Where does the label ‘Dq’ to denote the field split in coordination compounds come from?

I will expand my previous comment into an answer.
In the $O_h$ case the metal ion is surrounded by $6$ ligand ions, each with charge $q=-Ze$ located at $\langle\pm a,0,0\rangle$, $\langle0, \pm a,0\rangle$, and $\langle0,0,\pm a\rangle$.
The electrostatic potential of a point charge $q$ is
$$V=\frac{kq}R$$
where
$$k=\frac1{4\pi\epsilon_0}$$
is the Coulomb's Law constant and $\epsilon_0$ is the permittivity of free space and $R=\left\|\vec{r_f}-\vec{r_s}\right\|$ is the distance from the source point $\vec{r_s}$ to the field point $\vec{r_f}$.
Working for the moment with the potential $V_1$ due to the charge at $\langle a,0,0\rangle$,
$$R=\left\|\langle x,y,z\rangle-\langle a,0,0\rangle\right\|=\left\|\langle x-a,y,z\rangle\right\|=\sqrt{(x-a)^2+y^2+z^2}=\sqrt{a^2-2ax+x^2+y^2+z^2}=a\sqrt{1-\frac{2x}a+\frac{r^2}{a^2}}$$
Where $r^2=x^2+y^2+z^2$ is the squared distance from the origin. Then
$$V_1=-\frac{Ze}{4\pi\epsilon_0a}\left(1-\frac{2x}a+\frac{r^2}{a^2}\right)^{-1/2}$$
Now we need to expand $(1+x)^{-1/2}$ to fourth order in $x$. The Taylor series yields
$$(1+x)^{-1/2}=1-\frac12x+\frac38x^2-\frac5{16}x^3+\frac{35}{128}x^4+O\left(x^5\right)$$
We could have obtained this result via algebra alone by assuming that $(1+x)^{-1/2}=1+ax+bx^2+cx^3+dx^4+O\left(x^5\right)$, then squaring and multiplying by $(1+x)$ so that
$$\left(1+ax+bx^2+cx^3+dx^4\right)^2(1+x)=1+O\left(x^5\right)$$
and then finding the unknowns $a$, $b$, $c$, and $d$ from the conditions that the coefficients of $x$, $x^2$, $x^3$, and $x^4$ should be zero. Then we have
$$\begin{align}V_1= & -\frac{Ze}{4\pi\epsilon_0a}\left\{1-\frac12\left[-\frac{2x}a+\frac{r^2}{a^2}\right]+\frac38\left[\frac{4x^2}{a^2}-\frac{4xr^2}{a^3}+\frac{r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[-\frac{8x^3}{a^3}+\frac{12x^2r^2}{a^4}\right]+\frac{35}{128}\frac{16x^4}{a^4}\right\}+O\left(\left(\frac ra\right)^5\right)\end{align}$$
This is complicated, but now we simplify by combining the potentials due to all $6$ ligands. We want to find group operations from $O_h$ that map the first ligand to the others, and fortunately the powers of $S_6$, the rotoreflection by angle $\frac{\pi}3$ about the axis through the origin and $\langle1,1,1\rangle$ will do the trick.
Its powers cycle the original ligand point through all the others. We just need to find the mappings of the powers of $x$ by the powers of $S_6$ and sum over all powers of $S_6$:
$$\begin{array}{c|cccccc|c}
f(x,y,z)&Ef&S_6f&S_6^2f&S_6^3f&S_6^4f&S_6^5f&\sum\limits_{i=0}^5S_6^if \\
\hline r^2&r^2&r^2&r^2&r^2&r^2&r^2&6r^2 \\
x&x&-z&y&-x&z&-y&0 \\
x^2&x^2&z^2&y^2&x^2&z^2&y^2&2r^2 \\
x^3&x^3&-z^3&y^3&-x^3&z^3&-y^3&0 \\
x^4&x^4&z^4&y^4&x^4&z^4&y^4&2\left(x^4+y^4+z^4\right)
\end{array}$$
So now
$$\begin{align}V_{O_h}= & \sum\limits_{i=0}^5S_6^iV_1 \\
& =-\frac{Ze}{4\pi\epsilon_0a}\left\{6-\frac12\left[0+\frac{6r^2}{a^2}\right]+\frac38\left[\frac{8r^2}{a^2}-0+\frac{6r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[0+\frac{24r^4}{a^4}\right]+\frac{35}{4}\frac{x^4+y^4+z^4}{a^4}\right\}+O\left(\left(\frac ra\right)^5\right) \\
& =-\frac{Ze}{4\pi\epsilon_0a}\left[6-\frac{21r^4}{4a^4}+\frac{35\left(x^4+y^4+z^4\right)}{4a^4}\right]+O\left(\left(\frac ra\right)^5\right) \\
& =-\frac{6Ze}{4\pi\epsilon_0a}-\frac{35Ze}{16\pi\epsilon_0a^5}\left(x^4+y^4+z^4-\frac35r^4\right)+O\left(\left(\frac ra\right)^5\right)\end{align}$$
All this mathjax is exhausting. I can't finish in one sitting, but at least I have gotten to
$$D=\frac{35Ze}{16\pi\epsilon_0a^5}$$
EDIT: Now that the potential for $O_h$ geometry has been developed, the same will be done for the $T_d$ case where the metal ion is surrounded by $4$ ligand ions, each with charge $q=-Ze$, this time located at
$\left\langle\frac a{\sqrt3},\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle$, $\left\langle\frac a{\sqrt3},-\frac a{\sqrt3},-\frac a{\sqrt3}\right\rangle$, $\left\langle-\frac a{\sqrt3},\frac a{\sqrt3},-\frac a{\sqrt3}\right\rangle$, and $\left\langle-\frac a{\sqrt3},-\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle$.
The distance from the source point at $\vec{r_s}=\left\langle\frac a{\sqrt3},\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle$ to the field point at $\vec{r_f}=\langle x,y,z\rangle$ is
$$\begin{align}R= & \left\|\vec{r_f}-\vec{r_s}\right\|=\left\|\left\langle x-\frac a{\sqrt3},y-\frac a{\sqrt3},z-\frac a{\sqrt3}\right\rangle\right\| \\
& =\sqrt{\left(x-\frac a{\sqrt3}\right)^2+\left(y-\frac a{\sqrt3}\right)^2+\left(z-\frac a{\sqrt3}\right)^2} \\
& =\sqrt{a^2-\frac{2a}{\sqrt3}(x+y+z)+x^2+y^2+z^2} \\
& =a\sqrt{1-\frac{2(x+y+z)}{\sqrt3a}+\frac{r^2}{a^2}}\end{align}$$
Then the potential due to this charge is
$$\begin{align}V_2 & =\frac{kq}R=-\frac{Ze}{4\pi\epsilon_0a}\left(1-\frac{2(x+y+z)}{\sqrt3a}+\frac{r^2}{a^2}\right)^{-1/2} \\
& =-\frac{Ze}{4\pi\epsilon_0a}\left\{1-\frac12\left[-\frac{2(x+y+z)}{\sqrt3a}+\frac{r^2}{a^2}\right]+\frac38\left[\frac4{3a^2}(x+y+z)^2-\frac{4(x+y+z)r^2}{\sqrt3a^3}+\frac{r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[-\frac8{3\sqrt3a^3}(x+y+z)^3+\frac{4(x+y+z)^2r^2}{a^4}\right]+\frac{35}{128}\left[\frac{16}{9a^4}(x+y+z)^4\right]\right\}+O\left(\left(\frac ra\right)^5\right)\end{align}$$
This time the group operations that map the model point to all the others will be the members of the $D_2$ point group: the identity $E$, and the rotations by $180°$ about the x-axis $C_{2x}$, the y-axis $C_{2y}$, and the z-axis $C_{2z}$.
Again a table for combining the contributions from all point charges:
$$\begin{array}{c|cccc|c}
f(x,y,z)&R_1f=Ef&R_2f=C_{2x}f&R_3f=C_{2y}f&R_4f=C_{2z}f&\sum\limits_{i=1}^4R_if \\
\hline r^2&r^2&r^2&r^2&r^2&4r^2 \\
x&x&x&-x&-x&0 \\
xy&xy&-xy&-xy&xy&0 \\
xyz&xyz&xyz&xyz&xyz&4xyz
\end{array}$$
And the reader can readily verify that also
$$\sum\limits_{i=1}^4R_iy=\sum\limits_{i=1}^4R_iz=\sum\limits_{i=1}^4R_ixz=\sum\limits_{i=1}^4R_iyz=0$$
So the rule is that terms in the expression for $V_2$ with odd powers of any of the coordinates cancels except for $xyz$, which, along with all terms with only even powers of coordinates, adds.
Expanding powers of $(x+y+z)$ and summing over the four sources,
$$\begin{align}V_{T_d} & =\sum\limits_{i=1}^4R_iV_2=-\frac{Ze}{4\pi\epsilon_0a}\left\{4-\frac12\left[0+\frac{4r^2}{a^2}\right]+\frac38\left[\frac4{3a^2}(4r^2)-0+\frac{4r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[-\frac8{3\sqrt3a^3}(6)(4xyz)+\frac{4(4r^2)r^2}{a^4}\right]\right. \\
& \left.+\frac{35}{128}\frac{16}{9a^4}\left[4\left(x^4+y^4+z^4\right)+6(4)\left(x^2y^2+x^2z^2+y^2z^2\right)\right]\right\}+O\left(\left(\frac ra\right)^5\right) \\
& = -\frac{Ze}{4\pi\epsilon_0a}\left\{4+\frac{3r^4}{2a^4}+\frac{20}{\sqrt3a^3}xyz-\frac{5r^4}{a^4}\right. \\
& \left.+\frac{35}{72a^4}\left[4\left(x^4+y^4+z^4\right)+12\left(r^4-x^4-y^4-z^4\right)\right]\right\}+O\left(\left(\frac ra\right)^5\right) \\
& = -\frac{Ze}{4\pi\epsilon_0a}\left[4+\frac{20}{\sqrt3a^3}xyz-\frac{35}{9a^4}\left(x^4+y^4+z^5-\frac35r^4\right)\right]+O\left(\left(\frac ra\right)^5\right) \\
& = -\frac{4Ze}{4\pi\epsilon_0a}-\frac{5Ze}{\sqrt3\pi\epsilon_0a^4}xyz+\frac{35Ze}{36\pi\epsilon_0a^5}\left(x^4+y^4+z^4-\frac35r^4\right)+O\left(\left(\frac ra\right)^5\right)\end{align}$$
On comparison with $V_{O_h}$, the first term is the potential due to the net charge of all the ligands uniformly spread over the surface of a sphere of radius $a$.
The second term has no counterpart in $V_{O_h}$ and causes no splitting but does create mixing between the $d_{xy}$ and $p_z$ orbitals and would lower the energy of the former and raise the energy of the latter.
I don't know how significant that effect is, though. The splitting term is $\frac49D\left(x^4+y^4+z^4-\frac35r^4\right)$ and is in fact $-\frac49$ the value of the splitting term for $O_h$ geometry.
Still not done, but it's getting late.

EDIT:To get the $q$ part of $Dq$ we are going to have to do some integrals. I am going to use the notation
$$\oint f(x,y,z)d^2\Omega$$ to represent the integral of $f(x,y,z)$ over the surface of the unit sphere.
One integral we know right off the bat is
$$\oint 1d^2\Omega=4\pi$$
using the mensuration formula from geometry for the area of a sphere.
Actually, if we didn't know this formula, it would be no problem since if the answer were an unknown, say $\xi$, all the $\xi$'s would cancel out in the final results.
If the integrand is a monomial with an odd power of $z$, then
$$\begin{align}\oint x^ay^bz^{2c+1}d^2\Omega & =\oint \sigma_{hz}\left(x^ay^bz^{2c+1}\right)d^2\Omega=(-1)^{2c+1}\oint x^ay^bz^{2c+1}d^2\Omega \\
& =\frac{1+(-1)^{2c+1}}2\oint x^ay^bz^{2c+1}d^2\Omega=0\end{align}$$
Where $\sigma_{hz}$ is a reflection though the $xy$-plane. Clearly any proper rotation of the integrand about an axis through the origin doesn't affect the value of the integral because it just corresponds to viewing the same function from a different angle.
Also inversion of the integrand through the origin doesn't change the integral and any improper rotation including reflection can be written as inversion times a proper rotation. Thus it can be seen that if
$$\oint x^ay^bz^cd^2\Omega \ne 0$$
Then $a$, $b$, and $c$ must all be even. Also we can rotate coordinates so that we only need compute
$$\oint x^{2a}y^{2b}z^{2c}d^2\Omega$$
for $a \ge b \ge c$.
We can calculate
$$\oint r^2d^2\Omega=\oint(x^2+y^2+z^2)d^2\Omega=3\oint x^2d^2\Omega=4\pi$$
where we have rotated individual terms and then used $r^2=1$ on the surface of the unit sphere. Now,
$$\oint x^2d^2\Omega=\frac{4\pi}3=\oint x^2r^2d^2\Omega=\oint \left(x^4+x^2y^2+x^2z^2\right)d^2\Omega=\oint x^4d^2\Omega+2\oint x^2y^2d^2\Omega$$
Rotating by $\frac{\pi}4$ about the $z$-axis,
$$\begin{align}\oint x^4d^2\Omega & =\oint \left(\frac{x+y}{\sqrt2}\right)^4d^2\Omega=\frac14\oint\left(x^4+4x^3y+6x^2y^2+4xy^3+y^4\right)d^2\Omega \\
& =\frac12\oint x^4d^2\Omega+\frac32\oint x^2y^2d^2\Omega\end{align}$$
This gives us two eqautions in two unknowns and we can solve using Excel or Matlab or whatever to get
$$\oint x^4d^2\Omega=\frac{4\pi}5\text{ and }\oint x^2y^2d^2\Omega=\frac{4\pi}{15}$$
Again we have
$$\oint x^2d^2\Omega=\frac{4\pi}3=\oint x^2r^4d^2\Omega=\oint x^6d^2\Omega+6\oint x^4y^2d^2\Omega+2\oint x^2y^2z^2d^2\Omega$$
$$\oint x^4d^2\Omega=\frac{4\pi}5=\oint x^4r^2d^2\Omega=\oint x^6d^2\Omega+2\oint x^4y^2d^2\Omega$$
$$\oint x^6d^2\Omega=\oint \left(\frac{x+y}{\sqrt2}\right)^6d^2\Omega=\frac14\oint x^6d^2\Omega+\frac{15}4\oint x^4y^2d^2\Omega$$
Solving the last $3$ equations in $3$ unknowns we get
$$\oint x^6d^2\Omega=\frac{4\pi}7\text{, }\oint x^4y^2d^2\Omega=\frac{4\pi}{35}\text{, and }\oint x^2y^2z^2d^2\Omega=\frac{4\pi}{105}$$
Almost there :)
$$\begin{align}\oint x^2d^2\Omega & =\frac{4\pi}3=\oint x^2r^6d^2\Omega \\
& =\oint x^8d^2\Omega+8\oint x^6y^2d^2\Omega+6\oint x^4y^4d^2\Omega+12\oint x^4y^2z^2d^2\Omega\end{align}$$
$$\begin{align}\oint x^4d^2\Omega & =\frac{4\pi}5=\oint x^4r^4d^2\Omega \\
& =\oint x^8d^2\Omega+4\oint x^6y^2d^2\Omega+2\oint x^4y^4d^2\Omega+2\oint x^4y^2z^2d^2\Omega\end{align}$$
$$\oint x^6d^2\Omega=\frac{4\pi}7=\oint x^6r^2d^2\Omega=\oint x^8d^2\Omega+2\oint x^6y^2d^2\Omega$$
$$\oint x^8d^2\Omega=\oint \left(\frac{x+y}{\sqrt2}\right)^8d^2\Omega=\frac18\oint x^8d^2\Omega+\frac72\oint x^6y^2d^2\Omega+\frac{35}8\oint x^4y^4d^2\Omega$$
Solving these $4$ equations in $4$ unknowns,
$$\oint x^8d^2\Omega=\frac{4\pi}{9}\text{, }\oint x^6y^2d^2\Omega=\frac{4\pi}{63}\text{, }\oint x^4y^4d^2\Omega=\frac{4\pi}{105}\text{, and }\oint x^4y^2z^2d^2\Omega=\frac{4\pi}{315}$$
With these integrals in hand we want to normalize the polynomials $xy$ and $x^2-y^2$.
$$\oint (xy)^2d^2\Omega=\oint x^2y^2d^2\Omega=\frac{4\pi}{15}$$
So a normalized spherical harmonic that transforms as the $xy$ partner of the $T_{2g}$ representation of $O_h$ and as the $z$ partner the $T_2$ representation of $T_d$ is
$$Y_{xy}^2(x,y,z)=\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}$$
And we have
$$\psi_{xy}(x,y,z)=Y_{xy}^2(x,y,z)f_{xy}(r)$$
To find the electrostatic potential energy of an electron with such a wave function in the $O_h$ potential we integrate over its charge density to get
$$\begin{align}U_{O_h,xy} & =\int V_{O_h}(x,y,z)\rho_{xy}(x,y,z)d^3volume \\
& =\int\left[-\frac{6Ze}{4\pi\epsilon_0a}-D\left(x^4+y^4+z^4-\frac35r^4\right)\right]\times \\
& (-e)\left[\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}f_{xy}(r)\right]^*\left[\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}f_{xy}(r)\right]d^3volume \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}\oint\left|\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}\right|^2d^2\Omega\int_0^{\infty}\left|f_{xy}(r)\right|^2r^2dr \\
& +De\left(\frac{15}{4\pi}\right)\oint \left(\frac{xy}{r^2}\right)^2\frac{x^4+y^4+z^4-\frac35r^4}{r^4}\int_0^{\infty}r^4\left|f_{xy}(r)\right|^2r^2dr \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}(1)(1)+\frac{15De}{4\pi}\left(\frac{4\pi}{63}+\frac{4\pi}{63}+\frac{4\pi}{315}-\frac35\frac{4\pi}{15}\right)\left\langle r^4\right\rangle \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}-\frac{8De}{105}\left\langle r^4\right\rangle\end{align}$$
Where we have used the separate normalizations of the angular part of the wave function $Y_{xy}^2(x,y,z)$ and the radial part $f_{xy}(r)$ and the average value of $r^4$ for the radial distribution, $\left\langle r^4\right\rangle$.
Now we calculate
$$\oint \left(x^2-y^2\right)^2d^2\Omega=2\oint x^4d^2\Omega-2\oint x^2y^2d^2\Omega=2\left(\frac{4\pi}{5}\right)-2\left(\frac{4\pi}{15}\right)=\frac{16\pi}{15}$$
So a normalized spherical harmonic that transforms as the $x^2-y^2$ partner of the $E_g$ representation of $O_h$ and the $E$ representation of $T_d$ is
$$Y_{x^2-y^2}^2(x,y,z)=\left(\frac{15}{16\pi}\right)^{1/2}\left(x^2-y^2\right)$$
We could also have obtained this normalization by rotating $Y_{xy}^2(x,y,z)$ by $-90°$ about the $z$-axis.
Now we can get the electrostatic potential energy of an electron with wave function
$$\psi_{x^2-y^2}(x,y,z)=Y_{x^2-y^2}^2(x,y,z)f_{x^2-y^2}(r)$$
in the $O_h$ potential as
$$\begin{align}U_{O_h,x^2-y^2} & =\int V_{O_h}(x,y,z)\rho_{x^2-y^2}(x,y,z)d^3volume \\
& =\int\left[-\frac{6Ze}{4\pi\epsilon_0a}-D\left(x^4+y^4+z^4-\frac35r^4\right)\right]\times \\
& (-e)\left[\left(\frac{15}{16\pi}\right)^{1/2}\frac{x^2-y^2}{r^2}f_{x^2-y^2}(r)\right]^*\left[\left(\frac{15}{16\pi}\right)^{1/2}\frac{x^2-y^2}{r^2}f_{x^2-y^2}(r)\right]d^3volume \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}\oint\left|\left(\frac{15}{16\pi}\right)^{1/2}\frac{x^2-y^2}{r^2}\right|^2d^2\Omega\int_0^{\infty}\left|f_{x^2-y^2}(r)\right|^2r^2dr \\
& +De\left(\frac{15}{16\pi}\right)\oint \left(\frac{x^2-y^2}{r^2}\right)^2\frac{x^4+y^4+z^4-\frac35r^4}{r^4}\int_0^{\infty}r^4\left|f_{x^2-y^2}(r)\right|^2r^2dr \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}(1)(1)+\frac{15De}{16\pi}\left((2)\frac{4\pi}{9}+(4)\frac{4\pi}{105}-(4)\frac{4\pi}{63}-(2)\frac{4\pi}{315}-\frac65\frac{4\pi}{5}+\frac65\frac{4\pi}{15}\right)\left\langle r^4\right\rangle \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{4De}{35}\left\langle r^4\right\rangle\end{align}$$
Where we have assumed the average value of $r^4$ is the same as for $\psi_{xy}(x,y,z)$ because they both started out as $d$ orbitals and we haven't taken into account higher-order effects.
Since there are $3$ $T_{2g}$ orbitals and $2$ $E_g$ orbitals, the average potential energy over all orbitals is
$$U_{O_h,ave}=\frac15\left[3\left(\frac{6Ze^2}{4\pi\epsilon_0a}-\frac{8De}{105}\left\langle r^4\right\rangle\right)+2\left(\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{4De}{35}\left\langle r^4\right\rangle\right)\right]=\frac{6Ze^2}{4\pi\epsilon_0a}$$
The splitting in potential energies of the orbitals is
$$\begin{align}U_{O_h,x^2-y^2}-U_{O_h,xy} & =\left(\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{4De}{35}\left\langle r^4\right\rangle\right)-\left(\frac{6Ze^2}{4\pi\epsilon_0a}-\frac{8De}{105}\left\langle r^4\right\rangle\right) \\
& =\frac{4De}{21}\left\langle r^4\right\rangle\equiv10Dq\end{align}$$
Then we get
$$q=\frac{2e}{105}\left\langle r^4\right\rangle$$
and so
$$U_{O_h,x^2-y^2}=\left(\frac{6Ze^2}{4\pi\epsilon_0a}\right)+6Dq$$and$$U_{O_h,xy}=\left(\frac{6Ze^2}{4\pi\epsilon_0a}\right)-4Dq$$
This derivation was a little bloody because I chose to do everything with only algebra.
There is a much more elegant exposition possible that requires higher level math.
Is it worth adding to the end of this answer or posting as another answer?


Why derived name of nomenclature is used for Alkanes and alcohols only?

The derived system is really, really old. Even my textbook from the 20th century has IUPAC naming. My recommendation would be to just find IUPAC naming online, or just get a new textbook.
I can't seem to find much on the derived system, but here's another question about it; LINK
I tried searching about, and there really isn't much about it. I don't see why you'd want to learn about it though, as history of nomenclature isn't very useful or interesting. Besides, if you try to use the derived system, no one else is going to know what compound you're describing.


Is there an easy way to test for the amount of iron (Fe) in water?

In a rainwater tank in contact with air, dissolved iron is probably oxidized to iron(III) ($\ce{Fe^3+}$).

A simple and sensitive test for $\ce{Fe^3+}$ in water uses thiocyanate ions ($\ce{SCN-}$, also known as rhodanide), which form the blood-red coloured complexes $\ce{[Fe(SCN)(H2O)5]^2+}$, $\ce{[Fe(SCN)2(H2O)4]+}$, and $\ce{[Fe(SCN)3(H2O)3]}$.

$$\ce{[Fe(H2O)6]^3+ + SCN- <=> [Fe(SCN)(H2O)5]^2+ + H2O}$$

Take a precise volume of the water (e.g. $10\ \mathrm{ml}$), acidify with dilute sulfuric acid (e.g. $2\ \mathrm{ml}$), add a thiocyanate solution (e.g. $5\ \mathrm{ml}$ $\ce{KSCN}$ or $\ce{NH4SCN}$, $c =0.5{–}2\ \mathrm{mol/l}$), and fill up with distilled water to a precise volume (e.g. $25\ \mathrm{ml}$).

An orange to red colour should appear.

The colour of the solution can be compared to that of standard solutions that contain known concentrations of $\ce{Fe^3+}$, e.g. prepared from iron(III) nitrate ($\ce{Fe(NO3)3.9H2O}$).

In a laboratory, the intensity of the red colour could be measured using a photometer.

Similar simple test kits are commercially available, e.g. for aquariums, garden ponds, or other water or soil samples.


Does Suncream have any affect or help protect against radiation

The only radiation sunscreen protects you against is near UV light not radioactive emission

Sunscreens are designed to protect from the near UV radiation present in sunlight. Some block both UVA (315-400nm) and UVB (280-315nm) light but some don't work well for UVA. All UV light at ground level is harmful to skin to some extent and too much of it will cause skin damage that looks a little like a burn (though small amounts are good for encouraging vitamin D production).

Radioactivity can cause similar looking burns. But the cause of the burns is not near UV light but gamma rays (hard X-rays of much shorter wavelengths and much higher energy than UV light) and alpha and beta particles. All of these are much more dangerous than UV light (as the amount of energy deposited in tissues is far greater per photon or particle).

And sunscreen will offer no protection at all to radioactive emissions. Clothing can protect against alpha particles. But you will need several millimetres of aluminium to stop beta particles (and even then the secondary gamma rays from the interaction of those particles with the aluminium can be harmful). And gamma rays might need a thick chunk of lead to keep them out.


You're right: the silver is reacting with sulfur compounds in the food to form a tarnish of silver sulfide. This is most commonly observed, in my experience, using silver teaspoons with boiled eggs, which are pretty rich in sulfur.
There are a number of reactions that can take place depending on the sulfur-containing species - the abstract from this paper from the Journal of Chemical Education (unfortunately pay-walled) has a nice example:

Silver and silver-plated objects react with sulfur and sulfur compounds to produce silver sulfide $\ce{Ag2S}$ or tarnish. Contact with materials that contain sulfur compounds, such as hard boiled eggs, mayonnaise, mustard, and rubber bands can cause tarnish. In the air, a silver object can tarnish owing to the reaction of silver with hydrogen sulfide ($\ce{H2S}$). This is a gas found in the air as a result of some industrial processes and the decomposition of dead plants and animals. The reaction of silver with hydrogen sulfide to form tarnish is as follows:
$\ce{2 Ag(s) + H2S(g) -> Ag2S(s) + H2(g)}$

In Sterling silver specifically, you'll often also get copper sulfide formed.
The paper also mentions that you can remove the tarnish abrasively, or through reacting the tarnish with aluminium (via solution, electrochemically):

$\ce{3 Ag2S(s) + 2 Al(s) -> 6 Ag(s) + Al2S3(s)}$

But a more common approach used by many commercial products is to dissolve off the tarnish, usually in sodium thiosulfate or thiourea. This means you end up with slightly less silver on your item, but is otherwise pretty effective and very mild (it's commonly used for museum pieces).


Is hydrogen peroxide safe for cleaning porcelain?

While hydrogen peroxide is not normally used for cleaning surfaces, there is nothing in normal glazed porcelain that it would damage. Porcelain is about as unreactive to most things as glass, which is unreactive enough to be used for the bottles that store hydrogen peroxide.

But you should probably be careful with it as it can be dangerous if the solution is a strong one. Normally available hydrogen peroxide should not be as strong as the solutions used in the lab, though, which is often a 30% solution in water and should only be handled if you know what you are doing (see the safety data sheet). Consumer hydrogen peroxide is usually a 3-6% solution and isn't any more dangerous that hypochlorite bleach (and is a safer way to bleach hair).

Standard household (hypochlorite) bleach is usually better for cleaning, though.


It may pretty well be true, and here's why. Not only hydrogen peroxide is highly unstable by itself, but its decay can be catalyzed by a wide variety of compounds. And by wide, I mean really wide; at times it seems to me that those compounds that don't catalyze it are a minority. Now, I don't know what you have in your saline, but chances are that it works on the peroxide, too.

With deionized water, this should not be a problem... or should it? See, that link by @ron says that the 30% thing is stabilized. Guess what? It means that they've put some additive into it, which somewhat slows down the decomposition. When you dilute your peroxide, the stabilizer gets diluted too, and hence can't work as well as before. This is a possible reason why you may get shorter shelf-life even when diluting with pure water.

I'm not sure what is the stabilizer they use. In any case, if you find that out, then add it during the preparation, and don't add anything that can catalyze the decay, your 3% solution should be fine (in terms of shelf-life) - but I'm afraid these additions and subtractions may stand in the way of your primary objectives.


Bulging Li-ion battery – is it an immediate danger? What’s the likelihood of explosion?

The battery may have already ruptured due to thermal runaway, meaning that despite it being bulged, pressure from boiling electrolyte has already escaped.  In this case it is also likely that the lithium graphite electrode has already degraded through reaction with air.  If this is true, you are lucky that the previous thermal runaway did not result in a fire.

However, it is impossible to say that the the battery is completely dead without more information, information that would require making measurements that are difficult or impossible for most home chemists.  Thus there may well still be appreciable amounts of flammable electrolyte and air-reactive lithium graphite remaining in the battery. If heated or allowed to react with air, any such material could re-pressurize the system, resulting in a fire or small explosion.  In awareness of this possibility, you should still treat the battery as dangerous.


can you explain the structure of gemtuzumab ozogamicin? where are the linker boundaries?

The trick is probably the disulphide bridge which will be cleaved by reduction, once the construct has been internalized into the target cell, liberating the drug. So I'd consider everything proximal from it towards the antibody as linker structure.


How does one classify aqueous solutions of ions as acidic, basic and neutral?

Your approach thus far is correct. $\ce{Fe^3+}$ is acidic; $\ce{Fe(OH)3}$ is not a strong base. Indeed, if $\ce{Fe(OH)3}$ were a strong base, it would dissociate and form $\ce{Fe^3+}$, but, as you've mentioned, the opposite process happens.

Your confusion may arise from the fact that $\ce{Fe^3+}$ is a Lewis acid, and produces protons in a "different" way than Arrhenius acids:
\begin{align*} \ce{HCl &<=> H+ + Cl-}\\ \ce{Fe^3+ + H2O &<=> Fe(OH)^2+ + H+ }\end{align*}

Either way, your thought process (1.) will work for both cases.


How does steric hindrance cause sulfur hexafluoride to have almost no reaction chemistry?

As you can see:

 Sulfur (16) is significantly smaller than Se (34). That size difference
 means that there is more room between the fluorine atoms on Se than on
Sulfur. That is what steric hindrance means. Think of steric hindrance
 as limited access. The smaller the atom the less access there is.
Oxygen is so small that less than 6 fluorine atoms can crowd around it
because of steric hindrance... probably more like four.

So because there is more room between the fluorine atoms on Se there is
more opportunity for other atoms to attack the central atom Se, but
sulfur is completely shielded by inert fluorine.


It’s very hard to tell for some molecules a priori. For most, there are enough indicators.

Polyketides:

will typically have 1,3-functionalisation patterns, so hydroxy groups or keto groups are typically two carbons away from the next one. And most methyl groups also have a nonmethylated carbon in-between. Note that 1,3 can be expanded to 1,5; 1,7; 1,9; …
if cyclic, will often be only monocyclic or any second cycle will typically be something like a ketal (two oxygens bonded to a single carbon; formed by a hydroxy group attacking a ketone often five carbons away).
if cyclic are typically macrolactones, so large cycles (14, 20 or even more carbon atoms) and linked by an ester group.
can contain phenyl groups or other aromatic systems formed by elimination reactions. Often, these will be 3,5-bishydroxyphenyl groups.
typically contain an even number of carbon atoms in the chain (i.e. excluding any methyl groups mentioned above).

On the other hand, terpenoids:

very commonly have a carbon count that is a multiple of five (even often multiples of ten).
can very often be broken down into isoprene (2-methylbutane) units by paper chemistry.
typically have 1,5-methyl functionalisation (although this can change due to Wagner–Meerwein rearrangements in the biosynthesis).
can have 1,1-dimethylated carbons which polyketides cannot.
often form polycyclic structures composed of small all-carbon cycles (small being 5- to 7-membered rings).
rarely contain lactones or lactams (cyclic esters or amides).
are generally less oxygenated.
very rarely contain more than one or two double bonds, never contain phenyl rings to the best of my knowledge.

With that, you have some criteria at hand to attempt a classification. Artemisinin’s tricyclic structure with an all-carbon cycle looks a lot more like a terpenoid than a polyketide.


Does the methyl rotation in propene have a symmetric or asymmetric potential energy surface?

These rotations are the exact same physical processes for a molecule because we cannot fix either end in place. In order for rotation to mean anything, we must choose an origin. Mathematically, I must be able to draw a vector from all of the nuclei to some arbitrary point, but be consistent about this arbitrary point. Or, I must determine positions of nuclei by some measurement relative to other nuclei.

Imagine I pick the carbon of the methyl rotor as the origin. If you were standing at this origin and this rotation you describe were happening, you would see the so-called "asymmetric rotation". If, however, we put the origin on C1, we would see the "symmetric rotation." We could also define internal coordinates so that we simply changed a dihedral angle and observed the same rotation. In this case it's almost like both motions are happening. Clearly only one physical process is happening, so we might be picturing this wrong in our heads.

I believe the problem you're encountering is that if you had a macroscopic object shaped like this, you could imagine grabbing either end and you would have two separate rotations which we could choose to call symmetric and asymmetric if we so desired (though I'm not convinced of this naming). We know the motions really are different because we are spinning around different total masses. On a large scale where we hold half of this molecule in place, it would seem easier to do the symmetric rotation than the asymmetric rotation. Stated another way, for an equivalent impulse to each of the possible rotors, the initial rotational frequency of the methyl rotor would be larger than that of the $\ce{C=C}$ rotor. Note that we would still be resisting torque in the hand with which we are holding the object and the size of this torque would change depending on the rotation.

Molecules, however, cannot be thought of as fixed in this kind of way, and the resistance to rotation does not come from friction as it probably would for a macroscopic object. The "resistance" to rotation for the molecule comes from how much energy it takes to rearrange electrons for a change in nuclear positions. You can measure these nuclear positions relative to each other using whatever method you want, three of which we already discussed, and this will not change the minimum energy of the electrons for the set of nuclear positions. This minimum energy path (dare I call it a gradient?) is the path along which the rotation would happen. Whichever actual rotation you picture in your head does not matter at all. Because the molecule is free during this rotation, you can get to any of the many rotations by imagining yourself on a different atom, but this will not change the electronic energy because the relative nuclear coordinates will remain unchanged when you change the origin. In the language of one of your comments, because no part of the molecule is fixed, the inertial properties of the rotation are determined by the PES, so whatever symmetry you're thinking of is the same as the symmetry of the PES.

I think I addressed the potential pitfall here. It is indeed quite weird to imagine what the "actual" internal rotation of the molecule looks like, and maybe someone has a better intuition for what that means, but I think describing the actual motion is fairly arbitrary.


Nitration of pyrrole with sulfuric and nitric acids

Would 2,5-dinitropyrrole be a product of nitration of pyrrole with $\ce{H2SO4/HNO3}$?

I doubt it and would rather expect a lot of tar to be formed here. Pyrrole tends to undergo acid-catalyzed polymerization!

For decades, acetyl nitrate, formed by addition of fuming nitric acid to acetic anhydride, is the reagent of choice.

Mononitration of pyrrole predominantly gives 2-nitropyrrole, accompanied by minor amounts of 3-nitropyrrole. (K.J. Morgan and D.P. Morrey, Tetrahedron, 1966, 22, 57–62. DOI).

In the case of 1-methylpyrrole, the yield of the 3-nitropyrole is somewhat larger (H.J. Anderson, Canad. J. Chem, 1957, 35, 23–29). In you want even more of the 3-nitro isomer, a more bulky alkyl substituent (t-Bu) or N-silylation with $\ce{((CH3)CH)2SiCl}$ is a viable approach.

Nitration of 2-nitropyrrole at −15 °C with nitric acid in acetic anhydride  furnishes a 4:1 mixture of 2,4- and 2,5-dinitropyrrole with a total yield of 61%. (R. Alan Jones, Gerritt P. Bean, The Chemistry of Pyrroles: Organic Chemistry: A Series of Monographs, Vol. 34, Academic Press, p. 124)


Formation of radicals from C-H and C-C bonds

Note that most radical mechanisms do not explicitly feature homolysis of $\ce{C-H}$ bonds or $\ce{C-C}$ bonds. If one were to compare these two processes, say for ethane $\ce{CH3CH3}$, we would find that homlysis of $\ce{C-C}$ is favored over homolysis of $\ce{C-H}$. Obligatory bond dissociation energy references for the organics and here for the halogens.

  Homolysis

$\ce{C-C}$

$$\ce{H3C-CH3 -> 2H3C.}\ \ \Delta H^\circ=377\ \text{kJ/mol}$$

$\ce{C-H}$

$$\ce{H3CCH2-H -> H3CCH2. + H.}\ \ \Delta H^\circ=423\ \text{kJ/mol}$$

However, $\ce{C-C}$ and $\ce{C-H}$ homolysis steps are not common steps in radical mechanisms of alkanes. For example, the chlorination of ethane begins with the homolysis of $\ce{Cl2}$, which is much more favorable than homolysis of $\ce{C-C}$ or $\ce{C-H}$.

$$\ce{Cl2 -> 2Cl.} \ \ \Delta H^\circ = 243\ \text{kJ/mol}$$

The chlorine radicals then react with the alkane by abstraction, meaning that bonds are being formed as well as being broken.

  Abstraction

$\ce{C-C}$

$$\ce{Cl. + H3C-CH3 -> Cl-CH3 +H3C.}$$

$$\begin{array}{c|c|c|}
 & \text{Broken} & \text{Formed} \\ \hline
\text{bond} & \ce{H3C-CH3}  & \ce{Cl-CH3} \\ \hline
\Delta H^\circ & 377\ \text{kJ/mol} & -350 \ \text{kJ/mol} \\ \hline
\end{array}$$

$$\Delta_r H^\circ = +27 \ \text{kJ/mol}$$

$\ce{C-H}$

$$\ce{Cl. + H3CCH2-H -> Cl-H +H3CCH2.}$$

$$\begin{array}{c|c|c|}
 & \text{Broken} & \text{Formed} \\ \hline
\text{bond} & \ce{H3CCH2-H}  & \ce{Cl-H} \\ \hline
\Delta H^\circ & 423\ \text{kJ/mol} & -432 \ \text{kJ/mol} \\ \hline
\end{array}$$
$$\Delta_r H^\circ = -9 \ \text{kJ/mol}$$

Thus, abstraction of $\ce{H}$ by $\ce{Cl.}$ is more favored because it is coupled to the formation of the $\ce{H-Cl}$ bond.


The migrating group attacks trans to the leaving OH group:


What kind of materials are plastic syringes made from?

The syringes are advertised as “ISO compliant” on the website of the manufacturer. This is suspiciously vague and does not necessarily mean that the product complies with all requirements of all relevant ISO standards. (Also note that, concerning compliance and quality, the website mentions that the manufacturing facility is “ISO certified”, which is not permissible. However, the manufacturer actually has been certified to comply with ISO 9001:2008 and ISO 13485:2003 – that’s more like it.)

Ideally, such prefilled syringes are made in accordance with the applicable standards of the ISO 11040 Prefilled syringes series and other documents that are referenced in these standards (e.g. ISO 7886-1:1993 Sterile hypodermic syringes for single use – Part 1: Syringes for manual use).

However, the normative minimum requirements for materials are very general and not very restrictive. For example for the material of the barrel according to ISO 11040-6:2012 Prefilled syringes – Part 6: Plastic barrels for injectables:

  For the manufacturing of the barrels, suitable polymers shall be selected and used, based upon the intended application and processing techniques, which include, for example, the method of sterilization or the method of decontamination. The material shall exhibit the appropriate performance properties, e.g. oxygen and water permeability.

or the material of the plunger stopper according to ISO 11040-5:2012 Prefilled syringes –
Part 5: Plunger stoppers for injectables:

  The elastomeric material used shall meet the requirements specified in Clause 7.
  Plunger stoppers shall be made from the elastomeric formulation originally tested and approved by the end user. The manufacturer of the plunger stopper shall ensure the conformance of each delivery with the type sample and compliance with previously agreed functional and compendial requirements.

  The elastomeric material shall withstand two sterilization cycles when autoclaving in saturated steam at (121 ± 2) °C for 30 min without impairment of its function under conditions of normal use. If other sterilization methods are used, e.g. irradiation, the suitability of the material shall be evaluated.

The mentioned Clause 7 includes minimum requirements on hardness

  The hardness agreed between the manufacturer and the user shall not differ from the nominal value by more than ±5 Shore A when tested in accordance with ISO 7619‑1 on a special test specimen. Alternatively, the hardness can be tested on the plunger stoppers in accordance with ISO 48. If tested according to ISO 48, the microhardness shall not differ by more than ±5 IRHD from the type sample.

resistance to ageing

  The maximum time between the date of manufacture of the plunger stoppers and their pharmaceutical use should be agreed upon between the manufacturer and the user.

  The plunger stoppers shall maintain their performance characteristics throughout the entire shelf-life of the medicinal product, which is tested as part of the stability test by the user.

chemical requirements

  The aqueous extractables of the plunger stopper material shall not exceed the limits specified for elastomeric parts defined in ISO 8871‑1.

and biological requirements

  The requirements in ISO 8871‑4 shall apply.

In addition to the normative minimum requirements, ISO 11040-6:2012 also includes informative guidance on materials:

  Materials used in the construction of syringe barrels for use in pre-filled syringes should be suitable for the process to be used, the sterilization method to be applied and for the specific requirements as primary packaging for injectable drug products. Materials used in the construction of pre-filled syringes should not cause the syringes to be detrimentally affected, physically or chemically, by the normal use of injectable preparations and should be safe for the intended use.

  Certain grades of polypropylene (homo-polymer and/or co-polymer polypropylene), cyclo-olefin copolymer and cyclo-olefin polymer have been used for the barrels of pre-filled syringes in applications with a variety of injectable products (drugs and biologics).

  The organization responsible for the marketing approval of the injectable drug product is required to demonstrate the appropriateness and compatibility of the selected material and the suitability of the prefillable syringe as a container closure system. This includes, among others, the verification of drug/container interactions (leachable, extractable and sorption assessments) in reference to regulatory guidance (e.g. FDA and EMA), pharmacopoeias and other guidelines (e.g. ICH).

  To develop the necessary data as outlined above, a collaborative approach is recommended between the pharmaceutical company that holds the marketing authorization of the injectable product and the supplier of the packaging system, with reference to ISO 15378.

Furthermore, ISO 11040-6:2012 includes a list of polymer materials codification:

Cyclo-olefin copolymer (COC)
Cyclo-olefin polymer (COP)
Polypropylene (PP)
Other polymer (OTH)

Remarkably, the manufacturer mentions the materials glass, COP, and COC (without further context) for prefilled syringes on another page, whereas the package insert mentions “The luer lock syringes are made using polypropylene.”

For reliable information, you should directly inquire from the manufacturer of the prefilled syringes. Apparently, this company does not make the empty syringes but only fills syringes that are obtained from another supplier.


Question is solved. Poutnik taught me to do it using steady state approximation :)

Just adding the image for somebody who might find this in future.


How would you know if ... is H-bond or permanent dipole or induced dipole?

Hydrogen bonding is defined as an electrostatic attraction between a hydrogen atom bearing a partial positive charge in a molecule and an electronegative element of another molecule. The partial positive charge is the result of a dipole, caused by an electronegative element, such as F, O, or N. The other electronegative element to which the partially positive hydrogen is attracted to is also usually F, O, or N.

Here's the IUPAC definition; note that it does not exclude the possibility of hydrogen bonding in molecules such as HCl:

  A form of association between an electronegative atom and a hydrogen
  atom attached to a second, relatively electronegative atom. It is best
  considered as an electrostatic interaction, heightened by the small
  size of hydrogen, which permits proximity of the interacting dipoles
  or charges. Both electronegative atoms are usually (but not
  necessarily) from the first row of the Periodic Table, i.e. N, O or F.
  Hydrogen bonds may be inter-molecular or intramolecular.

http://goldbook.iupac.org/H02899.html


How to determine the equivalent weight of CO2

This is the danger of using formulae from books without fully understanding where the formulae came from.

So let's take it step by step.

We have a titration of an unknown concentration solution with a known (standardised) solution.

The reaction $$\ce{2 NaOH + CO2 -> Na2CO3 + H2O}$$ gives us the stoichiometric relationship between $\ce{NaOH}$ and $\ce{CO2}$, i.e. 2 moles of $\ce{NaOH}$ react with 1 mole of $\ce{CO2}$.

How much $\ce{CO2}$ Was used? Half the amount of $\ce{NaOH}$!

How much $\ce{NaOH}$ was used?

$$\begin{align}
c(\ce{NaOH})  &= \frac{\text{amount of substance}}{volume}\\
              &= \frac{n_{\ce{NaOH}}}{V}\\
n_{\ce{NaOH}} &= c(\ce{NaOH})\times V\\
n_{\ce{CO2}}  &= \frac{1}{2} n_{\ce{NaOH}} \\
              &= \frac{1}{2} c(\ce{NaOH})\times V\\
\end{align}$$

What is the concentration of $\ce{CO2}$ in sample?

$$c(\ce{CO2}) = \frac{n_{\ce{CO2}}}{\text{Volume of sample}} = \frac{1}{2}\frac{c(\ce{NaOH})\times V}{\text{Volume of sample}}$$

Concentration is also expressed as $\mathrm{mg/L}$. 1 mole of $\ce{CO2}$ weighs $44~\mathrm{g}$ or $44\times 1000$ mg so sticking this into the above

$$c(\ce{CO2})\mathrm{(mg/L)} = \frac{1}{2}\frac{c(\ce{NaOH})\times V}{\text{Volume of sample}} \times 44 \times 1000 = \frac{c(\ce{NaOH})\times V\times 22\times 1000}{\text{Volume of sample}}$$

That's where the $22$ came from.
This may not address exactly where the equivalent weight of $\ce{CO2}$ comes from but at least it shows you the derivation of how the $22$ got into the formula.


Excitation of an electron of H-atom from n=2 to any $n_2 $

The excitation of electrons are caused by photons while their de-excitation is causes the emission of photons of the same energy as the energy differemce between two states. The general expression for energy of n-th excited state of H-atom is given by
$$E_n=-\frac{13.6 eV}{n^2}$$

Hence by the conservation of energy, the energy released/absorbed is given by the formula you have written.
$$h\nu=\Delta E=E_{n_2}-E_{n_1}=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) eV$$

Hence your initial state is always $n_1$ and final state is always $n_2$. The energy will be positive if photon is absorbed and negative if it is released.

So, for specific values of $n_1$ and $n_2$ ($n_1<n_2$), if the photon released while going from $n_2$ to $n_1$ is in visible range, the photon absorbed for corresponding excitation will have same frequency, and hence it has to be in visible range.

Please note that there are selection rules for the transitions, determined by the
transition moment integral. Please go through the book in refernece for details.

References:

D.J. Griffiths, Introduction to Quantum Mechanics , Pearson, 2 ed,  2004. ISBN: 9332542899


Aldolisation reaction and chalcone formation

Well, actually you've made a reaction called the "Baeyer–Drewson indigo synthesis" (original article here), which was route developed in the nineteen century to obtain the indigo dye, nowadays replaced by other methods. I'm not writing the mechanism for this reaction because it is quite long and well-documented elsewhere.

Your blue compound is therefore the indigo dye represented here. As for the brown compound, it's difficult to say without more data (NMR spectra for instance), but the synthesis can give many colored intermediates or by-products such as isatin (bright red).

I hope this helps.


Bond strength of pi acceptors and metals with filled d(pi) orbitals

TL;DR Yes, you need low oxidation states unless there is a significant $\ce{M\bond{->}L}$ σ contribution to the bonding.

There are a few kinds of π acceptor ligands to distinguish between. These classifications are quite arbitrary and the line is not always clear, but their behaviour is sufficiently different and it is hopefully an instructive endeavour.

Firstly you have ligands which are mostly σ donors, but are also capable of acting as π acceptors, albeit weak ones. Two common examples are the cyanide anion and alkyl phosphines. Because the $\ce{M-L}$ bond is mostly donation from ligand to metal, you don't necessarily need particularly low oxidation states for these complexes to form: so $\ce{[Fe(CN)6]^3-}$ is OK despite having a relatively high oxidation state of iron, $\ce{Fe(III)}$.

Next up you have ligands which are poor σ donors and pretty good π acceptors. Carbon monoxide falls into this category: it is not really all that good a σ donor and backbonding is often required for a stable $\ce{M-L}$ bond to be formed (for some exceptions, google "non-classical carbonyl complexes"). Effective backbonding, in turn, requires high-energy d electrons on the metal. So, this usually precludes higher oxidation states. Many metal carbonyls have the metal in a negative oxidation state or zero, and $\ce{[Fe(CO)6]^3+}$ most certainly doesn't exist.

Lastly you have weird ligands such as $\ce{BF3}$ which are both σ and π acceptors. In the case of $\ce{BF3}$, it is a σ acceptor via the empty p orbital on boron, and a π acceptor via the $\ce{B-F}$ σ* orbitals. These are pretty obscure - I have never heard of them before - and are called Z-ligands (Wikipedia). Obviously any bond that forms must involve significant contribution of electron density from the metal. It turns out that as of 2011 there were no examples of unsupported $\ce{M-BF3}$ bonding,1 but there are quite a few examples of similar molecules $\ce{EX3}$ where $\ce{E}$ is a heavier Group 13 element.

In all these cases the metal fragment is very electron-rich and is again in a low oxidation state (zero or negative). I am sure the electron density on the metal required for these to form are even stricter than for carbonyls; otherwise we would undoubtedly have heard about $\ce{[M_pZ_q]}$ complexes, since there are already lots of $\ce{[M_p(CO)_q]}$ carbonyls known.

Reference

Amgoune, A.; Bourissou, D. σ-acceptor, Z-type ligands for transition metals. Chem. Commun. 2011, 47 (3), 859–871. DOI: 10.1039/C0CC04109B.


You're right about the expected order. From largest to smallest the expected order would be:

$\ce{_{53}I^-}\quad$ = [Kr] $4d^{10}\text{ }5s^2\text{ }5p^6$ = [Xe]

$\ce{_{34}Se^{2-}}$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr]

$\ce{_{35}Br^-}\text{  }$ = [Ar] $3d^{10}\text{ }4s^2\text{ }4p^6$ = [Kr]

$\ce{_8O^{2-}}\text{   }\text{ }$ = [He] $2s^2\text{ }2p^6$ = [Ne]

$\ce{_9F^-}\quad$ = [He] $2s^2\text{ }2p^6$ = [Ne]

You expressed the idea poorly though. $\ce{_8O^{2-}}$ and $\ce{_9F^-}$ have the same number of electrons. However $\ce{_9F^-}$ has an extra proton which should draw the electrons in closer.

There are a lot of ways to measure atomic/ionic size. No idea why the book would have $\ce{_8O^{2-}}$ and $\ce{_9F^-}$ flipped.


I will not consider hexagonal cell in this answer.

Unit cell is generally parallelepiped, which has three sides $a$, $b$ and $c$ and three angles between them $\alpha$, $\beta$ and $\gamma$. The sides can be f.e. of equal length and angles can be same and equal to $90^\circ$, this way you will have cubic unit cell. You can try to vary sides and angles and you will discover, there is only those seven types of unit cells.

When I way learning this I found useful third chapter in "Foundations of material science and engineering". At least there is nice table and images to show those cells. You can find it by typing "Crystal structures and crystal geometry" into Google.

Well, now answer last question. Let's say we have cubic base-centered (more correct than end-centered, I think) cell. From top it would look like this:

I draw only two cubic base-centered cells (I can add more if it will be more clear) next to each other. However, as you can see, there is one smaller cell (red) inside of our base-centered lattice (green) and that is simple tetragonal cell (If you draw it in three dimension, you will see, that the height is $\sqrt{2}$ times greater). As it has lower volume, this is the primitive cell, not that base-centered.


Does a freezing point of a substance always increase as its boiling point increases?

Yes and no.

Yes in the sense that boiling point and freezing point usually correlate. If there is a line of homologous compounds, than this properties rise in general, as both are increased with increase in number of dispersion interactions involved. However, it is likely to find several compounds with odd behavior, especially around beginning of the line.

As a counterexample. Toluene has higher boiling point than benzene, but much lower freezing point. This is because toluene has less symmetrical form and consequently has harder time packing into strictly ordered lattice.

Another counterexample. Gallium melts around 30 celsium, but boils well above 2000 celsium. However, Tallium melts around 300 celsium, but boils around 1700.

However, for similar compounds the values usually correlate in the series. Say, for alkanes both values steadily grow with increase of number of carbons... give or take occasional peaks due to details in crystal packing.

However, if you consider compounds of different families, it is not predictable without looking into structures of the compounds.


The meaning of different parts in the naming of glucose

Glucose can exist in the open form or cyclic form.

In the open form, there are only two types, D-glucose (illustrated in Fischer projection above) and L-glucose which is the mirror image of the above.

Experimentally, D-glucose rotates plane polarize light clockwise, hence the symbol "+", and L-glucose rotates plane polarize light counterclockwise, hence the minus symbol.  For more information on the meaning of + versus - see: http://en.wikipedia.org/wiki/Dextrorotation_and_levorotation

In the cyclic form, there is an additional chiral center, so there is alpha (where 3 OH groups are on the same side of the ring) vs beta (where 2 OH groups are on each side of the ring) as ilustrated in the drawing of the question.

Each of alpha and beta have a D and an L form, so there are 4 types in the cyclic form.

So in summary, there are only two types in the open form, but 4 types in the cyclic form.


Is it possible to separate/remove sugar from common drinks, such as juice?

Why not add yeast to your juice and allow it to ferment?  This will depend on the juice having pH and sugar concentrations; look at commercial wine making kits which provide the proper equipment and recipes. If you conduct the fermentation properly without introduction of air, the sugar will be converted to alcohol, which later can be boiled off (see above) to give the sugar-free version of your juice. Otherwise, with air present, the fermentation will go on to form acetic acid and give the juice a vinegary taste.


Do your chair parade.  See what the Kahn Academy has to offer.  Have more than one texbook.  If organic is not the best thing in the world pursued clothed, consider other options.  Employment at the bench will be under management that has other goals.  Grant funding considers discovery to be unacceptable risk.  Remember what they called the first designed degenerate Cope rearrangement - BULLVALENE.  It worked.  So did its two-step synthesis using no reagents.

LCAO for hybridization, geometry, resonance, and 1,3-shifts.  Nomenclature and chirality.  Woodward-Hoffmann comes later.  Carbocations, carbanions, radicals.  Carbonyl condensations.  Hard and soft acids and bases, Lewis and Brønsted acids and bases.  Reductants and oxidants; reagents in general.  Name reactions (that you already know as pedagogy).  Practice looking into stereograms.

NMR, IR, UV-Vis (not so much now), mass spec with database and software.  You should have some glassblowing under your belt - Schlenkware;  T-, butt, and ring seals; making an ampoule.  Own your own didymium glasses and Kennametal cutter.  Burn a pencil for its "lead."

Synthesis.  Work backwards via the synthon approach.  We used to read the new Aldrich catalog, and Fieser and Fieser, but both have has gotten out of hand.  Spend a weekend/month skimming the main journals - JACS, Angew. Chemie, JOC, Tetrahedron - and what interests you.  Never fill a reaction flask more than half full, because the universe hates you.  COTTON lab coats, because they don't melt onto your flesh.  Don't do anything stupid twice.  When in doubt - palladium.

Buy HyperChem Lite, oh yes indeed.  Remember to add hydrogens, assign atom types, then minimize well below the recommended cutoff for 10 times the recommended iterations, at least.  Note how the software is boggled when minimizing high symmetry cage ketones

If it isn't fun, you might be a theorist.  There is no shame in that - but you don't get to make crystals.


Lithium batteries a reality - why not sodium, pottasium batteries?

The sodium–sulfur battery is already in use for power leveling and other stationary applications, such as a 1.2 MWh installation in WV, USA. The Ford Ecostar did use a Na-S battery, but it did not have sufficient energy for distant travel (and some of the experimental batteries caught fire, but so do Li-ion cells on occasion. Oops!).

Sodium-ion batteries are also in development. In the JACS, the article Na-ion batteries get closer to replacing Li-ion batteries discusses some improvements in maximum current capacity and total energy storage.

That said, for mobile use, lithium has a higher energy/weight ratio than heavier metals, but for other use, sodium is less expensive and is readily available.


Which orbitals of the hydrogen atom are degenerate for n=3?

First of all isn't there only 1 electron in hydrogen?

yes

  And how could the s orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?

"degenerate" means having the same energy.  "Degenerate" refers to a set of orbitals.  It doesn't make sense to say one orbital is degenerate.

Solving the non-relativistic Schrodinger equation, all the orbitals for a given "n" are degenerate.  Energy only depends upon n.

More complete consideration including relativity, spin and quantum electrodynamics shows that they are not all degenerate however.


Why does temperature affect cell potential?

Depending on how you write the Nernst equation the temperature might increase or decrease the potential of the cell. This depends more on the reaction quotient than on anything else.

The temperature comes in to the equation as a scaling factor where RT/nF has units Volt. This essentially defines how much the voltage changes per a decade change in the reaction quotient.

The temperature dependence here is just a matter of the scaling factor.

Going back to the collision model, that is more about kinetics than thermodynamics and it is not at play for the Nernst equation.

Nernst equation describes the thermodynamics of the electrochemical system.

The temperature dependence can be explained by the temperature dependence of work-functions.

Any electrochemical reaction by definition is an electron transfer. The electron goes from the highest occupied MO of one species to the lowest unoccupied MO of the other. Both these levels are temperature dependent.


Open Babel does not generate "ball and stick 3D depictions" directly, because it's not a molecular visualization program.

That's why we started Avogadro.

There are a few formats supported in Open Babel to write out 3D depictions, e.g.:

POV-Ray
STL - for 3D printing
Point Cloud - outlining the Van der Waals surface

But you can make a pseudo-3D "ball and stick" depiction in the SVG format:

obabel [file] -O file.svg -xS


How can you calculate ring strain in beta lactams for different penicillin drugs?

For the calculation of ring strain for ring-opening metathesis polymerization, I used DFT calculation.
In generell you open your molecule at the position you want to and saturate the bonds. Then you compare the enthalpy of products and educts.
As an example you can look at cyclopentene.If it is opened during Metathesis and you saturate the doublebonds at the end of the chain with two $\ce{CH2}$ or one ethene molecule.

This has also been reported in literature for example:
https://onlinelibrary.wiley.com/doi/full/10.1002/pola.28695
I have also done a quick search and in this paper they have used the same technique (they just used ethane instead of ethene because they saturated single bonds instead of double bonds). They used post Hatree Fock B3LYP to calculate the enthalpy. I think you are fine to use either one.
https://chemistry-europe.onlinelibrary.wiley.com/doi/10.1002/cphc.201900173


Methods of splitting carbon dioxide into elements

I assume you meant convert $\ce{CO2}$ into $\ce{O2}$.

Lithium peroxide ($\ce{Li2O2}$) is a substance that does this. It absorbs $\ce{CO2}$ and converts itself into $\ce{Li2CO3}$ and $\ce{O2}$:

$$\ce{2Li2O2 + 2CO2 -> 2Li2CO3 + O2}$$

It was used as an air purifier in the spacecraft to scrub carbon dioxide. It offers it's bonus of releasing oxygen as well as being non-hygroscopic, so it won't remove the moisture content in the air.

But at any day, you would need a lot of lithium peroxide if you want to make Mars habitable.

Edit: The partial pressure of carbon dioxide in the Martian atmosphere is $\pu{575.8Pa}$. Using the ideal gas law, and temperature of Mars is $218K$, we get amount of substance of $\ce{CO2}$ as $\pu{0.32mol/m^3}$.

Such an amount would require $\pu{14.65g}$ of lithium peroxide per cubic meter of martian atmosphere. According to SigmaAldrich, that costs about ₹21.22 (33 US cents) per cubic meter.


Given the list of ingredients in Pat's comment, I would strongly doubt that a chemical reaction has occurred here, and suggest that most likely the volatile solvents ethanol and isopropanol have evaporated, leaving a cloudy residue of the remaining ingredients. Your concern of a possible hazard is appreciated, but I would say no worries.


Do Sulfuric Acid and Calcium Hydroxide Completely Neutralize One Another?

The first proton transfer is simple enough and you have no doubts.

$$\ce{H2SO4 + Ca^2+ + 2 OH- -> HSO4- + Ca^2+ + OH- + H2O}\tag{1}$$

Now, you wonder whether hydrogensulphate is a strong enough acid to combine with the remaining hydroxide. Well, hydrogensulphate may be a weak acid but hydroxide is a strong base. So even the second step will go to completion:

$$\ce{HSO4- + Ca^2+ + OH- + H2O -> SO4^2- + Ca^2+ + 2 H2O}\tag{2}$$

Frankly, it does not end there. Calcium and sulphate form a salt with a very low solubility so this salt will precipitate out of solution.

$$\ce{SO4^2- + Ca^2+ + 2 H2O -> CaSO4 v + 2 H2O}\tag{3}$$

So the only thing you are left with in reasonable amounts is water and a precipitate. This, of course, is a neutral solution.

However, you don’t really always have to go that far. Check if there are strong acids and bases (don’t forget hydroxides), cancel them away and then check what remains. If you had done that, you would have noticed that there are as many hydroxide ions as acidic protons, overall giving an approximately neutral solution.


Voltage needed for electrolysis of water

I'm assuming you mean this Wikipedia article on the hydrolysis of water in which the method of determining that potential difference is described farther down the page.

I'll address your other questions - what might cause this value to change.

  Volume

No, except see below on concentration. The reaction potential for this electrolysis is independent of the amount of water since it is actually a thermodynamic function. Electrolysis may occur at different rates and take varying time depending on the amount of water, but the potential needed remains the same.

  Conductivity

Conductivity is a measure of the ease of transmitting electrical current (electrons) through a substance. So, increased conductivity will increase the rate of electrolysis, but since it does not change the original or final substances in the chemical equation, does not change any thermodynamic values like the reaction potential.

However, there is a danger here. We typically increase conductivity by increasing electrolyte content. Some of these electrolytes could be electrolyzed at a reduced potential compared to water. For example, if you use sodium iodide as your electrolyte, the iodide ion is oxidized to the triiodide ion at -0.53 V:

$$\ce{3I- -> I3- + 2e-}\ \ E^\circ = -0.53\ \pu{V}$$

  Concentration

The reaction potential -1.23 V is the standard reaction potential, which is determined when all applicable species are at 1 molar concentration and 1 bar pressure. The two half reactions in this case are:

$$\ce{2H2O -> O2 + 4H+ + 4e-}\\
\ce{4H+ + 4e- -> 2H2}$$

Thus, this potential is only valid at $[\ce{H+}] = 1\ \pu{M}$ or $\pu{pH}=1$. As the reaction proceeds, water is consumed and the volume decreases. The amount of protons remains constant, so the pH increases. As the pH increases, so too does the potential required and the electrolysis eventually reaches equilibrium and progresses no further unless you are providing an overpotential.

Note that -1.23 V is equally valid at $[\ce{OH-}] = 1\ \pu{M}$ or $\pu{pH}=14$ thanks to Hess's Law.


Can an uncharged particle pass through a molecular bond?

The molecular bond is not a solid thing. It is merely an energetically suitable arrangement of atoms and their electrons. So another particle that happens to come close to that bond can only interact with those atomic nuclei and electrons. Whether it does interact depends.

Which particles can interact?

Considering electron-magnetic forces only:

Charged particles can interact.
Uncharged small particles can not interact.
Larger uncharged particles (e.g. other small molecules) could get polarized. This means they develop partial charges which then can interact.

How can charged particles interact with the bond?

Slow particles (with low kinetic energy) can get attracted or be repelled by the molecule, depending on the (partial) charges of the particle and molecule. If it gets attracted a chemical reaction can happen.
Fast particles (with high kinetic energy) might just pass through the molecular bond. The bond has actually much less electron density then any atom. Depending on how close it gets to an atomic nuclei, it might get deflect. Or it may even deflect the atom and break the bond. The particle could also grab or deflect an electron of the bond and thus ionize the molecule.
As a third option: there is always a slight chance of quantum tunneling.

Does this ever happen in real life?

Actually a lot: Neutrinos have no electric charge and almost no mass. As a result they can pass through matter without any consequences. They are emitted by stars and pass through earth by millions per second.

An example for charged particles would be using an electron gun to ionize molecules for mass spectrometry. The electron gun creates a beam of fast electrons, which can remove an electron from the targeted molecule upon collision (1 goes in, 2 go out).

Is there a physical size limit?

If the particle becomes larger than the molecular bond is long ... well I would not call it "passing through the bond" anymore. But the above concepts would still be the same.


Confused about strong/weak dibasic and tribasic acids

If you think about each dissociate state separately, than it's easy to answer your question. $\ce{H2SO4}$ is a very strong acid $(\mathrm{p}K_\mathrm{a}\approx -3.0),$ but $\ce{HSO4-}$ is not so strong $(\mathrm{p}K_\mathrm{a}\approx 2.0).$

Phosphoric acid starts with $\mathrm{p}K_\mathrm{a}\approx 2.14,$ which as as you said is not so strong, and $\ce{H2PO4-}$ and $\ce{HPO4^2-}$ have $\mathrm{p}K_\mathrm{a}\approx 7.2$ and $\mathrm{p}K_\mathrm{a}\approx 12.3,$ so definetly weak.

So yes, we say sulfuric acid is strong because we're talking about $\ce{H2SO4},$ and what comes after should be mentioned separately.


Oxygen Preventing the Formation of Large Organic Compounds?

If we take a look at glucose we can see there are oxygen atoms in it, just with extra presence of the hydrogen.
The assumption :

carbon bonds with oxygen, the electrons are too close to the oxygen, and therefore they cannot just string together and form large molecules like glucose.

is wrong.
Take a look at Polymeric carbon oxides

Carbon suboxide spontaneously polymerizes at room temperature into a carbon-oxygen polymer, with 3:2 carbon:oxygen atomic ratio. The polymer is believed to be a linear chain of fused six-membered lactone rings, with a continuous carbon backbone of alternating single and double bonds. Physical measurements indicate that the mean number of units per molecule is about 5–6, depending on the formation temperature.
Carbon monoxide compressed to 5 GPa in a diamond anvil cell yields a somewhat similar reddish polymer with a slightly higher oxygen content, which is metastable at room conditions. It is believed that $CO$ disproportionates in the cell to a mixture of $\ce{CO2}$ and $\ce{C3O2}$ ; the latter forms a polymer similar to the one described above (but with a more irregular structure), that traps some of the $\ce{CO2}$ in its matrix.
Another carbon-oxygen polymer, with C:O ratio 5:1 or higher, is the classical graphite oxide and its single-sheet version graphene oxide.


Mechanism for reaction of aryl diazonium with potassium iodide

It follows a radical mechanism (SET  = Single Electron Transfer)


Finding the concentration of each ion in saturated solution of lead(II) hydroxide

The mistake you made is that you assumed mass concentrations of the ions from the stoichiometric relation, which is only applicable to the amount of substance, and, as a consequence, amount concentration.

You have to convert mass concentration $γ$ to the molar concentration c first:

$$c(\ce{Pb(OH)2}) = \frac{γ(\ce{Pb(OH)2})}{M(\ce{Pb(OH)2})} = \frac{\pu{0.155 g L-1}}{\pu{241.21 g mol-1}} = \pu{6.43E-4 mol L-1}$$

And, since it's a saturated solution, then a nearly complete dissociation can be assumed:

$$[\ce{Pb^2+}] = c(\ce{Pb(OH)2}) = \pu{6.43E-4 mol L-1}$$

$$[\ce{OH-}] = 2 × c(\ce{Pb(OH)2}) = 2 × \pu{6.43E-4 mol L-1} = \pu{1.29E-3 mol L-1}$$


Why is ammonia still considered polar despite undergoing rapid inversion?

The ammonia molecule inverts itself roughly 25 billion times per second (https://en.wikipedia.org/wiki/Nitrogen_inversion , http://courses.washington.edu/phys432/NH3/ammonia_inversion.pdf). Thus averaged over a long time the dipole moment of an individual ammonia molecule will be zero, as for every moment there is a dipole pointing in one direction, this will be cancelled at a different time by a dipole in the opposite direction (as the system is charge neutral the dipole takes the same value independent of where you place the origin, so molecular motion won't affect this)

However note the word "long" in the above. Any process that happens on a short time scale will not experience this long time averaging of the property, but rather the instantaneous, non-zero, value of the dipole. So what is short and what is long in this context? Well 25 billion inversions per second means that 1 inversion takes 0.04ns. How far does a molecule move in that time? Well taking an order of magnitude value from https://en.wikipedia.org/wiki/Thermal_velocity of 400m/s a molecule moves 160 Angstroms in the time it takes one inversion of ammonia to take place. This is a HUGE distance for a molecule, thus the motion is dictated by the instantaneous, non-zero, value of the dipole. Thus we talk of Ammonia as a polar species.


What is the chemical composition of Turnbull's blue?

Though your textbook is right, Wikipedia isn't completely wrong as in aqueous solution, $\left(\ce{Fe3[Fe(CN)6]2} \right)$ gets converted and is in equilibrium with $\left(\ce{Fe4[Fe(CN)6]3} \right)$ (a redox change) and hence prussian blue and turnbull's blue are identical to look visually. But in a textual way, the naming is given as your textbook says, i.e. $\left(\ce{Fe4[Fe(CN)6]3} \right)$ is Prussian blue whereas $\left(\ce{Fe3[Fe(CN)6]2} \right)$ is Turnbull's blue.


What molecular features create the sensation of sweetness?

For humans, the "sweet taste buds" are the T1R2+T1R3 & T1R3+T1R3 G-protein coupled receptors.

Chemicals that are sweet are the ones that can bind to these receptors. We know some structures that tend to do this, namely ketones and aldehydes (carbonyl groups).

However, there are chemicals that are sweet that we do not know the mechanism for, such as Curculin, which is up to ~2000 times sweeter than sucrose by weight.


Reaction Mechanisms: Free Radical Halogenation

H. C. Brown has suggested that the halogen atom does add but, at high temperatures, is expelled before the second step of free radical addition can occur$^1$.

Addition of halogen atom gives this radical, which falls apart to regenerate the starting material if the temperature is high or if it does not soon encounter a halogen molecule to complete the addition. Hence, low concentration of halogen can be used with high temperature to favour substitution over addition$^1$.
The allylic radical, on the other hand, once formed, has little option but to wait for a halogen molecule, however low the halogen concentration$^1$.

For the formation of minor product you may look into this question.
To follow the Markovnikov's or anti-Markovnikov's rule, there should be difference of  electronegativity between the two atoms. Therefore, I do not understand what you are trying to ask.
Reference

Organic Chemistry- Morrison and Boyd (6E); Page 389-390


What might serve as an initial starting photocatalyst for this large water-splitting solar simulator?

Vitamin B-12 is a photocatalyst, see, for example, https://pubs.rsc.org/en/content/articlelanding/2018/ra/c7ra13037f#!divAbstract and also https://www.ncbi.nlm.nih.gov/pubmed/29271445 , but it has not been applied to water splitting!

What has been used is TiO2, see this source to split H2O into H2 and O2.

Mechanics: The action of light on TiO2 creates electrons (e-) and electron holes (h+). Some reactions:

$\ce{H+ + e- -> .H}$

$\ce{.H + .H -> H2}$

$\ce{OH- + h+ -> .HO}$

$\ce{.HO -> H+ + .O-}$

$\ce{.O- + h+ -> .O}$

$\ce{.O + .O -> O2}$

One of the less desirable reaction is:

$\ce{.H + .HO -> H2O}$


Why acidity of p-methoxybenzoic acid is more acidic than p-hydroxybenzoic acid?

Both -$\ce{OCH3}$ and -$\ce{OH}$ groups have exhibited two effects on the aromatic ring: (1) Electron donating resonance or mesomeric effect (+M) and (2) Electron withdrawing inductive effect (-I). For both acids in hand, the electron withdrawing inductive effect (-I) is almost same since both -$\ce{OCH3}$ and -$\ce{OH}$ groups are 4 carbons away from the acid center. However, when -$\ce{OH}$ group is attached to either the para- or ortho-position, it has a more tendency to delocalize its lone pair electrons towards the aromatic ring than that of -$\ce{OCH3}$ group (Order of activating:  -$\ce{O-} \gt$  -$\ce{OH} \gt$  -$\ce{OCH3}$; for additional learning, read articles on Hammett plots).

As a result, the electron density on the carbon para to -$\ce{OH}$ substitution group increases more than that on the carbon para to -$\ce{OCH3}$ substitution group. Thus, the electron density on carbonyl carbon of carboxylic group attached to those para-carbons also increases accordingly, and consequently, the polarity of $\ce{O-H}$ of carboxylic group decreases as a result. That polarity decrease causes the reduced acidity ($\mathrm{p}K_\mathrm{a}$s are $4.58$ and $4.47$, when para-substitutions are $\ce{OH}$ and $\ce{OCH3}$, respectively) compared to benzoic acid ($\mathrm{p}K_\mathrm{a}$ is $4.20$ when para-substitution is $\ce{H}$). The $\mathrm{p}K_\mathrm{a}$s of p-hydroxybenzoic acid is higher than that of p-methoxybenzoic acid here, because as explained above, p-$\ce{OH}$ group caused reduced polarity on $\ce{O-H}$ of carboxylic group than that caused by p-$\ce{OCH3}$ group.


TL;DR There is twice  the amount of $\ce{SO3}$ produced compared to oxygen, thus the volume will be twice as much as oxygen. Therefore $8.4$ litres of $\ce{SO3}$ will be produced.

Since $\ce{SO2}$ is in excess, the amount of $\ce{SO3}$ produced will be twice as the amount of moles of oxygen. So if there is originally x moles of oxygen gas, 2x moles of $\ce{SO2}$ is produced. It is important to note that temperature and pressure remain constant throughout the reaction.

Note one of gas laws state, the same number of moles of any gaseous substance will have the same volume at constant temperature and pressure. So x moles of $\ce{SO3}$ will hold the same volume as x moles of oxygen gas, so 4.2 litres.

Now there is also another gas law which states that the volume of a gas is directly proportional to the number of moles of the substance. So 2x moles of $\ce{SO3}$ will have twice the volume of x moles of $\ce{SO3}$. Therefore the volume of 2x moles of $\ce{SO3}$ will be 8.4 moles.

Therefore the answer is $8.4$ litres


Your selection of the main ring and the bridges that results in the bridge lengths ‘[9.2.2.2]’ and your counting of the total number of skeletal atoms that results in the name ‘heptadecane’ are correct.

For your proposed name ‘bicyclo[9.2.2.2]heptadecane’, you obviously followed the nomenclature of similar bicyclic systems, such as bicyclo[9.2.2]pentadecane.

This is not a bad idea since the nomenclature of polycyclic hydrocarbon ring systems is actually based on the nomenclature of bicyclic systems.

However, the structure given in the question is not a bicyclic ring system because it includes one additional secondary bridge; therefore, the structure actually is a tricyclic system.

The additional steps for the naming of tricyclic hydrocarbons are described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows.

  P-23.2.5 Naming and numbering tricyclic alicyclic hydrocarbons

  P-23.2.5.1 Tricyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system (…). Rings not described by the bicyclic system are defined by citing the number of atoms in the independent secondary bridge as an arabic number. The locants of the two attachment points of the independent secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower number is cited first) separated by a comma.

  The name of the tricyclic system is then constructed by citing:

  (a) the prefix ‘tricyclo’, in place of ‘bicyclo’, indicating the presence of three rings in the polyalicyclic system;

  (b) numbers indicating the bridge lengths, starting with the two branches of the main ring (…), followed by the main bridge (…), and the secondary bridge (with superscript locants separated by commas indicating its points of attachment to the main ring), all separated by full stops and placed in brackets,

  (c) the name of the acyclic hydrocarbon having the same total number of skeletal atoms.

Therefore, the correct name for the structure given in the question is tricyclo[9.2.2.21,11]heptadecane.


Why there is a temperature change in adiabatic process?

Remember adiabatic process implies process in which no heat exchange takes place between the system and the surrounding i.e. neither heat is supplied nor heat is loosed.

Temperature of gases changes only when internal energy of the gases changes.

Net heat of the system constant  means dq = 0

dq= du+dw, to maintain dq = 0, you can have many possible set of values of du and dw. Like whatever amount work you do, say dw, to maintain addition of du and dw always as zero, du = -dw. So any amount of work you do , du becomes negative of that work you perform to maintain net heat of the system as constant (i.e. dq as zero).
So even while maintaining addition of du and dw as zero, you can have infinite possible values of du and dw ( i.e. do any aribitary work and du becomes the negative of the work you performed and hence addition becomes always zero)

Temperature of the gas is related to du by du = nCvΔT. Hence for all those infinite possible values of du, you will have infinite possible temperatures, still maintaining constant net heat of the system .i.e dq as zero.

So in an adiabatic process, if you performed work, then du has to change to a value which is negative of the amount of work you performed (otherwise you won't be able to maintain their addition to be zero) , and as du has to change, hence temperature also changes.

So every time you perform work in an adiabatic container, temperature  has to change.


Numbering for the IUPAC name of this benzene derivative

The main goal of IUPAC names is to keep them simple and understandable by anyone. So, you basically want to keep numbers the lowest possible. Hence, you shall start numbering where it will give the lowest numbers for your substituents.
For instance, if you began numbering from the carbon bonded with the chlorine, your molecule would have been named 1-chloro-2-methyl-5-nitrobenzene or 1-chloro-6-methyl-3-nitrobenzene. Or, if you chose your numbering to begin from the other group, you'd have named it 3-chloro-4-methyl-1-nitrobenzene or, even worse, 5-chloro-4-methyl-1-nitrobenzene. With these examples, you can notice that the simplest name, i.e. the one that has the lowest numbers, is the one PubChem gave you : and that's the reason one should start numbering by the methyl.


Relative acidity of aqueous LiF, SnCl2, and YCl3

Salt solutions are acidic because of hydrolysis.

LiF is obviously neutral, because it's a salt of very strong acid and very strong base.

Ironically, Wikipedia is wrong about Sn2+: it's much more acidic in water than Ca2+ or Zn2+, and more acidic than Y3+,
because Sn2+ forms very stable and non-soluble hydroxo compounds and hence readily undergoes hydrolysis.

The answer is: LiF < YCl3 < SnCl2

Table data:

YCl3 has pH 4.5-5.8 (source),
SnCl2 has pH 1-2 (source 1, 2).


Addition reaction of hydroiodic acid to 2-bromo-3-chloro-2-butene

TL;DR - Sometimes you need to do the experiment.

Like all questions involving Markovnikov's rule, you should compare the structures of the two carbocations:

  At C2 (Bromo)

$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-CH(Cl)-C+(Br)-CH3}$$

The two issues to consider are resonance and induction.

Resonance

You could draw a resonance structure to show that the bromo group is stabilizing the carbocation:

$$\ce{R2C+-Br <-> R2C=Br+}$$

However, since the bromine atom is much larger than the carbon atom, it cannot form as effective $\ce{p}-\pi$ overlap with carbon.

Induction

The carbocation has two stabilizing alkyl groups. What about that bromo group? Bromine is more electronegative than carbon, but not by much (2.96 vs 2.55 on the Pauling scale). Plus, bromine has a lot of electron density around it to donate, if it could. Bromine thus is a mild electron-withdrawing group by induction.

  At C3 (chloro)

$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-C+(Cl)-CH(Br)-CH3}$$

We once again examine resonance and induction.

Resonance

$$\ce{R2C+-Cl <-> R2C=Cl+}$$

Chlorine is smaller than bromine and closer in size to carbon. Thus, chlorine is a better resonance stabilizer than bromine.

Induction

The same two alkyl groups are present, so we will discount them. Chlorine is more electronegative than bromine (3.16 on Pauling), and it has less electron density around it to begin with. Chlorine is more powerful electron-withdrawer by resonance than bromine.

Compare:

                  Chlorine              Bromine
Resonance         Stronger donor        Weaker donor
Induction         Stronger withdrawer   Weaker withdrawer

In many cases, resonance usually wins, so I might expect carbocation formation preferentially at C3 with the chloro group, but I would expect the other product to form also.

Now... what else might happen.

Regardless of the initial regioselectivity, the carbocation also can be stabilized by forming a halonium ion. Bromine is better at this since it is larger (and less electronegative so it can stabilize the positive charge better).

This halonium would be attacked at the less hindered position (C2), resulting in transfer of the bromine to C3.

So in effect, you can get three products:

$$\ce{CH3-CHCl-CIBr-CH3}$$
$$\ce{CH3-CICl-CHBr-CH3}$$
$$\ce{CH3-CClBr-CHI-CH3}$$

You would then want to actually do the experiment to determine which of these is really the major product.


Combination of several unit symbols when finding new pressure from ideal gas law

Section 5.1 Unit symbols in SI Brochure [1, p. 130] has an answer:

  In forming products and quotients of unit symbols the normal rules of algebraic multiplication or division apply. Multiplication must be indicated by a space or a half-high (centred) dot (⋅), since otherwise some prefixes could be misinterpreted as a unit symbol. Division is indicated by a horizontal line, by a solidus (oblique stroke, /) or by negative exponents. When several unit symbols are combined, care should be
  taken to avoid ambiguities, for example by using brackets or negative exponents. A solidus must not be used more than once in a given expression without brackets to remove ambiguities.

  $\pu{m kg/(s3 A)}$,
  or $\pu{m kg s–3 A–1}$,
  but not $\pu{m kg/s3/A}$,
  nor $\pu{m kg/s3 A}$.

I personally prefer to use powers of the units — subjectively, this makes the expressions clearer and also reserves a slash "/" for separating the symbol of the quantity from the units on plots and in the table headers. However, in your case you don't need any of that. According to the ideal gas law

$$pV = nRT$$

and

$$n_iR = \text{const} \quad \implies \quad \frac{p_iV_i}{T_i} = \text{const},$$

so you have a simple proportion

$$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \quad \implies \quad p_2 = \frac{V_1T_2}{V_2T_1}p_1.$$

Obviously, there is no need to juggle around with the units as all of them except for the pressure units are cancelled out (note how I also grouped homogeneous quantities within the fraction one under another, and wrote pressure separately for clarity):

$$p_2 = \frac{V_1T_2}{V_2T_1}p_1 = \frac{\pu{35.0 L}\cdot\pu{316 K}}{\pu{36.5 L}\cdot\pu{288 K}}\cdot\pu{235 kPa} = \pu{247 kPa}.$$

And yes, generally it is considered a good practice not to do any intermediate calculations unless you are explicitly asked so, and solve the problem algebraically in the first place.

References

BIPM. Le Système International d’unitès / The International System of Units (“The SI Brochure”), 8th ed.; Bureau international des poids et mesures: Sèvres, 2006.


Are there compounds that melt under their own weight?

Sure. Even discounting non-equilibrium conditions, any substance which expands upon freezing will by symmetry melt under pressure, even if the pressure is caused by supporting its own weight. There's a simple example: water! A sufficiently tall self-supporting block of ice standing on a flat surface will create enough pressure at its bottom that it could melt in temperatures between approximately $-21.9\ ^\circ \mathrm{C}$ and $0\ ^\circ \mathrm{C}$, according to the phase diagram for water. This has very important implications for the study of glaciers, for example. However, this process requires a very large amount of pressure, and thus a very tall block of ice. Using an approximate value for the freezing point depression under pressure of $0.0072\ ^\circ \mathrm{C\ atm^{-1}}$, it would take an ice block about $7.5\ \mathrm{km}$ tall for the base to melt at a temperature of $-5\ ^\circ \mathrm{C}$ or higher.

I thought indium was among the substances which expanded upon freezing, but it actually turns out to contract by 2.5% when frozen, so this mechanism would not work (though there could be something else to consider).  Some other materials which could theoretically melt under their own weight in some regime of temperatures and pressures are gallium, bismuth and silicon. There might be some very specific conditions under which this effect could be achieved in a column of around $20\ \mathrm{cm}$, though pressure usually only slightly alters the freezing point, so it's likely that the material would have to be kept only slightly under its melting point at standard pressure for this to happen.


What is the correct approach to obtain PES using CASSCF?

Yes in principle that is a correct approach, but you should be aware of a few pitfalls here and there.

The basis set you chose for your optimisation might not be large enough. I assume you are not going to use STO-3G for your CASSCF calculation. It is better if you run your initial optimisation with the same basis set that you are using for the CASSCF calculation. You will most likely be closer to the minimum and will safe a few cycles when optimising the molecular structure on the larger level of theory.
Picking the right active space might not be trivial. You may have to run a couple of single points to find the right balance of what you must include and what you may neglect.
The SCAN produces a rigid scan of the molecule, i.e. the backbone of the structure does not react with the induced change. That might not lead to the correct behaviour. More sophisticated (and more costly) is a relaxed scan via the ModRedundant keyword.

Be aware, that CASSCF calculations are quite resource demanding and you probably won't get results overnight.


Why is melting point of Vanadium higher than Iron?

@Jon Custer makes a good point in his comment so lets look beyond the electronic shell configuration of V and Fe to help understand the different in melting points.
Consider the following:
Pauling’s Electronegativity Fe > V  1.8 vs 1.6
Vander walls radius Fe < V  0.126 nm vs 0.134 nm
And consistent with Fe having a higher electronegativity the first ionization energy of Fe [761 kJ per mol] is greater than that of V [ 649.1 kJ per mol]
Given that we think of melting as the breaking of lattice forces holding atoms together, Fe being a smaller and more electronegative atom [vs V] is consistent with the relative melting points of these two elements.
Ref:
https://www.lenntech.com/periodic/elements/fe.htm
https://www.lenntech.com/periodic/elements/v.htm


The reason behind this is the hydrophobic effect. Everyone has seen it if they pour a spoonful of vegetable oil into a pot of water, e.g. to cook pasta. As long as nothing is disturbing the vegetable oil, it will collect itself together in one big bubble rather than form many small bubbles.

Polar solvents will always be arranged in a way that positively polarised areas are close to negatively polarised areas in the neighbouring solvent molecule. Hydrogen bonds — especially in water — are nothing but the extension of this concept to even more polarisation. Polar molecules, as you stated, will fit together with this scheme well. Unpolar compounds mixed in do not. As soon as you add an unpolar compound to a polar solvent, you are creating a type of artificial boundary and only the areas of the solvent molecules that are neither particularly positive or negative will be happy to be in the near vicinity. That means that you have a much higher ordering of the solvent molecules where they hit an unpolar compound, because one direction is basically doomed to be neither positive nor negative.

This ‘solvent wall’ will form no matter whether the unpolar island is large or small. However, it will also have almost the same thickness, no matter how big the contained part is. Therefore, if multiple unpolar molecules clump together, the overall number of polar molecules constrained in that wall is lower — an entropic gain. Thus, unpolar compounds tend to either not dissolve or precipitate out of polar solvent solutions.

This also works in the opposite direction. A polar molecule will prefer, for energetic reasons, to have a polar neighbour. However, an unpolar solvent cannot provide that polar environment. A neighbouring undissolved polar molecule, however, can. Here again, the polar compounds rather stay clustered together as it allows the central molecules to be more disordered while only a small layer on the border must take care to interact as well as possible with the unpolar solvent.

Note that this answer assumes some kind of black/white dichotomy. In reality, most compounds are somewhere on a scale from absolutely unpolar to very polar and are able to adapt to a wide range of organic solvents. Conversely, some organic solvents have the reputation of dissolving almost anything organic; most notably dichloromethane. However, the solubilities may greatly vary.


Questions about van't Hoff factor in colligative property

In general, 'n' is the number of species the compound dissociates into, and such formulas are applicable only when the acid itself is monobasic, i.e.only one dissociaton is involved.For sulphuric acid,this is not applicable as there are two dissociations with two dissociation constants.The best way to find the van't hoff factor is to find the ratio of the final number of particles to the initial number of particles.


Calculate Michaelis-Menten constant of enzyme catalyzed reaction

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

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

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


Balancing a chemical equation, but with an incorrect mole ratio

The balanced reaction is $$\ce{2AgNO3 +Cu -> 2Ag +Cu(NO3)2}$$

Firstly, it's not necessary that all the reactants are depleted in the reaction.

There is the concept of a "limiting reagent" -- the reagent that gets depleted first and drives the extent of the reaction.

In this case, the limiting reagent is $\ce{AgNO3}$, and you will have half a unit of $\ce{Cu}$ left over.

In the general case, the limiting reagent need not be the one in the least quantity. For example, for $\ce{2A +B ->C +D}$ , if $\ce{A:B}$ is 3:2, then $\ce{A}$ is still the limiting reagent as it will get depleted first.


How many moles of the substance are in 2.5 g of lead?

Do not mix amu with mass of mol of substance! In your last formula you've forgot about Avagadro number.

One Pb atom weighs $\pu{207.2amu}$ or $\pu{207.2amu} \cdot \pu{1.66\times10^{-24}g}$.

But one $\text{mol}$ of Pb weighs $=\pu{207.2amu}\cdot \pu{1.66\times10^{-24}g}\cdot6.02\cdot10^{23}=\pu{207.2g}$


Your question is about as broad as chemistry itself. Like you said, you may try dissolving them in water, if one is soluble and the other is not. If it's not working with water, chances are it will work with ethanol, or acetone, or... see, there are many solvents on the shelf. No luck? Maybe by now you've found something that dissolves them both, so you may try fractional crystallization. Still no luck? Maybe the compounds differ in density or selective adhesion of something, so flotation might help. Still no luck? Chromatography, by any chance? No? Distillation, like you said (provided they both would survive until boiling point), or maybe sublimation (good for $\ce{NH4Cl + NaCl}$)? No? Well, there is a multitude of chemical processes; some of them might change (that is, destroy) one compound or both, but probably in such a way that they can be regenerated later.

And if everything else fails, you can always look at the mixture under a microscope and separate the crystals manually.


When naming hydrocarbons, when do you number the branch or double bond?

When there are 2 carbons in a chain, the double bond will be at the number 'one' position. Therefore it's not always mentioned in the naming.

For example:  Ethene
It can be written as 1-ethene or eth-1-ene. Since the position of the double bond can be at no other places than the first, the position in the naming you saw, wasn't mentioned.

When 3 or more carbons are present, the double bond can be between any 2 of the carbons, ie. either the first position or the second. (but-1-ene/butene or but-2-ene)

Position of double bonds and that of branches, must always be mentioned while naming organic compounds.


The amount of citric acid for golden syrup

Golden syrup is a type of inverted sugar syrup, like maple syrup, treacle and molasses.

It is made by splitting the disaccharide sucrose (sugar) into a mixture of glucose and fructose components, typically by carefully controlled heating, sometimes with a small amount of citric or potassium bitartrate (cream of tartar).

For example, if you mix 1 kg of fine granular sugar with 2 cups (500 mL) water and 1 g of citric acid, bring to the boil and simmer for about 30 mins, you make inverted sugar syrup. If you simmer this for another 2 - 2.5 hrs or so until it caramelises, you end up with 'golden syrup'.