts to the ortho position as negative charge. And it also answers your doubt that why ortho product is major.
You can refer to this mechanism. After reacting with OH- then the tautomerization takes place but before it no effect where as in case of phenoxide it can perform the same reaction even without OH-
So activation order
Anisole


To remember:

Hyperconjugation is the stabilising interaction that results from the interaction of the electrons in a σ-bond (usually C-H or C-C) with an adjacent empty or partially filled p-orbital or a π-orbital to give an extended molecular orbital that increases the stability of the system. Further information.

The proposed systems to take into account.

Looking at the resonance structures (which is a synonym of mesomeric effect) of the propene

(source: tutorvista.com)

and acetic acid

(source: nonsibihighschool.org)

Therefore, in hyperconjugation analysis you need to look at the position of the electron density. Not the vicinal hydrogen.

Moreover and with the aim to give a complete answer, the hyperconjugation of the acetic acid can be very well understood at here.


Would the lone pair be in the equatorial plane or the axial plane for bromine pentafluoride?

As @IvanNeretin said in the comments, "There are no axial and equatorial positions here". This is a symmetric 3-D structure. Think of this structure as two planes cutting each other perpendicularly and you can rotate it any way you want. This is well reprsented by this image.

So your line "I don't get why the lone pair gets placed at the top perpendicular to the equatorial plane instead of the equatorial plane, where it's more stable." does not make any sense. No matter where you place your lone pair it will still have no effect on the stability of the compound because everytime you think you are shifting your lone pair to a different position, it is forming the same molecule again.

Angle changes from 90 degrees to 84.8 degrees due to the lone-pair present. This creates a lone pair bond-pair repulsion and therefore slightly pushes the 4 Floruines, perpendicular to the lone pair, away from it making an 84.8 angle and not a 90.

Again in the case for $\ce{ICl4-}$ there is no axial or equatorial plane. Here's a diagram that should be helpful:

Keep in mind that (a), (b), (c), (d) can form a plane as well and according to your theory, it will become the equatorial plane. The main key point here is that the lone pairs must be furthest away from each other to be symmetric.


Is there a chemical treatment for pipes to reduce or eliminate leaching of lead into tap water?

In some cases the Langlier Saturation Index can be adjusted so a small carbonate scale barely forms on a pipe to prevent leaching.  The opposite may have happened to the index making the water corrosive when the water supply was switched.

This adjustment must be done by qualified professionals because if this index is too high the pipes will clog.


Why do highly branched alkanes have higher octane numbers than their corresponding linear isomer?

The reaction desired in an automobile piston is a rapid burning, not an explosion (very very rapid reaction). If the air + gasoline vapor burns much faster than the camshaft is turning, the explosion causes a ping, or high pressure rise in a smaller volume (because the piston has not moved far enough). This can damage the piston, or at least add an unusual stress.

Burning is a radical reaction and the radicals produced from linear alkanes are quite reactive (i.e., fast reactors). Branched alkanes might lose a hydrogen, even break apart, but the free electron is more stable, and the reaction rate is slower.

There are compounds with a higher octane number than iso-octane. E.g., toluene has an octane number ~120, and was highly sought after for airplane fuel during WW II. The benzylic radical which is the likely first product of a radical reaction is quite stable (but not inert).


The most important nuclear reactions for the generation of anthropogenic P-32 are
Cl-35 (n, α) P-32,
S-32 (n, p) P-32, and
P-31 (n, γ) P-32.
Since especially chloride, sulfate, and phosphate can occur as impurities in water, typical P-32 activity concentrations in the reactor coolant of light water reactors are in the range of several kBq per litre. (Sources for impurities in the reactor water of BWR’s can be leaking condensers; in the primary coolant of PWR’s, phosphorous compounds in the added boric acid. Sulfate might also be released from cation exchange resins.)

In the laboratory, P-32 can be produced (even as an experiment with students) using the S-32 (n, p) P-32 reaction with fast neutrons. A sample of pressed elemental sulfur is placed close to a neutron source for several days. After that, the sulfur is washed with boiling water. The conversion of the α-sulfur allotrope to β-sulfur at 95.3 °C helps to release the phosphorous compounds from the solid sulfur matrix to the water. Thus, the resulting P-32 solution can be made carrier-free.


It's really strange and contradictory info, I don't know what the ii paragraph is talking about. Referencing the source as “my book” makes is hard to verify and e.g. check if there is some illustratory graphics.

To the last paragraph: The cited information would lead to e.g. “D(+)-glucose”, which is not consistent with typographical details of IUPAC nomenclature of carbohydrates (or IUPAC nomenclature of organic chemistry). The correct name would be (+)-ᴅ-glucose.[1]

  2-Carb-4.5.  Optical rotation

  If the sign of the optical rotation under specified conditions is to be indicated, this is done by adding (+)- or (−)- before the configurational prefix. Racemic forms are indicated by (±)-.

  Examples:

  ᴅ-Glucose or (+)-ᴅ-glucose
  ᴅ-Fructose or (−)-ᴅ-fructose
  ᴅʟ-Glucose or (±)-glucose

Fructose is an example of laevorotatory ᴅ- sugar (therefore is also called levulose).

The stereochemical configuration (especially when it is of one of many chiral atoms, and in sugars, sometimes it's the one that forms the acetal group of the major cyclic form, in some sugars not) is not directly related to the physical optical rotation.

References:

Nomenclature of Carbohydrates (Recommendations 1996), 4.5. Optical rotation


Why isn't neptunium used in nuclear reactors in nuclear power plants?

There are several reasons that neptunium may not be used.

The first is the abundance of neptunium. From Wikipedia, it states the best source is from spent fuel rods. Basically, someone would need to handle very radioactive material to obtain a small amount of neptunium.

Next would be the likelihood of fission when struck with a neutron. If it is the same or less than uranium, I would go with uranium. It would have to be substantially more than uranium to offset the cost of refinement.

Also, there is the energy output per gram. I don't think it would be much greater than uranium, probably less.

The isotopes of neptunium that have long half-lives are $\ce{{}^{236}Np}$ and $\ce{{}^{237}Np}$. These are not produced in great abundance from $\ce{{}^{235}U}$ or $\ce{{}^{238}U}$.

To make plutonium from uranium, the uranium must absorb neutrons without fission. This is much more likely to happen then to make $\ce{{}^{236}Np}$ or $\ce{{}^{237}Np}$


Are enzymes with superior catalytic properties more sensitive to temperature changes?

There is no obvious direct connection between catalytic properties and stability of an enzyme.

You are trying to make a connection, but let's consider your statement:

  Then the Arrhenius equation can be used to describe how the rate constant of an enzymatic reaction changes with temperature.

According to the Arrhenius equation, all reactions (catalyzed by an enzyme or not) become faster as temperature increases. The higher the activation energy, the more pronounced the increase. However, if the enzyme unfolds, it will no longer catalyze the reaction.

  But how can I use this information to decide whether the more catalytic enzyme is more or less temperature dependent?

The biggest temperature dependence occurs at the temperature where the enzyme unfolds. There is the idea that enzymes have to be flexible to be effective catalysts, so there might be a correlation between optimal temperature for the activity and the unfolding temperature. The best way to figure this out would be to compare enzymes from mesophiles and thermophiles in terms of optimal temperature for catalysis vs melting temperature.


Why are conformational isomers considered as isomers?

Gaurang's answer deals well with the theory part I just wanted to give a visual aid.

(Taken from Atropisomerism of naphthyl alcohol)

Let's call the rotating substituent on the naphtahlene a fan.
The fan has two black fins one red fin.
Now let us consider two conformations

  Red fin at 3'o clock

And

  Red fin at 9'o clock

If you were to take snapshots of both the conformations they look different.
Now these two isomers are different and naming them as 9'o clock and 3'o clock  is obviously very inefficient.
That's why conformational isomers and their nomenclature exists, for us to point out specific configurations of a compound.


If you add solid sodium chloride to a saturated sodium chloride solution, it is still at equilibrium. If you increase the sodium ion concentration or the chloride ion concentration (by e.g. adding a high concentration of NaOH or LiCl), then the solution is supersaturated and will not be at equilibrium.

  What exactly is the requirement (in terms of saturation) for a reaction to be at equilibrium?

A reaction is at equilibrium when the reaction quotient is equal to the equilibrium constant. For a precipitation reaction, the solid species is not part of the equilibrium constant expression, so adding more solid does not change the situation. Changing the concentration of the ions in solution, though, will change the reaction quotient, and the reaction will no longer be at equilibrium.


Determine volume of methane used to heat x water to x degrees?

A great place to start would be the heat capacity $C$ of water. It is defined to be the "amount" of heat change divided by the resulting temperature change. In symbols,

$$C=\frac{q}{\Delta T}.$$

Heat capacity is an extensive property; the size of $C$ depends on how much matter there is. We take this into account by dividing the equation by mass $m$.

$$C_s=\frac{q}{m\Delta T}$$

$C_s$ is called specific heat (capacity). Taking $\rho_{\ce{(H2O)}}\approx1$,

$$C_s \approx \frac{q}{V\Delta T}.$$

Can you take it from here? (Be careful with units.)

Note: It is probably assumed that

the combustion of $\ce{CH4}$ is complete, yielding only $\ce{CO2}$ and $\ce{H2O}$;
density of water does not change over the temperature range;
heat capacity is constant as $\Delta T$ is rather small;
vaporisation is neglible;
no heat lost to surroundings;
decreasing solubility of $\ce{CO2}$ does not contribute to the overall amount of gaseous carbon dioxide (joking...)


This seems more like folklore than a tested method. However, gold is extremely soft (malleable), and it must be alloyed with other metals to make it hard enough to survive wear as jewelry. Simply rubbing a gold item with a cloth, with or without "ammonia" (ammonium hydroxide, $\ce{NH4OH}$), is enough to polish the gold, i.e. relocate some gold particles to fill in the scratch.

Household ammonia does not solidify, nor will it fill scratches. In fact, it simply evaporates into the air. It is an effective cleaner, though, and would help remove anything adhered to the gold, which might appear like a scratch.

For the same reason, ammonium hydroxide is an effective glass cleaner, but don't expect it to remove any scratches. The suggestion at your second reference to use toothpaste, though, is a good idea, particularly for those brands containing silica.

BTW, "milk of magnesia", $\ce{Mg(OH)2}$, is an effective mildly abrasive scratch remover for some plastics, if rubbed very well with a soft cloth.


Using covalent or ionic radius when calculating dipole moment

Personally, I think "50%" as a cut-off is very arbitrary to begin with. It's not like I can suggest anything better, since proper dipole moment calculations rely on quantum mechanics.

But, to answer the question:

The whole premise of my answer is that you want to get the sum of your radii to match the actual bond length in the compound. And to do that, you have to look at the context. For example, for $\ce{BF3}$, you might want to use covalent radii since the atoms are covalently bonded. And for a ionic compound like $\ce{KF}$, you'd better use ionic radii or you might run into some problems.

Now, note that the formula you're using is already extremely inaccurate in itself - if you take the covalent radii for B and F, you get $d = (82+71) \text{ pm} = 153 \text{ pm}$. And... the actual bond length is very short - it's $130 \text{ pm}$. Clearly, there is something about that compound that the crude model of adding covalent radii cannot account for. Still, it's better than if you were to use the ionic radii: $(41 + 119)\text{ pm} = 160 \text{ pm} > 153 \text{ pm}$ (sum of covalent radii) $> 130 \text{ pm}$ (actual bond length).

  "Hey @orthocresol, using $\ce{BF3}$ isn't fair! That compound has crazy strong $\pi$ bonding between B and F!"

Okay. Let's look at $\ce{BCl3}$, $\ce{BBr3}$ and $\ce{BI3}$, where the $\pi$ bonding is weaker.

$$\begin{array}{|c|c|c|c|}
\hline
\ce{X} & \text{Actual B-X bond length / pm} & \text{Sum of covalent radii / pm} & \text{Sum of ionic radii / pm} \\
\hline
\ce{Cl} & 175 & 82 + 99 = 181 & 41 + 167 = 208\\
\ce{Br} & 187 & 82 + 114 = 196 & 41 + 182 = 223 \\
\ce{I} & 210 & 82 + 133 = 215 & 41 + 206 = 247\\
\hline
\end{array}$$

The covalent radii are closer every time.

  "That's still not fair... Can't we just use a compound without $\pi$ bonding?"

Fine. Let's talk about the tetrahedral $\ce{BF4-}$ anion. There's no $\pi$ bonding here and as a result, the B-F bond length is increases to $145 \text{ pm}$. But the covalent radii still win.

You can extend the discussion to something like $\ce{SiX4}$. But, honestly, I think that representing $\ce{SiX4}$ as $\ce{Si^4+}$ and $\ce{X-}$ is not that a great an idea, to say the least.

Also, I'm lazy to dig up bond lengths in $\ce{AlCl3}$, so I'll stop here.

  "Okay, now what about ionic compounds?"

You'd better use those ionic radii, because guess how those ionic radii are tabulated? They're calculated by measuring the distance between cations and anions in an ionic lattice and assigning a certain part of that to the cation and another certain part to the anion. So if you add them back up, they'll of course give you the internuclear separation that you want.

Let me use the simple example of $\ce{NaCl}$. You'll have to trust that I'm doing the right things here; if you're interested, you can read up on the rock salt structure. The unit cell parameter of $\ce{NaCl}$, denoted by $a$, is $564 \text{ pm}$. This corresponds to two times the distance between $\ce{Na+}$ and $\ce{Cl-}$ ions, so we could say that the actual "bond length" is $564 \text{ pm} / 2 = 282 \text{ pm}$.

Now, the ionic radii of 6-coordinate $\ce{Na+}$ and $\ce{Cl-}$ are $102\text{ pm}$ and $181\text{ pm}$ respectively. That sums up to $283\text{ pm}$ - a beautiful match. And the atomic radii? They're $191\text{ pm}$ and $199\text{ pm}$ for $\ce{Na}$ and $\ce{Cl}$ respectively. That sums up to $290\text{ pm}$ - pretty close but not quite as close as the sum of the ionic radii.

Data is taken from this website, Wikipedia:Ionic radius, Shriver & Atkins, Inorganic Chemistry, 6th ed., and Greenwood & Earnshaw, Chemistry of the Elements, 2nd ed. Apparently, the numbers in that Wikipedia figure are taken from R. D. Shannon, Acta Cryst., 1976, A32, 751, which is in fact one of the most widely-used sources of ionic radii (all inorganic textbooks reference it) - so it shouldn't be that bad.


how does electronic conductance of a metal change with density?

You cannot change the density of a metal without also changing other properties.

Brass is a mixture of several elements. It can have different densities depending on the mixture of the elements. I haven't researched it, but the electronic conductance would likely be different for each type of brass.

Do you have a source for your statement?

  The electronic conductance of a metal depends on its density.


Converting SMILES string into valid sub-strings

While this might work in some cases, I am not sure how well it will work in general. The problem here might be reflected in this statement:

These sub-strings should be subset of the main SMILES string

Not every substring will be a subgraph of the parent molecule. As mentioned in the comments, this seems essentially a fragmentation problem. Take this one of your examples:

smi_sub1= 'C1c1ccccc1Cl'

The reason this won't parse is that it starts with a 'C1' but has no corresponding 'C1' to finish the ring. So it should be 'Cc1ccccc1Cl', which is 1-chloro-2-methylbenzene.
Theoretically, you could take substrings, then post-process them to remove 'disconnected' ring numbers. However, it would make more sense to use proper fragmentation on the molecular graph, and then you would have correct subgraphs anyway, without the need to convert them from SMILES.


Calculating the density of a magnesium sulphate solution

Each of parameters A-D is a polynomial function of t:
$$A = \text{"m"} + \text{"mt"}.t + \text{"mt2"}.t^2 + \text{"mt3"}.t^3 + \text{" mt4"}.t^4$$
$$B = \text{"m3/2“} + \text{"m3/2t"}.t + \text{"m3/2t2"}.t^2$$
$$C = \text{"m2“} + \text{"m2t"}.t$$
$$D= \text{"m5/2"}$$
The terms in quote means the equally denoted numerical values from the provided table.
The complete final function $1000 \cdot \Delta \rho=f(m,t)$ has for each term $a_{b,c} \cdot m^a \cdot t^b$ the value for $a_{b,c}$ equal to the particular table value named "$m^bt^c$".
So one can the complicated function use in 2 ways:

The full blown function $1000 \cdot \Delta \rho=f(m,t) = \sum {\left( a_{b,c} \cdot m^a \cdot t^b \right)}$ without the need of coefficients A-D.
Temperature specific $1000 \cdot \Delta \rho=A \cdot m + B \cdot m^{3/2} + C \cdot m^2 + D \cdot m^{5/2}$ with precalculated A-D values for given temperature.

BTW, it looks like a very interesting numerical approximation. Maybe even better could be rational functions like Padé approximant, that have smaller deviations and better  extrapolation behaviour, compared to polynomials.

Feedback for the Padé approximation:
For approximating of analytical functions, see e.g Math SE site - how-to-compute-the-pade-approximation. For approximation of experimental discrete data, the best approach is probably numerical mathematics, the least square method combined with function minimum search algorithm. They probably used the same technique for their numerical model with half-integer powers.
Using half-integer power model is unusual. It may work better than polynomials in many cases, but just for positive numbers. Fortunately, they do not calculate with negative Celsius temperature :-) ) Note that Math SE site has even tag pade-approximation . Also, it is included in MATLAB libraries, if applies. See Octave as both online/offline free functional clone of MATLAB.
See also my answer in this Q: Is there a quick way to approximate a logarithm in Nernst equation?.


How does the silver nucleate on glass in Tollen's reagent mirroring?

"If you make a surface treatment using silane groups, I think that you promote the presence of OH groups at your silica surface". This is NOT correct. You siliconize glass to cap the free SiOH groups

2 silica_Si-OH + n SiCl2Me2 + HO-Si_silica -> silica_Si-[O(SiMe2)]$_n$-O-Si_silica. As a result you decrease the number of active OH groups and make it highly non-polar.

Before siliconization silica is surrounded by a hydration sphere. Ag$^+$ can bind it well, but we need to precipitate neutral Ag. In case of non-siliconized glass we have two effects:

(1)  It is easier for silver to grow on the surface that already has a silver nano-crystal
(2) Before treatment glass have inhomogeneities, so nucliation in some places is heavily favored.

Both effects are negated by siliconization.

Note that I didn't account for surface charges which change as a result of siliconization and might play a role.


For covalent compounds, what is the microscopic difference between a solid and a liquid?

Let's break this down, and keeping it to simple covalent compounds...

  Are different forces dominant in between solid and liquid molecules? By "forces" I mean ionic bonds, dipole-dipole interactions, London forces, etcetera.

All those forces (if applicable) will still exert their expected effect and there is no 'dominant force' which increases in strength when temperature rises. Just that when we raise the temperature, the molecule may have enough kinetic energy to overcome London Dispersion but not ionic forces, etc.

  What is the difference in the geometric structure of the molecules' interactions? Solids are fixed and gases are mobile, but what precisely are liquids? Solid structures with enough degrees of freedom to pivot fluidly? Do liquids generally have the same structure (simple cubic, hexagonal, etcetera) as their solid counterparts when they do interact?

So let's take a look at a typical phase transition diagram, that of water:

At the flat stages, water is changing discreetly and exists either as solid (rigid lattice) or liquid (completely formless and structure-less). And if you look at ice melting, that is exactly what you see and experience.

But the source of this question is probably the fact that you've also seen certain solids soften under heat without melting completely. These covalent materials are called thermoplastic and have a property called the "Glass Transition Temperature". The Stiffness-vs Temperature graph looks like this:

In this state, the melting solid is able to overcome the forces that hold it to a lattice, but is not able to overcome the forces that hold the molecules together and so the molecules cannot 'flow'. The reason this exists for some compounds and not others is largely a property of how their molecules hold each other together.


Determining the torsion energy profile of butadiene by molecular dynamics

I really only want to make a comment, but it is too long...

Your equation $n_i = \exp^{-E_i/RT}/N$ does not make sense to me.

I agree with
\begin{equation}
P_i = \frac{\exp ^{-E_i/RT}}{Z}
\end{equation}

where $Z = \sum \exp ^{-E_i/RT}$ is the configurational partition function.

I can also accept that using a histogram, you could say $P_i = n_i/N$ where $n_i$ is number of samples in that bin, and N is total samples.

I do not see however how

\begin{equation}
n_i N = \exp ^{-E_i/RT}
\end{equation}

This does not make sense to me. Perhaps you could swap $P_i$ for $n_i/N$ so that you have
\begin{equation}
P_i =\frac{n_i}{N} = \frac{\exp ^{-E_i/RT}}{Z}
\end{equation}

which could lead to
\begin{equation}
n_i = \frac{N}{Z}\exp ^{-E_i/RT}
\end{equation}

Which could be rearranged to get
\begin{equation}
E_i =-RT \ln \left(\frac{n_i }{N}Z\right)
\end{equation}

If you are comparing to a reference energy then you could get

\begin{eqnarray}
E_i - E_{\rm ref} = \Delta E_{\rm i,ref} &=&-RT \ln \left(\frac{n_i }{N}Z\right) +RT \ln \left(\frac{n_{\rm ref} }{N}Z\right) \\
&=& -RT\ln \left(\frac{n_i }{n_{\rm ref}}\right)
\end{eqnarray}

It would be interesting to see if this works? I don't have the raw data, so I cannot try it (I also don't have the time, only the interest).


Is chloronitrobenzene and nitrochlorobenzene same?

In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of citation of substituent prefixes and the numbering of the corresponding locants are laid down in two different rules.

Simple substituent groups that are named by means of prefixes (such as ‘nitro’ and ‘chloro’) are arranged alphabetically.

  P-14.5 ALPHANUMERICAL ORDER

  Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved

  Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.

  (…)

  P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order for the compound given in the question is x-chloro-y-nitrobenzene (not x-nitro-y-chlorobenzene).

The locants x and y are used to indicate positions of the parent structure at which modifications represented by suffixes occur. The locant ‘1’ is omitted in monosubstituted homogeneous monocyclic rings (e.g. chlorobenzene or nitrobenzene); but in substitutive names of disubstituted benzene, numerical locants have to be used to distinguish the 1,2-, 1,3-, and 1,4-isomers.

Note that the letters o, m, and p have been used in place of ortho, meta, and para, respectively, to designate the 1,2-, 1,3-, and 1,4-isomers of disubstituted benzene (e.g. o-chloronitrobenzene). According to current IUPAC recommendations, this usage is strongly discouraged and it is not used in preferred IUPAC names.

The relevant rules concerning the numbering of locants for substituent prefixes are:

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

and

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule (f) takes precedence over Rule (g).

In accordance with Rule (f), the compound given in the question could be named as 1-chloro-2-nitrobenzene as well as 2-chloro-1-nitrobenzene since both names correspond to the locant set ‘1,2’. However, according to Rule (g), this example is named as 1-chloro-2-nitrobenzene rather than 2-chloro-1-nitrobenzene since chloro is cited first as a prefix in the name.


No.  Because remember, the oxygen is being consumed too.  If we assume it's reacting via that reaction (realistically, you're going to get a complex mix of products) then we start with something like: 30% NH3 56% N2 14% O2 (the mix is 70% air, which is an 80/20 mix of N2 and O2).  For every 4% reduction in Ammonia, you're getting a 3% decrease in oxygen.  So O2/NH3 reactions will be rarer (slower), and slow down heat generation.  Also, the amount of inert gas in the vessel will be increasing, which also need to be heated up, so less of the enthalpy of reaction will go into increasing the temperature.

That said, above the autoignition temperature, it's going to be a pretty fast reaction anyway...  What we can be sure of is it will stop reacting at about 10% NH3 when all the oxygen is gone.


What happens to copper(II) hydroxide when it's heated? what's inside its decomposition?

Well,this is a kind of dehydration reaction,and you will come across similar such reactions plenty of times as you read about the effect of heating on various compounds.

I am going to approach this in a sligthly different manner: how the copper(II) hydroxide was generated from the copper(II) oxide.
According to Wikipedia

  Copper(II) oxide belongs to the monoclinic crystal system. The copper atom is coordinated by 4 oxygen atoms in an approximately square planar configuration

So we can think of it this way,that when water was introduced in the copper(II) oxide powder,the water molecules entered into the voids(in a stoichiometric fashion)of the oxide lattice.Mind you,this is just a way of visualization,but it can help a lot. So now,when you are heating the hydroxide in a controlled manner,then we can imagine that the water molecules residing in the voids escape out,leaving the oxide in the same oxidation state. (Also,now you should keep in mind that most hydroxides lose one or more molecules of water to form their respective oxides)

Note: if heating is done excessively,then the oxide itself can decompose to a variety of products


In the covalent (L) method, $\mu$-hydrides should be treated as a normal hydride (1-electron ligand) for one metal. The $\ce{M-H}$ bond itself should be treated as a 2-electron ligand for the other method. This explains the two entries for hydride in the link you provided:

Cyano-ligands are isoelectronic to carbonyl ligands, so I'll leave that to you as an exercise to figure out.


How is taste of chemical compounds determined?

When a compound is synthesised or isolated, it needs to be characterised. This characterisation serves to determine whether something has been synthesised/isolated before and as a template for later generations to compare their synthesised/isolated compounds to the one you have.

Nowadays, one reports NMR and mass spectrometry data; for chiral compounds specific rotation is measured. Sometimes, elemental analysis is included. Depending on the journal’s requirement you might also need to measure IR and/or melting points. Furthermore, you typically describe your compound (is it a yellow oil or white needle-like crystals?) although the description is considered secondary and the NMR/mass data is what you look at for matches.

In the old days, these analytical methods were not available. Instead, analyses would rely heavily on what a compound looks like and how it behaves. Thus, it was attempted to generate various crystalline forms and reactions with various standard reagents were investigated. Additionally, melting points were measured and chemists used all available senses to characterise. Thus, when the paper says that boron phosphate is a white, odourless and tasteless compound, the chemist in question not only sniffed the flask but also put a bit on his tongue to assess taste.

The problem with this approach is that you have no idea whether something is toxic/harmful or not. As there are many mechanisms by which a compound can interfere with the tightly controlled reactions in our body and cells or the macromolecular structures therein, many (if not most) compounds will be harmful in at least one way. While the common approach was to taste as little as possible, the risk remains as sometimes the $\mathrm{LD_{50}}$ is in the $\pu{\mu g/kg}$ range so even the smallest sample could kill you (although compounds that toxic are thankfully rare). Furthermore, a lot of compounds are able to induce DNA mutations which likely led to increased cancer rates and an overall lower life expectancy for chemists.

This is, quite obviously, unacceptable from a workplace safety point of view. Thus, tasting chemicals was prohibited practically everywhere. Furthermore, to prevent accidental contamination or confusion of glasses, not even the consumption of food and drink is permitted in most laboratories. Therefore, for anything that has been synthesised or isolated for the first time after these measures were in effect, we typically do not know taste any more. In fact, unless we are certain that we have evidence a compound is harmless, it is to be treated as highly toxic in the lab with all according security measures.

A compound’s odour is a slightly different story. It is said that the chemist working ideally will never get a wiff of their compounds because they will make sure that none access the common lab air. In practice, that is practically unattainable especially since lab scales are often in the middle of the room and most solids are in glass bottles that need to be opened fully to take out a sample. Thus, it is unavoidable that chemists pick up the smell of a number of (typically particularly volatile or particularly odourous) compounds. Systematic odour examinations are no longer performed but where the odour is characteristic, it is sometimes included in a compound’s characterisation (e.g. ‘sulphurous odour’). Some chemists take this a step further: a former colleague once sniffed at the flask of another former colleague and exclaimed ‘it smells like your product’, explaining that tertiary alcohols apparantly have a distinctive odour.


When fugacity and z-compressability equations instead of Gibbs free energy equations?

Elastic solids (like gelled drilling muds and elastomers) can not be treated in the same way as liquids and gases, using the z factor and fugacity (unless the loading on the solid is purely isotropic).  This is because, at equilibrium, a solid is capable of storing elastic energy (involving anisotropic loading and strains) after it is deformed.  As a result, the equilibrium state of an elastic solid cannot be described in terms of only two intensive variables (like it can for a liquid or gas).  In the case of an isotropic elastic solid, you need to specify either (a) temperature and three principal stresses or (b) temperature and 3 principal strains).  So, rather than 2 intensive variables, you need to specify 4 intensive variables.  All the thermodynamic functions, U, H, S, A, and G are functions of these 4 intensive variables.


Mechanism of Oxymercuration-Demercuration reaction

Here is the mechanism I wrote out for you


Which element has a greater second electron affinity, sulfur or oxygen?

The most important fact here is that the electron added to $\ce{S-}$ goes into an $n=3$  orbital ($3p$), whereas the one added to $\ce{O-}$ goes into an $n=2$ orbital ($2p$).

The larger $n=3$ orbitals are further from the nucleus and hence experience a weaker attraction to the nucleus and are much higher in energy.  This dependence with quantum number scales approximately $E\sim - n^{-2}$ so is very important [1].

The dependance of orbital energy also scales quadratically with effective nuclear charge, however effective nuclear charges falls significantly compared to the true nuclear charge due to shielding and repulsion from other electrons.  A common first order approximation for the effective nuclear charge is given by Slater's rules [2]

Combining these two approximations we can calculate very approximate orbital energies.

$$
\begin{align}E_{2p}(\ce{O^2+}) & {} = - R_\infty h c \frac{Z_{eff}^2}{n^2} \\
& {} = - R_\infty h c \frac{(8-4.15)^2}{2^2} \\
& {} = - 3.7 \text{ Ry} \\
E_{3p}(\ce{S^2+}) & {} = - R_\infty h c \frac{Z_{eff}^2}{n^2} \\
& {} = - R_\infty h c \frac{(16-11.25)^2}{3^2} \\
& {} = - 2.5 \text{ Ry}
\end{align}
$$

Where 1 Ry is the electronic energy of a hydrogen atom.

As you can see, to first order, the energy of the new electron is much lower when added to $\ce{O-}$ than when added to $\ce{S-}$, leading to a more favourable electron affinity for $\ce{O-}$ than $\ce{S-}$, driven by the fact that the orbital occupied has a lower principal quantum number $n$

[1] - See the energy of a hydrogenic ion - Wikipedia.

[2] - Slater's rules - Wikipedia


Which quantum chemistry approximation to use?

A good guide comes from this series of slides by Jan Řezáč:

It seems that semiempirical methods (of which PM6-D3H4 in MOPAC is probably most accurate for the types of systems encountered in biology) are the only methods able to handle structure optimization of a whole protein (or whole domain of a larger protein).  DFT can handle an order of magnitude fewer atoms, and MP2 or CCSD(T) are mainly used for creating benchmark data sets to test the accuracy of methods, such as S66.


Why/how is blood red? (colours of hemoglobin)

The $\ce{Fe}$-porphyrin complex is a vastly conjugated $\pi$-system, in which the oxidation state of the $\ce{Fe}$ plays a central role. The wavelength of light absorbed by hemoglobin exactly corresponds to the the difference in energy between the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) of the heme $\pi$-system, and is what gives hemoglobin its color. These values are exact and quantifiable. Changes in this color, therefore, are the direct result of changes in the energies of the molecular orbitals of the hemoglobin $\pi$-system.

To understand the color change, it is important to think about the geometry of the molecule. When oxygen is not bound, the iron atom is in the +2 oxidation state. It's slightly too big to fit into the hole in the center of the plane of the immediately-surrounding "heme," so it actually rests just on top of the heme plane. Here is the molecular orbital of the iron species with no $\ce{O2}$ bound:

When $\ce{O2}$ does bind, the $\ce{Fe}$ and $\ce{O2}$ orbitals bond and their energies change. In particular, the $\ce{O2}$ $\pi$* orbitals interact with the $\ce{Fe}$ $\mathrm{xz}$ and $\mathrm{z^2}$ orbitals. A side effect of the $\pi$* accepting properties of all three of these ligands is that iron formally changes from being in the +2 oxidation state to the +3 oxidation state. You can think of it as having "given up" one of its electrons to the ligand via the ligand's $\pi$* orbitals. As a side effect of losing one of its outer electrons, the iron atom is now small enough to align perfectly with heme plane.

Almost counterintuitively, however, formation of oxyhemoglobin is actually energetically disfavorable, which can be explained by its color. Deoxyhemoglobin appears purplish-blue, meaning that it must be absorbing yellow light (lower energy),and is therefore more stable. Oxyhemoglobin appears red, meaning that it must be absorbing at blue-green light (higher energy) and is therefore less stable. This is actually vital to oxyhemoglobin's functionality. Because it is a higher energy, and thereby less stable structure, it easily releases the bound oxygen in the presence of hypoxic cells.

[Source]


How to measure concentrations using emission spectrum?

You are correct in your assumption. The scaling factor is determined experimentally, this is typically called the 'calibration curve'. Basically you measure the intensity of known concentrations and fit a line through the points. Perhaps this paper can help:
http://science.sciencemag.org/content/202/4364/183


How to calculate the hybridization of ethane with H=½[V+M-C+A]?

The mistake that you have made is incorrect understanding of M. From what I remember, M was number of single bonds, which would be only with a monovalent atom, in the context you learnt tbe formula. This was taught in chemical bonding where only atoms are bonded to central atom.
Ex- PCl5, PCl3
You should modify this definition to get the answer  here. Any monovalent group (-CH3) should be there in M.


According to the InCHI FAQ, the 2nd block encodes "encodes stereochemistry and isotopic substitution information". So if you want to generate the corresponding InChIKey for the case where there is no stereo present nor isotopes specified, then it looks like this is possible. If you also want to force the neutral form, you could try setting the final character to 'N' (it may instead be P or O).

Your mileage may vary. To be honest, I wouldn't do this if I were you. I would generate a SMILES without isotopic or stereo information (or remove this programmatically), and then get the InChIKey of that. This would avoid any nasty surprises.


Yes, it has a lot to do with mass.  Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus (see "Note" below for more detail on this point). The smaller orbital radius for the deuterium electron translates into a shorter (and stronger) $\ce{C-D}$ bond length.

A shorter bond has less volume to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) over resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond.

The net effect is that the shorter bond with deuterium increases the electron density at carbon, e.g. deuterium is inductively more electron donating than protium towards carbon.

Similar arguments can be applied to tritium and it's even shorter $\ce{C-T}$ bond should be even more inductively electron donating towards carbon than deuterium.

Note: Bohr Radius Detail

Most introductory physics texts show the radius of the $n^\text{th}$ Bohr orbit to be given by

$$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 m_\mathrm{e}}$$

where $Z$ is the atom's atomic number, $k_\mathrm{c}$ is Coulomb's constant, $e$ is the electron charge, and $m_\mathrm{e}$ is the mass of the electron.  However, in this derivation it is assumed that the electron orbits the nucleus and the nucleus remains stationary.  Given the mass difference between the electron and nucleus, this is generally a reasonable assumption.  However, in reality the nucleus does move too.  It is relatively straightforward to remove this assumption and make the equation more accurate by replacing $m_\mathrm{e}$ with the electron's reduced mass, $\mu_\mathrm{e}$

$$\mu_\mathrm{e} = \frac{m_\mathrm{e}\times m_\text{nucleus}}{m_\mathrm{e} + m_\text{nucleus}}$$

Now the equation for the Bohr radius becomes

$$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 \mu_\mathrm{e}}$$

Since the reduced mass of an electron orbiting a heavy nucleus is always larger than the reduced mass of an electron orbiting a lighter nucleus

$$r_\text{heavy} \lt r_\text{light}$$

and consequently an electron will orbit closer to a deuterium nucleus than it will orbit a protium nucleus.


Determining which compound is more 'ionic'

Tl, dr -- upon further review we can't really tell whether potassium sulfide or beryllium fluoride is more ionic without some quantitative details.

There is more involved in the balance between ionic and covalent bonding than electronegativity differential.  You have to consider molecular orbital structure, too.  One of the answers to this question describes how molecular orbital structure can drive highly ionic bonding when electronegativity differences are modest (as in salts with aromatic ions) or even the "wrong way" (as in alkalides).

With single atoms, as pointed out in the other question, you cannot expect 100% ionic bonds even when molecular orbitals favor it.  Yet there can still be an impact which we can see by comparing organolithium compounds with organic fluorides.  The former acts as a ready source of anionic carbon, while the latter are notoriously unreactive towards the displacement of fluoride ions, as if the carbon-lithium bond had much more ionic character.  Yet the electronegativity differentials in the carbon-lithium and carbon fluorine bonds are about the same (oppositely directed with respect to carbon, of course).  The difference comes from lithium having a diffuse valence orbital that overlaps poorly with carbon, whereas fluorine is more compact and overlaps well, thus promoting electron sharing between both atoms in the bonding orbital.

Similarly, potassium has a diffuse valence orbital, even more so than lithium, which overlaps poorly with most nonmetals.  The ionicity of potassium sulfide, already considerable based on electronegativity, is thus enhanced even more.  Beryllium and fluorine, with good covalent overlap, would have a more covalent interaction than what might be expected from their electronegativity difference.

Since potassium sulfide has poor overlap and beryllium fluoride has a somewhat higher electronegativity difference, the two effects oppose each other when the compounds are compared.  We cannot tell from these qualitative arguments which one is more ionic.


If the standard state symbol means that the substance is pure (and at 1 bar) how is it possible to have a standard REACTION enthalpy?

The standard reaction enthalpy you have describe is only for a pure substance if you are talking about a standard enthalpy of formation.  Otherwise the equation wouldn't be valid for reactions that have more than one product, right?  In the specific case of a standard enthalpy of formation, you are talking about the enthalpy change associated with the creation of one mole of a pure substance from its component elements in their standard states.  This gets at your second question ... the zero on the enthalpy scale is established by convention (and this is the reason why the definition of standard enthalpy of formation works) ... the standard enthalpy of formation of an element in its standard state is DEFINED to be zero.

Now, you can estimate the standard reaction enthalpy for any chemical reaction by subtracting the summed standard enthalpies of formation of the reactants from the summed standard enthalpies of formation of the products, taking into account the appropriate stoichiometric coefficients from the balanced chemical equation.  That process is similar to what you have written, but not precisely the same, since you omitted the stoichiometry information (and the chemical equation, for that matter).  The standard enthalpy estimate obtained will pertain to reactions carried out at the given conditions .. standard enthalpies are defined for 1 atm pressure but variable temperature, so you may need to correct tabulated standard enthalpies of formation to match the given reaction conditions.

One last point is that enthalpy is absolutely defined as a thermodynamic quantity .. it is the internal energy plus the product of pressure and volume, $H=U + pV$.  The issue is that the internal energy of a substance is hard to define absolutely .. are you going to include the binding energies of the core electrons to their atoms?  How about the strong forces holding the nuclei together?  Thus for chemical processes it almost always makes sense to deal with just changes in enthalpy, since most of the really low level stuff I mentioned doesn't change between reactants and products.  I think that may have been part of what you were getting at with your second question.


Yes.

The compound is soluble in water to at least 25 mg/ml - which is (obviously) the same as g/l. Presumably they didn't test at higher concentration.

Alternatively, it is soluble to 50 mM, i.e. mmol/l, in pure DMSO.

Why they would use different units is beyond me,  but I would guess this has to do with the units conventionally used in the two experiments they've run.


Why does a low concentration of soap in water behave as a strong electrolytes?

Could you define associated colloids as it appears in your text? This is not a modern colloid chemistry term. A quick search of Google Books shows that this terminology is popular in Indian test preparation books. Micelle is a modern term. IUPAC defines micelle as "Particle of colloidal dimensions that exists in equilibrium with the molecules or ions in solution from which it is formed" Above a critical concentration, called the critical micellar concentration (CMC), the conductivity of solution drops very quickly. I will let you read more about micelle and conductivity of micellar solutions from the web.

You can watch this video for starting purposes: https://www.youtube.com/watch?v=M6LS4XZswNg. Basically micelle is formed by molecules which have a polar (ionizable end) and a long alkyl chain. Just like any electrolyte they conduct electricity in water. However above CMC, these molecules rather suddenly form a structure called a micelle as shown below (from Google Images). For these reasons, there is sudden change in lot of properties such as surface tension, conductivity and so on. It is not only conductivity.


What does CH4 form when dissolved in aqueous solution?

Methane dissolved in water is simply methane. It does not react with the water under normal conditions.


Why is fluorine a gas, but lithium isn't?

A lithium atom has one valence electron, easily lost (shared), so it is connected to other atoms by a metallic bond. This is a  bit like the shell game where a pea (electron) is hidden under a walnut shell... the uncertainty of where it is at any instant implies, in a quantum sense, that all the atoms share it, and are held together. This bond is so strong that the next element, beryllium, has a boiling point of $\pu{3243 K}$.

Fluorine, on the other hand, has an almost complete shell and each atom forms a tight covalent bond with just one other fluorine atom. These molecules of $\ce{F2}$ keep their electrons to themselves and do not associate with each other.

The difference in bonding is like the difference between bricks mortared together in pairs, or mortared together in a three-dimensional structure, and is responsible for the difference in mp and bp.


Please explain the shapes of the orbitals

I'm guess that you have read about orbital shapes in the wikipedia article, or done a google search on the term.

In general, such "orbitals" shown are typically calculated for a lone electron, not any "real" multi-electron configuration.

Think of an orbital like a single loop of cotton fiber wound into a cotton ball. The planetary model of the electron-nucleus pair indicates that the electron is a solid ball following the fiber. The quantum model of the electron-nucleus pair indicates that the electron has no fixed position, but that it is essentially at all places on the loop at the same time.

So how big is the cotton ball in diameter? It has no limiting diameter, it depends on how hard you squeeze! So if I catch a Hydrogen atom with my magic tweezers, there is a finite probability according to the Schrödinger wave equation that its electron can be in orbit even further away than the dwarf planet Pluto, or inside the nucleus! So we we talk of an the "size" of an ion it is somewhat of an artificial abstraction, and the Schrödinger wave equation can't be "the gospel truth." (I don't mean to badly disparage the Schrödinger wave equation for it is very useful.)

Not to leaving you totally confused, the size of an ions does have some real physical significance. For instance consider table salt, $\ce{NaCl}$. This is really $\ce{NaCl}$. Using x-ray diffraction we can measure how close the $\ce{NaCl}$ are, so we know their "size."

Starting with the Aufbau principle for the atoms, chemists can predict the electron structure of an atom, say carbon. Given the structure of bonds in the atoms, say carbon and hydrogen, then chemists can predict how carbon and hydrogen will bond to form molecules. The fact that chemists can make predictions about molecular structure is the "proof" that the models work.

Howver weird things do have with real electron orbitals. For instance if there was a single "normal electron configuration" for chromium, then there would only be one chromium chloride. However chemists have synthesized three.

Chromium(II) chloride, also known as chromous chloride.
Chromium(III) chloride, also known as chromic chloride or chromium
trichloride
Chromium(IV) chloride

So there must be weird configurations that are stabilized by some sort of "hybridization." As another example all the C-H bonds in methane, $\ce{CH4}$, are the same because the four carbon orbitals (one 2s, and three 2p) orbitals of the carbon atom hybridize into four $sp^3$ orbitals that are equivalent.

To put such orbital shapes into perspective, a chemist thinks a unicorn is simply a hybrid of a horse and a rhinoceros. So a model helps making predictions, but you can't have a fixated belief that the model is the whole
truth.

Another example would be showing you a picture of a blueberry pie. Would just such a picture "explain" a blueberry pie? In order for the picture to have meaning you must have some underlying knowledge.

You know in general what a pie is.
You have know that pies contain fruit pieces, so this kind is made
from blueberries.
You know that pies are sweet.


Ideally, it takes 1.23 volts to split water. Less voltage than this will give essentially no current, no electrolysis. If you want more H2 + O2, you raise the potential. (Of course, you use an electrolyte like H2SO4 or NaOH with electrodes that will not corrode. Platinum is OK, also gold. Oh, yeah, also stainless steel for ordinary people.) If the surface becomes covered with bubbles of gas, the effective area of the electrode is reduced, so the current will drop. So you raise the voltage, and bubbles come pouring off. Then the electrolysis uses more energy (volts x amps). 1 amp for 96,500 seconds (27 hours) will give you 1 gram of hydrogen and 8 grams of oxygen. The product is dependent on the current, and the current is dependent on how high you raise the voltage.

It's not as slow as watching grass grow, but even at furious bubbling, it's surprising how slowly the volumes build up.


The bond between $\ce {Pt} $ and ethene is not purely a $\sigma $ bond and also not a pure $\pi $ bond. The bonding  actually consists of a combination of both bonding and back-bonding effect.

If you look into the Molecular Orbital Diagram of Ethene, the picture will be more clear. For the formation of bond between the metal and the ligand, electrons from filled $\pi $-bonding MO of ethene ($\pi_b $) is donated to the vacant orbitals of Platinum. And electrons from the filled orbitals of $\ce {Pt} $ is back-donated to the vacant $\pi^*$ orbitals of ethene. Thus the bonding would have been a $\sigma $ bond if the electrons are donated from a $\sigma_b $  or $\sigma_{nb} $ orbitals. But here the bonding mainly is $\pi $ donation and $\pi$ acceptance ( known as Synergic effect).

Thus, it would be better not to consider the bond as $\sigma $ bond, but may be considered more of a $\pi $ type but also not a pure one due to the back-bonding.


Why can't alcohols turn blue litmus red?

Litmus is a weak acid, with a $\mathrm{p}K_\mathrm{a}$ of about 6.5.

Alcohols are extremely weak acids, with $\mathrm{p}K_\mathrm{a}$'s (typically) around 17.

The litmus itself is a much stronger acid than the alcohol. Dissociation of the alcohol won't produce enough protons to shift the equilibrium between the two colored forms of the indicator, so you'll see no color change.


How do I find out if ions are oxidised/ reduced based on half cell potentials?

Look at this table of half-reactions: http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm

I'll explain how it contradicts your answer to the first question; you can then rework your answers and update your question with justifications.

  Is $\ce{Fe^{2+}(aq)}$ capable of reducing $\ce{Cr^{3+}(aq)}$ to $\ce{Cr^{2+}(aq)}$?

One entry of the table says $$ \ce{Cr^{3+}(aq) + e- -> Cr^{2+} (aq)} \quad –0.424.$$

The $-0.424$ is what's called the standard potential, expressed in volts.  It is a measure of how favorable the half-reaction is (compared with a standard half-reaction).  The higher the number, the more favored the products will be.

Now you have to find a half-reaction involving $\ce{Fe^{2+}}$.  Since you want Fe to reduce something, it will itself be oxidized.  The table shows no half-reaction where $\ce{Fe^{2+}}$ is oxidized, but that's because by convention only reduction half-reactions are shown; to get the opposite, oxidation, half-reactions, just swap the terms and negate the potential.  That is, take this line from the table:
$$
\ce{Fe^{3+}(aq) + e^{–}  -> Fe^{2+}(aq) } \quad 0.771
$$
and turn it into
$$
 \ce{Fe^{2+}(aq) -> Fe^{3+}(aq) + e^{–}} \quad {-}0.771
$$

Now add this, term by term, to the Cr half-reaction. The result is
$$
\ce{Fe^{2+} + Cr^{3+} -> Fe^{3+} + Cr^{2+}} \quad {-}1.195
$$
If the potential in the final reaction is negative, the reaction is impossible (more precisely, equilibrium lies to the left --- in the case very far to the left, becase 1 volt goes a long way).

That's the basics.  There is a subtlety involved when the number of electrons exchanged is not the same in the two half-reactions; you'll find it in your textbook.  You could try answering the other 5 questions now and if you get stuck, tell us precisely where.


Using osmotic pressure to find the molar mass of a compound

I couldn't resist. The osmotic pressure equation is

$$Π = cRT,$$

where $Π$ is osmotic pressure in atmospheres, $c$ is molar concentration, $R = \pu{0.082057366080960 L atm mol-1 K-1}$ (exactly) and $T$ is temperature in Kelvin.

In this problem, $Π = \pu{0.236 torr}$ times exactly $\pu{1 atm}/\pu{760 torr}$ and $T = \pu{(273.15 + 19.0) K}.$ So $c = \pu{1.2953E-5 M}$. But $\pu{12.5E-6 g}$ of saxitoxin was dissolved in $\pu{3.10 mL}$ of water, which is $\pu{4.032E-3 g/L}$. Hence $\pu{1 mol}$ of saxitoxin has a mass of $\pu{311 g}.$


Questions about air-quality testing in high school

First off the acceptance levels the first time are way lower then the second time, you should ask why there is a large difference. (I have no clue, it's quite important since the second time those limits are NOT met!)

I found NO acceptance level for the chemicals (long term exposure.)
THE most accepted guidelines are usually set by the world health organisation and even they do not dare to set limits since it is difficult see http://www.euro.who.int/en/health-topics/environment-and-health/Housing-and-health/publications/2010/who-guidelines-for-indoor-air-quality-selected-pollutants
For common benzene they wrote a small book.

But to put it in perspective, I think that you inhale way more of those chemicals during filling your car. Then staying in the classroom for a week ;) [Citation needed!]
Edit: the concentrations are actually all more elevated then the averages mentioned by literature. Can we please have a copy of the raport? Next time it might be better to measure air quality according to the WHO guidelines since we simply do not know what we are looking at.

"First, the first round of tests analyze specific areas in the school, but the second does not identify any area.
Both rounds seem to have measured almost exactly the same levels."
I agree, this is weird. You should ask where and how this is sampled. But there are regulations for the sampling, let's assume they are professionals and based on previous results this is a worst case measurement.

"The recommended NIOSH residential levels are measured in ug/m3 in round one testing, but ppm in round two. In round one, for example, the recommended maximum for benzene is 0.360, which according to this tool, translates to 0.00036 ppm. However, on the second test, has a recommended maximum of 0.1 ppm. Also, the same calculator translates benzene’s if/m3 of 8.9 to 0.0089 ppm, but in the Test two results, it is translated to 0.0028."
PPM is a horrible unit without actual dimensions just like percentages. Are we talking about volumes, weights , molars or a mix? Anyway, just use the ug/m3 since that's what really Is measured. (I you like you can see the right conversions inside the WHO document http://www.euro.who.int/en/health-topics/environment-and-health/Housing-and-health/publications/2010/who-guidelines-for-indoor-air-quality-selected-pollutants but that's only if you like math ;)

"Finally, parents are simply unfamiliar with these chemicals and what the levels imply, and how they affect teenagers. It is hard to fight with you can’t see or smell."
I totally agree! But don't underestimate the nose ;)
I GUESSE, looking at the numbers there is no extreme dangerous exposure. The second report concludes the the same. The world health organisation does not exactly know what levels are really exceptional dangerous, therefore it's by "law save" as long as there are no observed health problems. But just ask the professionals ;)
This is where my field ends, it's up to you to tell the right people the right information and ask the right questions, (you are probably way better at that then I am ;) good luck!!


How to calculate the parts-per-million concentration from a percent concentration?

You're using $2.5\ \mathrm g$ of solute out of $100\ \mathrm g$ of solution, which is $2.5\ \%$, but the initial question is asking about a $0.025\ \%$ solution, which is why your answer is two orders of magnitude too large. You should be using $0.025\ \mathrm g$ of solute and $99.975\ \mathrm g$ of solvent.

As a general helpful hint for answering multiple choice questions, when you notice that one of the answers has the same coefficient but is off by a couple orders of magnitude (in this case the $64\ \mathrm{ppm}$ answer) go and double check all your initial concentrations and volumes, you probably forgot to account for a "$\%$" or a "$\mathrm k$" somewhere.


Why can't sodium salts be produced through a reaction between a metal and an acid?

Normally, reacting a metal and an aqueous acid will produce a salt and hydrogen gas, as in the case of iron:

$$\ce{Fe (s) + 2HCl(aq) -> FeCl2 (aq) + H2 (g)}$$

However, you should not use the same method to produce, for example, sodium chloride:

$$\ce{2Na (s) + 2HCl(aq) -> 2NaCl (aq) + H2 (g)}$$

because sodium reacts violently with water.


Reaction of Grignard reagent with primary amines

Here's the thing about Grignard reagents.  Though we typically want them to do nucleophilic addition to something, these powerful bases tend to go for protons -- a faster reaction -- if there is just about anything that might make the proton vulnerable.  Attaching the proton to an electronegative atom like nitrogen is enough, so primary and even secondary amines will react like that with Grignard reagents.  You just deprotonate the amine, the hydrocarbon part of the Grignard reagent ends up an an alkane (or an arene if there are aromatic rings), then on acid workup you get just the amine (or ammonium ion) back.  Yup, this is not a very productive synthesis.

Choose your substrate carefully with Grignard reagents to avoid this deprotonation and get the addition you want, or consider less strongly basic alternatives like an organocadmium compound (which itself can be made from a Grignard reagent plus a cadmium salt).


The bonding of alkali cations to water, to EDTA, and to crown ethers is strictly electrostatic in all three cases, at least based upon QTAIM analysis$^{[1]}$ of representative systems.

For this analysis, I ran gas-phase quantum chemical optimizations on $[\ce{Li}(\text{12-crown-4})]^+$, $[\ce{Na}(\text{EDTA})]^{3-}$, and $[\ce{Na}(\ce{H2O})_6]^+$ using ORCA v3.0.3. (I can post the computation inputs/outputs if anyone is interested.) I also optimized $\ce{NaCl}$ as an additional system with uncontroversial ionic bonding. While the gas-phase results will poorly represent a variety of aspects of the dissolved species, I think they should capture well enough the characteristics of the wavefunction necessary to study the types of bonding involved. Other gas-phase calculations I've run on, e.g., $[\ce{Cu}(\ce{H2O})_6]^{2+}$ have optimized smoothly to the expected octahedral ligand arrangement, lending credence to this assumption.

Some notes on the optimizations:

Geometry optimization of $[\ce{Li}(\text{12-crown-4})]^+$ converged readily.
For $[\ce{Na}(\text{EDTA})]^{3-}$, despite starting in a prototypical octahedral chelation geometry, the EDTA molecule immediately began to 'loosen its grip' on the $\ce{Na+}$, clearly heading toward a less compact configuration. In fact, one of the oxygen atoms actually discarded its bonding interaction with the $\ce{Na+}$ in favor of an intramolecular hydrogen bond.  I performed the QTAIM analysis on the last geometry obtained before ORCA reached its iteration limit in the geometry optimzation procedure.
Similarly, for $[\ce{Na}(\ce{H2O})_6]^+$, despite starting the optimization in a prototypical octahedral coordination geometry, the water ligands 'wandered' throughout the optimization process, in a fashion loosely reminiscent of a molecular dynamics run. As with the EDTA complex, I just chose the final geometry after the optimizer halted as the one on which I ran the QTAIM analysis.

For illustration, I generated GIF animations of the $[\ce{Na}(\text{EDTA})]^{3-}$ and $[\ce{Na}(\ce{H2O})_6]^+$ optimizations. The individual frames were produced with the 'Movie Maker' module of VMD 1.9.1 and the final GIFs were generated using EZgif.com.

Several metrics generated by QTAIM analysis argue for strictly ionic bonding of the alkali metal atoms in these systems:

Low electron density $(\rho \substack{<\\\sim}0.1)$ and positive density Laplacian $(\nabla^2\rho>0)$ at the relevant line critical points between the alkali cation and the coordinating oxygen (or nitrogen) atoms
High localization index $(\mathrm{LI}\substack{>\\\sim}0.9)$ in the alkali atom basin
Low delocalization index $(\mathrm{DI}\substack{<\\\sim}0.75)$ between the alkali atom basin and those of the coordinating atoms

Strictly speaking, in addition to $\#1$ above, the terms of the Laplacian perpendicular to the bond path must also be small in magnitude in order to diagnose ionic bond character$^{[2]}$, but I have omitted those values here.

The table below presents these metrics for the above representative chemical systems, as well as some reference molecules possessing uncontroversial bonding types.  Due to the fairly regular geometry of $[\ce{Li}(\text{12-crown-4})]^+$, the metrics didn't vary much throughout the system and I averaged them for each bond and atom type.  Despite the lack of a deep energy well, the metrics were also similar for the six oxygens in $[\ce{Na}(\ce{H2O})_6]^+$, so I averaged the values here as well.  For the irregular geometry of $[\ce{Na}(\text{EDTA})]^{3-}$, the values differed enough that I chose to report them indivdually; the numbering of the coordinating atoms is arbitrary.  All non-literature data were generated by MultiWFN v3.3.7 on Windows 7 using the default settings and the "Medium" grid for generation of the atomic basins for the $\mathrm{DI}$ and $\mathrm{LI}$ calculations. All values are in atomic units: $\rho\equiv{e^-\over\mathrm{Bohr}^3}$ and $\nabla^2\rho\equiv{e^-\over\mathrm{Bohr}^5}$ ($\mathrm{DI}$ and $\mathrm{LI}$ are dimensionless).

$$
~ \\
\textbf{QTAIM Results} \\
\begin{array}{ccccccccc}
\hline
\mathbf{Species} & ~ & \mathbf{Bond} & \rho_\mathbf{LCP} &
    \mathbf{\left(\nabla^2\rho\right)_\mathbf{LCP}} & \mathbf{DI} & ~ &
    \mathbf{Atom} & \mathbf{LI} \\
\hline
 & &  \ce{Li-O} & 0.036 & +0.26 & 0.077 & & \mathrm{Li} & 0.922 \\
[\ce{Li}(\text{12-crown-4})]^+ & & \ce{O-C} & 0.25 & -0.50 & 0.856 & &
     \ce O & 0.870 \\
 & & \ce{C-C} & 0.26 & -0.70 & 0.949 & & \ce C & 0.656 \\
 & & \ce{C-H} & 0.28 & -1.01 & 0.901 & & \ce H & 0.420\\
\hline
& & \ce{Na-O}1 & 0.015 & +0.086 & 0.064 & & \ce{Na} & 0.974 \\
& & \ce{Na-O}2 & 0.021 & +0.131 & 0.096 & & \ce{O}1 & 0.899 \\
[\ce{Na}(\text{EDTA})]^{3-} & & \ce{Na-O}3 & 0.020 & +0.121 & 0.089 & &
     \ce{O}2 & 0.900 \\
& & \ce{Na-N}1 & 0.014 & +0.074 & 0.056 & & \ce{O}3 & 0.900 \\
& & \ce{Na-N}2 & 0.018 & +0.094 & 0.065 & & \ce{N}1 & 0.766 \\
& &  & & & & & \ce{N}2 & 0.769 \\
\hline
[\ce{Na}(\ce{H2O})_6]^+ & & \ce{Na-O} & 0.014 & +0.086 & 0.057 & & \ce{Na} & 0.975 \\
& & & & & & & \ce{O} & 0.916 \\
\hline
[\ce{NaCl}]^0 & & \ce{Na-Cl} & 0.035 & +0.197 & 0.321 & & \ce{Na} & 0.978 \\
& & & & & & & \ce{Cl} & 0.975 \\
\hline
[\ce{LiF}]^{0\,\ddagger} & & \ce{Li-F} & 0.079 & +0.749 & 0.179 & & \ce{Li} & 0.957 \\
& & & & & & & \ce{F} & 0.991 \\
\hline
[\ce{CH4}]^{0\,\ddagger} & & \ce{C-H} & 0.291 & -1.168 & 0.980 & & \ce{C} & 0.661 \\
& & & & & & & \ce{H} & 0.469 \\
\hline
^\ddagger~_{\text{Values from Ref. [2]}}
\end{array}
$$

As can clearly be seen, the interactions between the alkali metals and the ligating atoms are uniformly of ionic character: low electron density and positive Laplacian at the LCP; low $\mathrm{DI}$ between the paired basins, and quite high $\mathrm{LI}$ for the alkali metal basins. This is consistent with the ionic species $\ce{LiF}$ and $\ce{NaCl}$ in the table, and is in marked contrast to the covalent systems included.

$^{[1]}$ More information can be found on Wikipedia and at RFW Bader's page at McMaster University.  A more readable (but unfortunately paywalled) description is provided in Ref. [2].
$^{[2]}$Bader & Matta, Found Chem 15: 253 (2013). doi:10.1007/s10698-012-9153-1.


What is the net ionic equation for the reaction between Ba(OH)2 and H2SO4?

The reaction between $\ce{Ba(OH)2, H2SO4}$ is known as acid-base neutralisation, as $\ce{Ba(OH)2}$ is a relatively strong base and $\ce{H2SO4}$ the strong acid.
The molecular reaction is given as,
$$\ce{Ba(OH)2 + H2SO4 -> BaSO4 (s) + 2H2O}$$

Before moving to net-ionic reaction, let's see the reaction of $\ce{BaCl2, Na2SO4}$ [1]:
$$\ce{BaCl2 + Na2SO4 -> BaSO4 (s) + 2NaCl}$$
So, breaking up into ions, we see:
$$\ce{Ba^{2+} + 2Cl- + 2Na^{+} + SO4^{2-} -> BaSO4 (s) + 2Na^{+} + 2Cl^{-}}$$
$\ce{Na^{+}, Cl^{-}}$ remains completely dissociated in the aqueous solution as they're salts of strong acid and strong base (i.e., they're strong electrolytes). Hence, $\ce{Na^{+}, Cl^{-}}$ are spectator ions as they don't participate in ionic reaction. Let's cancel them out, and we get the net ionic reaction:
$$\ce{Ba^{2+} + SO4^{2-} -> BaSO4 (s)}$$

Now, let's see the reaction given in the question. In this, $\ce{BaSO4}$ precipitates out, so $\ce{Ba^{2+}, SO4^{2-}}$ can't be spectator ions. Also, H2O has a very low dissociation contant ($\ce{K_W = 10^{-14}}$, at 25°C), so it can't dissociate into ions. Hence, $\ce{H+, OH-}$ are too not the spectator ion. Therefore, this reaction has no spectator ion and hence ${H_2O}$ will be included in the ionic reaction, which is given as:
$$\ce{Ba^{2+} + 2OH- + 2H+ + SO4^{2-} -> BaSO4 (s) + 2H2O(l)}$$

Another example:
[1]: Complete ionic and net ionic equations, Khan Academy
Note: As mentioned by andselisk in comments, $\ce{BaSO4 (s)}$ is the recommended way to write the precipitate, as per IUPAC. Old school symbol for precipitate was "↓".


Cleaning an overhead water tank using Alum or other chemicals?

Aluminum is remarkable safe for human consumption, also considering that it is present in high concentration essentially everywhere, and it is not known to have any significant specific biological roles. Wikipedia happens to have a nice well researcher section on aluminum's possible effects on human and animal health: Aluminium: health concerns. Alum is also routinely added to water in low concentrations for elimination of solid particulate, as you mentioned.

My main concern for your application is the fact that typical methods of water purification from solid particulate, just like addition of alum, will tend to form flocs in your water tank that will then need to be removed! Depending whether this scum will float on top of the water, or sink to the bottom of the tank, the removal may be more or less easy, also depending on what kind of access you have to this tank. I doubt if any of those solutions would count as not "doing much effort"...


A double bond between 2 carbon atoms means there is a $\pi$ bond with the side orbital overlaps in 1 plane, in parallel to the axial $\sigma$  bond.

The consequence is, free rotation  would need to break this $\pi$ bond, making the double bond just a single bond, what would require a strong torque and a lot of energy.

By other words, particular rotation positions around a single bond have ( almost ) the same energies. But for double  bonds, the molecule energy strongly depends on on this orientation. The $\pi$ bond can exist, only if the bystander groups on both carbon atoms   are about in the same plane. If they wete on about perpendicular planes, the  orbitals of both atoms cannot overlap to form the bond.


The carbon chain of a fischer projection is drawn vertically.

Yes, the vertical and horizontal lines depict 'into the page' and 'out of the page' respectively.

However, when you have a carbon chain with more than 3 carbons, each down carbon on the backbone which would normally be projected away from you (according to the above rule) has to be rotated. Essentially, this creates the 'zig-zag' you would see in other projections (e.g. Natta Projection).


3,4-Methylenedioxyamphetamine (MDA) basic extraction

MDA is an alkaloid, i.e., basic. When you talk about its $\mathrm{p}K_\mathrm{a}$ value, you're really talking about the $\mathrm{p}K_\mathrm{a}$ of the protonated ammonium ion.

The ammonium ion is water soluble, allowing us to potentially separate it from the other organic components of the reaction. After making the aqueous solution basic, the ammonium ion is deprotonated to form the free amine, which is much more soluble in the organic phase than the aqueous phase. Then, it is back extracted into clean organic solvent. This isolates (hopefully) just the free amine in the organic layer during the extraction.

So in answer to your question, what you have in the organic layer is the free amine form of MDA, not the ion.