What is the correct IUPAC name of this compund?

According to P-14.4 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-14.4 Numbering
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
[...]
(e) saturation/unsaturation: [...]
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;

In this compound, unsaturation, in the form of the double bond, has the priority for low locants. Since both numbering schemes (shown below) assign the locant '2' to the double bond, criterion (e) is not sufficient for the purposes of deciding between the two options.

However, criterion (f) does allow for a choice to be made. The detachable prefixes in this name are 'bromo' and 'methyl'. Since the locant set (1,2) is lower than the locant set (3,4) (note that both sets are ordered in increasing numerical order as stipulated), the preferred IUPAC name (PIN) will use the locant set (1,2).
Note, however, that your double bond also contains configurational information. According to the Cahn–Ingold–Prelog rules (see P-92 of the Blue Book), the double bond in your molecule as drawn is (E)-configured. Stereodescriptors such as 'E' are cited together with the locant describing the position of the stereogenic unit (see P-91.3).
Hence, the PIN of the molecule as depicted in your question is (2E)-1-bromo-2-methylbut-2-ene.

Note that contrary to existing answers, there is no inherent preference between "bromo" and "methyl" for low locants, unless the application of criterion (f) does not lead to a difference. In that case, criterion (g) must be invoked. Since 'bromo' is earlier than 'methyl' in alphanumerical order, it will be the prefix cited first in the name, and will therefore have priority for a lower locant. For example:


How to be able to tell if a molecule will react via SN1 or SN2?

Factors affecting SN1 mechanism:

  Weak nucleophile / base and Good leaving group.
  More steric hindrance.
  Polar protic solvent should be used.

Factors affecting SN2 mechanism:

  Strong nucleophile / base and Good leaving group.
  Less steric hindrance.
  Polar aprotic solvent should be used.

Coming back to your problem. Actually, both of the compounds can go through SN1 or SN2 mechanism as the carbon linked with $\ce{Cl}$ is a secondary carbon. But, as you mentioned that you're using $\ce{EtOH}$, and you should know that $\ce{EtOH}$ is little basic, almost neutral, therefore, we can say that now, both of the compounds will go through SN1 mechanism (Remember, the first point of SN1 mechanism, i.e., weak base).

Now, the question is if both the compounds will follow SN1 mechanism, then which will react faster. For this, we will have a look at the SN1 mechanism:

  Step 1: Leaving of the (-X) group

  Step 2: Weak nucleophile attacks from either side

  Step 3: Substitution product formed (Inversion or retention, based on
  the step 2)

So, our concern is only the step 2.

Note that, rate of reaction in step 2 depends on steric hindrance in your case. So, let's have a look at the structure of (1-chloro-2,2-dimethylpropyl)benzene and (1-chloro-2-methylpropyl)benzene:

The Carbon with red dot is the point where $\ce{EtOH}$ will attack after the (-Cl) will leave.

Now, on comparing the both the compounds, you will notice that in the first compounds there is greater steric hindrance as compared to the second compound. And, that's the reason why second compound will react faster than the first one.

Conclusion:

  (1-chloro-2-methylpropyl)benzene will react faster than
  (1-chloro-2,2-dimethylpropyl)benzene due to less steric hindrance
  around the carbon that is directly attached to aromatic ring (i.e.,
  the carbon with red dot).

  And, both of the compounds will go through SN1
  mechanism due to presence of $\ce{EtOH}$.


The reason behind this is the mainly inert pair effect .

In $\ce{BiCl3}$, due to the much electronegativity difference between $\ce{Bi}$ and $\ce{Cl}$, Chlorine atom forms bonds with almost pure $p$ orbitals of $\ce{Bi}$, and the lone pairs on $\ce{Bi}$ are of almost pure $s$ character. Thus, $\ce{Bi}$ atom doesn't actually utilise much of its $s$ orbital electrons in forming the bonds which is energetically much more preferable. ( If it seems non-obvious, recall Bent's rule which says that more electronegative atoms prefer to form bonds with orbitals with more $p$ - character.)

On the other hand, if $\ce{Bi}$ is present in $\ce{BiCl5}$, it has no other option other than utilising its $s$ orbitals for bonding. The equatorial bonds will mainly consist $s$ character and $p_x, p_y$ orbitals will also participate in forming the equatorial bonds. So, $s$ character in the equatorial bonds will be slightly lesser than $33$% (again due to Bent's rule). And the axial bonds will consist of mainly $p_z $ and $d_{z^2}$ orbitals and little bit of $s$ mixing may also be there as it is spherically symmetric. So, overall central atom $\ce{Bi}$ has to use its $s$ electrons in bonding which is difficult energetically.

Now, the participation of $s$ electrons is difficult in bonding is due to the relativistic contraction of $s$ orbitals. It becomes closer to the nucleus and more stable in case of heavy elements like $\ce{Au, Hg, Tl, Pb, Bi}$ etc. Hence, the central atom needs to pay more energy to involve those stable electrons in bonding. That's why, the $s$ electrons become kind of inert and thus these heavy elements don't prefer higher oxidation states to avoid participation of these inner $s$ orbitals in bonding.


Fulvene consists of a cyclopentadiene ring and a $\ce{{}=CH2}$ substituent group. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the prefix ‘methylene’ is no longer recommended to designate the $\ce{{}=CH2}$ group. The preferred prefix is ‘methylidene’; thus, the systematic name of fulvene without locants is methylidenecyclopentadiene.
Concerning the numbering of locants, the current IUPAC recommendations read as follows.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Therefore, low locants are given first to the double bonds in the ring according to (e) and then to the prefix ‘methylidene’ according to (f). Thus, the correct systematic name is 5-methylidenecyclopenta-1,3-diene rather than 3-methylidenecyclopenta-1,4-diene since the locant set ‘1,3’ for the double bonds is lower than ‘1,4’.

Note that according to IUPAC nomenclature since 1993, locants are placed immediately before that part of the name to which they relate (e.g. cyclopenta-1,3-diene rather than 1,3-cyclopentadiene).


Temperature change and gas evolution for vinegar and baking soda experiment

Although your observations make it pretty clear that you are using enough sodium bicarbonate to consume the acetic acid, even with $1$ teaspoon of sodium bicarbonate, we can also do a rough calculation to validate this hypothesis. It's a "rough" calculation because we don't know the concentration of acetic acid in your vinegar.

Vinegar is typically about $\pu{5\%}$ acetic acid by volume, but can be as high as $\pu{20\%}$. For our calculation, we will estimate that there is $\pu{5 g}$ of acetic acid in $\pu{100 ml}$ of vinegar, and use $\pu{6 g}$ as the mass for $1$ teaspoon of sodium bicarbonate, per your comment. Then, using the molecular weights of acetic acid ($\pu{60 g/mol}$) and sodium bicarbonate ($\pu{84 g/mol}$), we can calculate the following molar quantities of the reagents:

acetic acid: $\pu{0.083 mol}$ per $\pu{100 ml}$ vinegar
sodium bicarbonate: $\pu{0.071 mol}$ per teaspoon of baking soda

If our estimate of the acetic acid concentration is correct, it is close enough to a $1:1$ molar ratio with the sodium bicarbonate that you may not have noticed the difference in your measurements. In this case there is $\pu{17\%}$ more acetic acid than sodium bicarbonate. This means that the addition of excess sodium bicarbonate (i.e. $2$ teaspoons) should result in a $\pu{17\%}$ greater temperature difference, which would be $\pu{4.7^oC}$ rather than $\pu{4^oC}$. Since you are only reading your thermometer to $\pu{+/- 1^oC}$, these could be said to be the same and it is perfectly reasonable that you wouldn't have noticed the difference. It is not stated how the volume of evolved carbon dioxide was measured, but it would likely be even more difficult to notice this small change by the change in observed carbon dioxide volume.

Summary, TL;DR:
Your observations indicate that a single teaspoon of baking soda is consuming all of the acetic acid present in the vinegar, so when additional baking soda is added, there is no more acetic acid to react with and thus there is no indication of further reaction. Making some simple assumptions about the concentration of vinegar and mass of sodium bicarbonate in one teaspoon allows us to confirm that this is the case, within the detection limits of your method (measurement of temperature change and volume of carbon dioxide evolution).


Why is enthalpy and not heat released by system used while calculating entropy of surroundings?

Because $q=\Delta H$ only if only expansion work is done and the pressure of the system is constant. In that case, for an isothermal process, in which T is the same for system and surroundings, the entropy of the surroundings can be equated with the change in enthalpy divided by T. Which is why the applicability of the integrated formula $\Delta G = \Delta H -T \Delta S$ is restricted to constant p and T. You could write $q_p$ in place of $\Delta H$ the equation but only when work is restricted to pV work and when T and p are constant. It would not hold generally since sometimes work other than expansion work is performed. In general $\Delta H = q_p+w_\textrm{non-pV}$ at constant T and p.
To show this more explicitly, for a small (differential) change:
$$dU=dq+dw$$
$$dU=dq+dw_{\textrm{non-pV}}+dw_{\textrm{pV}}$$
$$dH = dU+d(pV) = dq+dw_{\textrm{non-pV}}+dw_{\textrm{pV}} + pdV+Vdp$$
At constant pressure of the system $dw_{\textrm{pV}}=-pdV$, $Vdp=0$ and
$$dH = dq+dw_{\textrm{non-pV}}$$
Further derivation is possible* to show how this relates to $G$, by assuming constant T and considering reversibility, but this should suffice to show that $dH$ and $dq$ (or $\Delta H$ and $q$) are not in general the same.

*For a reversible process $dq=TdS$ and
$$dH =  TdS+dw_{\textrm{non-pV,rev}}$$
Finally the differential expression for $G$ at constant $T$ follows (since then $dG=dH-d(TS)=dH-TdS$):
$$dG=dH - TdS = dw_{\textrm{non-pV,rev}} $$


What is the cost of 1.00 g of calcium ions as provided by this brand of dry milk?

First you calculate the value $v$ of the powder per gram:
$$ 25.06~\mathrm{oz} = 710.44~\mathrm{g} \; \hat{=} \; 5.46~\$ \Rightarrow v=0.007685~\mathrm{\$\, (g\, Powder)^{-1}}$$

Then you calculate the mass of powder ($m$) that contains one gram of calcium ions:
$$ m=\frac{1.483~\mathrm{g\, Powder}}{0.0196~\mathrm{g\, \ce{Ca^{2+}}}} = 75.66~\mathrm{\frac{g\, Powder}{g\, \ce{Ca^{2+}}}} $$

Finally, you multiply the two results for the price $p$:
$$ p = v\times m = 0.58~\mathrm{\$\, (g\, \ce{Ca^{2+}})^{-1}}$$


How does one find the probability of tunneling along a reaction coordinate computationally?

If the reaction coordinate $\xi$ can be expressed as the change in the relative position between two groups of the molecule (e.g. for instance as a change in a bond length so that $\xi$ is given in units of length) the problem may be treated as one dimensional and one may use computational software to calculate the potential energy $V(\xi)$ along this path. You may then proceed solving the time-independent Schroedinger equation using numerical techniques such as Numerov's integration method. However, in many cases WKB theory provides an accurate estimate.

In terms of WKB, the tunneling probability is given by
$$
T=e^{-2G}
$$
where $G$ is the Gamow factor given by
$$
G=\frac{2}{\hbar}\int_{\xi_a}^{\xi_b}{ \sqrt{2\mu_\text{red}\left [V(\xi) - E \right ]}}d\xi
$$
where $\mu_\text{red}$ is the reduced mass along the reaction coordinate and $E$ is the energy of the incoming particles. The integral runs over the clasically forbidden region from $\xi_a$ to $\xi_b$. The Gamow factor thus represents the square root of the integral over the "reaction barrier" multiplied with the effective mass that tunnels. Of course, you should take zero point energy into account and in the case of light reactants also the vibrational and rotational structure of the products as this may lead to resonances.


Why don't equivalent hydrogens cause splitting in NMR?

I will provide a full quantum mechanical explanation here.[1] Warning: rather MathJax heavy. Hopefully, this lends some insight into how the diagrams that long and porphyrin posted come about.

Finding the states between which transitions occur

The Hamiltonian for two coupled spins is

$$\hat{H} = \omega_1\hat{I}_{\!1z} + \omega_2\hat{I}_{\!2z} + \frac{2\pi J_{12}}{\hbar}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{1}$$

where $\omega_1$ and $\omega_2$ are the Larmor frequencies[2] of the two nuclei and $J_{12}$ is the coupling constant (in Hz) between the two nuclei. (The factor of $2\pi/\hbar$ is simply there to bring it into energy units.) $\hat{I}_{\!i}$ is the operator for the spin angular momentum of nucleus $i$ and $\hat{I}_{\!iz}$ is the operator for its projection along the $z$-axis.

However, in NMR, it is customary to work in frequency units instead of energy units. Since $E = h\nu$, we simply need to divide through by $h$. Bearing in mind that $\omega = 2\pi\nu$ and $h = 2\pi\hbar$, we get:

$$\hat{H}_\text{freq} = \frac{\nu_1}{\hbar}\hat{I}_{\!1z} + \frac{\nu_2}{\hbar}\hat{I}_{\!2z} + \frac{J_{12}}{\hbar^2}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{2}$$

On top of that, to make the maths easier, it is also quite common to set $\hbar = 1$. Therefore, we have

$$\hat{H}_\text{freq} = \nu_1\hat{I}_{\!1z} + \nu_2\hat{I}_{\!2z} + J_{12}(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{3}$$

We will deal with two spin-$1/2$ nuclei here, where $|\alpha\rangle$ and $|\beta\rangle$ represent the "up" and "down" spins (recall we set $\hbar = 1$ so it doesn't appear in the eigenvalues):

$$\begin{align}
\hat{I}_{\!iz}|\alpha_i\rangle &= \frac{1}{2}|\alpha_i\rangle & \hat{I}_{\!iz}|\beta_i\rangle &= -\frac{1}{2}|\beta_i\rangle & (i = 1,2) \tag{4}
\end{align}$$

Furthermore since we are dealing with equivalent nuclei we can simply set $\nu_1 = \nu_2 = \nu$ and drop the subscript in $J_{12}$ just to make it a bit cleaner:

$$\hat{H} = \nu(\hat{I}_{\!1z} + \hat{I}_{\!2z}) + J(\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}) \tag{5}$$

Now, we need to find the eigenstates and eigenvalues of $\hat{H}$. To do so, we will adopt the basis set of product functions $(|\alpha_1\alpha_2\rangle, |\alpha_1\beta_2\rangle, |\beta_1\alpha_2\rangle, |\beta_1\beta_2\rangle)$. From equation $(4)$ we have

$$\begin{align}
\hat{I}_{\!1z}|\alpha_1\alpha_2\rangle &= \frac{1}{2}|\alpha_1\alpha_2\rangle & \hat{I}_{\!2z}|\alpha_1\alpha_2\rangle &= \frac{1}{2}|\alpha_1\alpha_2\rangle \tag{6} \\
\hat{I}_{\!1z}|\alpha_1\beta_2\rangle &= \frac{1}{2}|\alpha_1\beta_2\rangle & \hat{I}_{\!2z}|\alpha_1\beta_2\rangle &= -\frac{1}{2}|\alpha_1\beta_2\rangle \tag{7} \\
\hat{I}_{\!1z}|\beta_1\alpha_2\rangle &= -\frac{1}{2}|\beta_1\alpha_2\rangle & \hat{I}_{\!2z}|\beta_1\alpha_2\rangle &= \frac{1}{2}|\beta_1\alpha_2\rangle \tag{8} \\
\hat{I}_{\!1z}|\beta_1\beta_2\rangle &= -\frac{1}{2}|\beta_1\beta_2\rangle & \hat{I}_{\!2z}|\beta_1\beta_2\rangle &= -\frac{1}{2}|\beta_1\beta_2\rangle \tag{9}
\end{align}$$

The action of the scalar product $\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}$ is more complicated. We need to introduce the shift operators (or ladder operators).

$$\begin{align}
\hat{I}_{\!i+} &= \hat{I}_{\!ix} + \mathrm{i}\hat{I}_{\!iy} & \hat{I}_{\!i-} &= \hat{I}_{\!ix} - \mathrm{i}\hat{I}_{\!iy} \tag{10}
\end{align}$$

from which we can obtain

$$\begin{align}
\hat{I}_{\!ix} &= \frac{\hat{I}_{\!i+} + \hat{I}_{\!i-}}{2} & \hat{I}_{\!iy} &= \frac{\hat{I}_{\!i+} - \hat{I}_{\!i-}}{2\mathrm{i}} \tag{11}
\end{align}$$

So, finally, we can write

$$\begin{align}
\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}} &= \hat{I}_{\!1x}\hat{I}_{\!2x} + \hat{I}_{\!1y}\hat{I}_{\!2y} + \hat{I}_{\!1z}\hat{I}_{\!2z} \tag{12} \\
&= \frac{\hat{I}_{\!1+}\hat{I}_{\!2-} + \hat{I}_{\!1-}\hat{I}_{\!2+}}{2} + \hat{I}_{\!1z}\hat{I}_{\!2z} \tag{13}
\end{align}$$

where in going from $(12)$ to $(13)$ one simply substitutes in $(11)$ and does a fair bit of algebraic manipulation. The action of the shift operators are

$$\begin{align}
\hat{I}_{\!i+}|\alpha_i\rangle &= 0 & \hat{I}_{\!i+}|\beta_i\rangle &= |\alpha_i\rangle \tag{14} \\
\hat{I}_{\!i-}|\alpha_i\rangle &= |\beta_i\rangle & \hat{I}_{\!i-}|\beta_i\rangle &= 0 \tag{14} \\
\end{align}$$

This allows you to work out the effect of $\hat{\vec{I}_{\!1}}\cdot\hat{\vec{I}_{\!2}}$ on our basis states. The actual maths is left to the reader and I will simply quote the results:

$$\begin{align}
\hat{H}|\alpha_1\alpha_2\rangle &= \left(\nu + \frac{J}{4}\right)|\alpha_1\alpha_2\rangle \tag{15} \\
\hat{H}|\alpha_1\beta_2\rangle &= -\frac{J}{4}|\alpha_1\beta_2\rangle + \frac{J}{2}|\beta_1\alpha_2\rangle \tag{16} \\
\hat{H}|\beta_1\alpha_2\rangle &= \frac{J}{2}|\alpha_1\beta_2\rangle - \frac{J}{4}|\beta_1\alpha_2\rangle \tag{17} \\
\hat{H}|\beta_1\beta_2\rangle &= \left(-\nu + \frac{J}{4}\right)|\beta_1\beta_2\rangle \tag{18} \\
\end{align}$$

Therefore in this basis the Hamiltonian matrix is

$$\mathbf{H} =
\begin{pmatrix}
\nu + J/4 & 0 & 0 & 0 \\
0 & -J/4 & J/2 & 0 \\
0 & J/2 & -J/4 & 0 \\
0 & 0 & 0 & -\nu + J/4
\end{pmatrix} \tag{19}$$

Finding the eigenvectors and eigenvalues of this matrix is again left to the reader (it is not a difficult task) and they are (eigenvalues denoted $E_i$)

$$\begin{align}
|1\rangle &= |\alpha_1\alpha_2\rangle & E_1 &= \nu + \frac{J}{4} \tag{20} \\
|2\rangle &= \frac{1}{\sqrt{2}}(|\alpha_1\beta_2\rangle + |\beta_1\alpha_2\rangle) & E_2 &= \frac{J}{4} \tag{21} \\
|3\rangle &= \frac{1}{\sqrt{2}}(|\alpha_1\beta_2\rangle - |\beta_1\alpha_2\rangle) & E_3 &= -\frac{3J}{4} \tag{22} \\
|4\rangle &= |\beta_1\beta_2\rangle & E_4 &= -\nu + \frac{J}{4} \tag{23} \\
\end{align}$$

The form of the eigenstates should be familiar: they are simply the triplet and singlet states of two spin-$1/2$ particles. These states arise from the coupling of two sources of angular momenta, $I_1$ and $I_2$, to form one overall angular momentum denoted $I$.

$$\vec{I} = \vec{I}_{\!1} + \vec{I}_{\!2} \tag{24}$$

The allowed values of $I$ are determined by the Clebsch-Gordan series:

$$I = I_1 + I_2, I_1 + I_2 - 1, \cdots, |I_1 - I_2| \tag{25}$$

Since $I_1 = I_2 = 1/2$, $I$ can take the values $1$ and $0$. The values of $M_I$, the projection of the total angular momentum along the $z$-axis, are as usual

$$M_I = I, I-1, \cdots, -I \tag{26}$$

so the states with $I = 1$ ("triplet") have $M_I = 1, 0, -1$ and the state with $I = 0$ ("singlet") has $M_I = 0$. One can use more quantum mechanics to work out which state is associated with which quantum numbers, but I will not do it here. They are:

$$\begin{array}{ccc} \hline
\text{State} & I & M_I \\ \hline
|1\rangle & 1 & 1 \\
|2\rangle & 1 & 0 \\
|3\rangle & 0 & 0 \\
|4\rangle & 1 & -1 \\ \hline
\end{array}$$

Selection rules

We have four different states, which leads to ${4\choose 2} = 6$ different possible transitions. However, not all of these transitions are allowed.

The intensity of the transition is proportional to the square of the matrix element $\langle \psi_\mathrm{f} | \hat{H'} | \psi_\mathrm{i} \rangle$ (the so-called "transition dipole moment"), where $\hat{H'}$ is the Hamiltonian for the process that induces the transition. In the case of NMR transitions, the transition arises due to a magnetic field aligned along the $x$-axis.[3] The corresponding Hamiltonian is therefore

$$\hat{H'} = \omega'(\hat{I}_{\!1x} + \hat{I}_{\!2x}) = \omega'\hat{I}_{\!x} \tag{27}$$

Exactly what $\omega'$ represents is not important here because we are only really concerned about whether the transition dipole moment is zero or not.[4] Making use of the relations established in equations $(10)$ and $(11)$,[5] one can find that the selection rules are

$$\Delta I = 0; \Delta M_I = \pm 1 \tag{28}$$

which means that the allowed transitions are $|4\rangle \leftrightarrow |2\rangle$ and $|2\rangle \leftrightarrow |1\rangle$. Transitions to and from the singlet state $|3\rangle$ are forbidden. The energies of the transitions are

$$\begin{align}
E_{4\leftrightarrow2} &= \frac{J}{4} - \left(-\nu + \frac{J}{4}\right) = \nu \tag{29} \\
E_{2\leftrightarrow1} &= \left(\nu + \frac{J}{4}\right) - \frac{J}{4} = \nu \tag{30} \\
\end{align}$$

i.e. the two transitions are degenerate and only one line in the spectrum at frequency $\nu$ is observed. This is exactly what is depicted in the diagrams posted in the other answers.

Notes and references

[1] I am assuming the reader has some knowledge of the quantum mechanical treatment of angular momentum, which is a topic that is treated thoroughly in most quantum mechanics textbooks. See, for example, chapter 4 of Atkins's Molecular Quantum Mechanics (5th ed.).

[2] The Larmor frequency is given by $\omega = -\gamma B_0$, where $\gamma$ is the magnetogyric ratio of the nucleus in question and $B_0$ is the strength of the external magnetic field. It represents the frequency with which a magnetic moment precesses about a magnetic field. See any textbook on magnetism for further details.

[3] I am glossing over some details here. The so-called magnetic field in the $x$-axis is a component of the radiofrequency pulse applied in the $xy$-plane. If you are interested please consult a textbook on the vector model of NMR. In particular I recommend Keeler's Understanding NMR Spectroscopy (2nd ed.).

[4] It is related to the strength of the magnetic field in the $x$-axis, $B_1$, by $w' = |\gamma|B_1$. The usual symbol is $\omega_1$, but I chose not to use this here to avoid potential confusion. Again, please consult a textbook on the vector model of NMR if you wish to find out more.

[5] A full proof can be found in J. Chem. Educ. 1982, 59 (10), 819. There is also some discussion of the selection rules in Gunther's NMR Spectroscopy (3rd ed.), p 156 onwards.


I found your numbers confusing, but I think the first equation you wrote is not correct. This is how I'd solve it:

Let's say we have $\mathrm{n}$ equivalents of metal in the sample, and let's call the equivalent weight of the metal $\mathrm{e}$. For the hydroxide sample, we have:

$n(e+17)=10$

where $\mathrm{e+17}$ is the molar mass of the hydroxide $\ce{MOH}$.

For the oxide, we have:

$n/2(2e+16)=8$

where $\mathrm{2e+16}$ is the molar mass of the oxide $\ce{M_2O}$. Note that $\mathrm{n}$ becomes $\mathrm{n/2}$ because of stoichiometry:

$\ce{2MOH->M2O + H2O}$

So you have a system of two equations with two unkowns ($\mathrm{n}$ and $\mathrm{e}$), which you can solve normally. For instance, if you divide the equations:

$\cfrac{n(e+17)}{n/2(2e+16)}=\cfrac{10}{8}$

$\cfrac{e+17}{e+8}=\cfrac{10}{8}$

$8e+136=10e+80$

$2e=56$

$e=28$

So the equivalent weight of the metal is 28g.

Makes sense?


What are the mechanisms for the reduction of organic compounds employing red phosphorus and iodine?

I found Red P and Iodine reduce many different organic compounds.

Actually, it is typically used for the deoxygenation in benzylic positions, using benzylic alcohols or benzoins as starting materials.

Aliphatic alcohols yield iodoalkanes, propargylic alcohols often undergo elimination under these conditions whereas allylic alcohols are typically converted to complex mixtures without synthetic value.

The deoxygenation is often carried out with catalytic amounts of iodine but an excess of red phosphorous. In principle, catalytic amounts of $\ce{NaI}$ plus one equivalent of $\ce{H3PO3}$ in water would work as well.

When using iodine and red phosphorous, hydrogen iodide is generated in situ via the formation and subsequent hydrolysis of $\ce{PI3}$.

The formal replacement by a proton is not fully understood and is disputed whether the formation of a benzylic iodide is really involved.

In princpiple, elimination from water from the protonated starting material would yield a stabilized benzylic cation. This might get reduced by a iodide to a benzylic radical and, in a second step to a very shortlived benzylic anion which is instantaneously protonated.

Both reduction steps yield iodine radicals from iodide and in summary regenerate iodine.


Is it possible to burn things in a halogen atmosphere?

They certainly can. Put some water in the bottom of a beaker, fill the beaker with chlorine gas, and drop in bits of calcium carbide. The carbide will produce bubbles of acetylene, which will spontaneously burst into flame when they hit the chlorine.

Flame colors depend on many things, even in oxygen. In general, flames in a chlorine atmosphere will probably look slightly more greenish than flames in air or oxygen, simply because the gas itself is green. Fluorine isn't as strongly colored, and flames from combustion in fluorine will be VERY VERY BRIGHT.


Disclaimer: this is a preliminary answer based on the current state of the discussion with OP; it might be the subject to major changes in the near future.

So.

OP (original poster) is trying to reproduce some relative energies previously computed at ωB97X-D/6-31+G(d,p) at a different M06-2X/6-31G(d,p) level of theory. OP has exact same results for closed-shell molecular species in question, the meaning of which is probably that the resulting M06-2X/6-31G(d,p) relative energies in kcal/mol or kJ/mol are exactly equal to the previously published ωB97X-D/6-31+G(d,p).
OP also mentioned that he has some troubles in reproducing the relative energies for open-shell species, by which he probably means that for open-shell species M06-2X/6-31G(d,p) spits out different numbers in kcal/mol or kJ/mol than ωB97X-D/6-31+G(d,p).

Before we have some numbers let me say few things which hold in generality.

The difference in few kJ/mol in relative energies computed using two different functionals is pretty normal (even at exact same geometry). One should not even expect exactly same relative energies with any two different functionals in the first place.
OP presumably run geometry optimization at M06-2X/6-31G(d,p) but did not mention how different are the equilibrium geometries obtained at this level from those at ωB97X-D/6-31+G(d,p) level published previously. In fact, this would be difficult to do, because I could not find the Cartesian coordinates of the species in question in the Supporting Information for the work published previously. But at least some quick visual comparison has to be done to make sure that equilibrium geometries are at least not totally different.
OP did not mention why M06-2X functional was chosen. Even if the ultimate goal is not just to reproduce the previously published work, it is mentioned there that:

  ωB97X-D, a range-separated double-hybrid functional, has been shown to describe reaction energies accurately.45

But, first of all, ωB97X-D is not a double-hybrid functional. And secondly, the paper quoted there is fairly well-known benchmark of density functional methods done by Lars Goerigkab and Stefan Grimme, and in fact, Goerigkab and Grimme found M062X-D3 and not ωB97X-D to be "statistically the best of all hybrids". Note though that M062X-D3 differs from M062X: the former includes the empirical dispersion correction while the later does not.
With respect to the empirical dispersion correction, today it is usually recommended to include it in DFT calculations, since it is available at almost no (computational) cost but can substantially improve the results. Note also that dispersion correction is already included in case of ωB97X-D functional but has to be "manually" added for the case of M06-2X. In fact, the absence of it might explain some differences in equilibrium geometries and relative energies.


Nomenclature of Halogen substituted Alcohol and longest chain

Your rationale is correct.

The proposed name “4-chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol” is unambiguous and describes the correct compound; however, it is not in accordance with IUPAC nomenclature.

The important simplified criteria for the choice of a principal chain in such compounds are:

greater number of suffixes (here: “-ol”)
longest chain
lower locants for suffixes
greater number of prefixes (e.g. “chloro-”)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

Therefore, the preferred IUPAC name for the compound that is given in the question is 3-(chloromethyl)-2-(propan-2-yl)pentan-1-ol.

The position of the chloro group or other substituent groups that are cited as prefixes is not relevant for the choice of a principal chain in this case. Only if there is still a choice, such substituent groups are taken into account according to P-45.2.


How to determine if a given reaction requires heating to initiate?

The ease of initiation of a reaction depends on the activation energy required. In order for reactions to proceed (at a reasonable rate), the forces binding the individual reactants must be overcome, which may require energy, solvent, catalyst or other input.

For example, $\ce{H2O2}$ is quite stable at room temperature, but exposure to horseradish peroxidase, silver oxide or other catalyst causes rapid decomposition. Iron filings do not oxidize quickly, but apply a bit of salt water and they rust so quickly they make an effective heater. $\ce{NI3}$ has an extremely low activation requirement!


Conversion of isonitrile (isocyanide) group into cyanide group?

Isonitriles will, given the correct conditions, spontaneously transform into nitriles. The reaction mechanism is very interesting; it is basically a cationic [1,2]-migration of the $\ce{R}$ group under retention with a three-membered cyclic transition state. To reach this transition state, the $\ce{C-N}$ bond must be broken and the alkyl residue must form a cation-π interaction with the $\ce{C#N}$ triple bond. This is a non-classical carbonium cation, as shown in scheme 1.

Scheme 1: Mechanism of the [1,2]-migration and sp³-π interaction in the intermediate carbonium ion.

The reaction rates were extensively studied by Meier et al.[1] They note that the typical migration preference of $\text{tertiary} > \text{secondary}>\text{primary} > \text{methyl}$, indicative of cationic carbenium intermediates and observed e.g. in Criegee, Beckmann and Lossen rearrangements, does not apply here. Instead, a migratory aptitude similar to the Curtius or pinacol rearrangements is observed. Another interesting observation is ‘the almost complete lack of a bridgehead effect on the [reaction] rates’, further indicating that a planar carbenium transition state cannot be part of the mechanism and that bond angles and lengths will probably remain unchanged en route to the transition state.[1] When phenylic isonitriles react, next to no substituent effect is observed, indicating that the cation is not delocalised into the phenyl ring.

All of this evidence indicates towards the reaction being ‘properly prototypical for a sigmatropic 1,2-rearrangement via a nonpolarized, tight, hypervalent three-centered transition state.’[1] The authors go on to state that this is in perfect agreement with ab initio calculations, but do not provide an appropriate source.

Isonitriles were known in the 19th century; I am not able to find a definite first description but A. Gautier noted their reactions with acids in 1867.[2] In 1868, he goes on to be the apparant first to describe an isonitrile–nitrile rearrangement.[3] Weith, however, was probably the first to actually observe the reaction as such; he notes:[4]

  Zu dem Ende wurde nach der  H o f m a n n ’schen Methode dargestelltes und durch Destillation möglichst von Anilin befreites Cyanphenyl 2—3 Stunden lang im geschlossenen Rohr auf 200–220° erhitzt. Nach dem Erkalten fand sich eine beträchtliche Menge dunkel gefärbter nadlicher Krystalle vor, der Röhreninhalt besass nicht mehr den Geruch des Cyanphenyls, dagegen den des Benzonitrils. [sic! to all apparant errors]

  Thus, phenylisonitrile prepared by the Hofmann method and purified from aniline by distillation was heated in a sealed tube to 200–220 °C for 2–3 h. After cooling, a significant amount of dark-coloured, needle-like crystals were found; the tube’s content no longer smelt like phenylisonitril but like benzonitrile.

Therefore, for the rearrangement to take place temperatures above $150~\mathrm{^\circ C}$ appear to be required — which also explains why isonitriles are kinetically inert enough to be isolated under standard conditions and even with 19th century methods. (Gautier used $\ce{AgCN}$ to synthesise them from iodoalkanes.)[3] An activation energy as high as this makes sense, since, as noted above, the $\ce{C-N}$ bond must be broked and a significant rearrangement must take place for the carbonium ion to form.

Some isonitriles, namely those which upon dissociation form very stabilised cations such as benzylisonitrile, can also react according to a different path that is basically an $\mathrm{S_N1}$ rearrangement or can be considered in a chain-like structure. This, however, is a rather special case as Meier et al. note.[1]

References:

[1]: M. Meier, B. Müller, C. Rüchardt, J. Org. Chem. 1987, 52, 648. DOI: 10.1021/jo00380a028.

[2]: A. Gautier, Liebigs Ann. Chem. 1867, 142, 289. DOI: 10.1002/jlac.18671420304.

[3]: A. Gautier, Liebigs Ann. Chem. 1868, 146, 124. DOI: 10.1002/jlac.18681460108.

[4]: W. Weith Ber. Dtsch. Chem. Ges. 1873, 6, 210. DOI: 10.1002/cber.18730060180.


Activity of Polonium in half-time in Becquerels

Well, first of all, I should point out that your answer and the would be the answer ($1.7$ times your answer) are both wrong by the fact that after 552 days (4 half-lives), the activity should be $\displaystyle\frac{\pu{2E6 MBq}}{16} = \pu{1.25E5 MBq}.$ Since this is smaller than both your answer and the would be answer, they both would be incorrect. Accordingly, correct answer would be smaller than $\pu{1.25E5 MBq},$ because the sample would be further decays for another six days. Let’s do the proper calculations:
The half-life $\left(t_\frac{1}{2}\right)$ of a radioactive element is defined as
$$t_\frac{1}{2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}.\label{eqn:1}\tag{1}$$
The $t_\frac{1}{2} = \pu{138 d}$ in the equation \eqref{eqn:1} for your case. Thus from the equation \eqref{eqn:1}, you can calculate for $\lambda:$
$$\lambda = \frac{\ln 2}{t_\frac{1}{2}} = \frac{0.693}{\pu{138 d}} = \pu{5.02E-3 d-1}. \tag{2}$$
Also, the radioactive decay is according to the equation:
$$A_t = A_0\mathrm e^{-\lambda t}.\label{eqn:3}\tag{3}$$
where $A_0$ is initial activity and $A_t$ is the activity of the same same sample after time $t$ (Note: $t_\frac{1}{2}$ and $t$ should have same units). Thus, applying the $\lambda$ value on the equation \eqref{eqn:3}:
$$A_t = \pu{2E6 MBq}\times\exp\left(-\pu{5.02E-3 d-1}\times\pu{558 d}\right) = \pu{1.21E5 MBq}.$$
Note: I disregard the correct significant figures because I believe the initial activity should be given in three significant figures.
For fun, you can also checked the validity of the answer using my initial remark. You can use the activity of the sample after four consecutive $t_\frac{1}{2}$s (after $138 \times 4 = \pu{552 d})$ as its initial activity and find the activity after another six days using the equation \eqref{eqn:3}:
$$A_t = \pu{1.25E5 MBq}\times\exp\left(-\pu{5.02E-3 d-1}\times\pu{6 d}\right) = \pu{1.21E5 MBq}.$$


Why are more easily oxidized elements more reactive than less easily oxidized elements?

Ease of oxidation is how easily it loses electrons. Electrons are the basis of chemical reactions. If chemical A is more easily oxidized than B, then B is a relative oxidizer to A, so in terms of the activity series, the elements that are more easily oxidized can react with more chemicals, since they will be able to act as reducing agents to more chemicals.


Your reasoning is not quite correct. Since electron density on $\ce{N=O}$ bond is more around $\ce{O}$ than $\ce{N}$ (electronegativity of $\ce{O}$ is greater than that of $\ce{N}$), $\ce{N}$ should have a slight $\delta+$ charge, which demands electrons from adjacent $\pi$-system. As a result, the electron density on the phenyl nucleus of nitrozobenzene is less compared to that of benzene. This phenomena is shown as a ring deactivating factor. However, this fact may be hard to prove practically because of the instability of nitrozobenzene.

Yet, in computational studies, one set of descriptors of electron structure of aromatic compounds may come from the pEDA/sEDA approach (Ref.1). The pEDA and sEDA descriptors are defined as populations of the $\pi$- and $\sigma$-orbital electrons, respectively, in a given planar molecule or its planar part (Ref.1 & Ref.2):

The abstract of Ref.2 states that:

  An application of quantum chemical modeling allowed us to investigate a substituent effect on a $\sigma$ and $\pi$ electron structure of a ring and the nitro group in a series of meta- and para-$\ce{X}$-substituted nitrobenzene derivatives ($\ce{X =}$ $\ce{NMe2}$, $\ce{NHMe}$, $\ce{NH2}$, $\ce{OH}$, $\ce{OMe}$, $\ce{Me}$, $\ce{H}$, $\ce{F}$, $\ce{Cl}$, $\ce{CF3}$, $\ce{CN}$, $\ce{CHO}$, $\ce{COMe}$, $\ce{CONH2}$, $\ce{COOH}$, $\ce{NO2}$, and $\ce{NO}$). The obtained pEDA and sEDA parameters (the $\pi$- and $\sigma$-electron structure characteristics of a given planar fragment of the system obtained by the summation of $\pi$- and $\sigma$-orbital occupancies, respectively) of the $\ce{NO2}$ group and the benzene ring allowed us to reveal the impact of the substituents on their mutual relations as well as to analyze them from the viewpoint of substituent characteristics. The decisive factor for dependence of pEDA on sEDA of the ring is electronegativity of the atom linking the substituent with the ring; in subgroups an increase of sEDA is associated with a decrease of pEDA. The obtained mutual relation between pEDA($\ce{NO2}$) and pEDA(ring) characteristics documents strong resonance interactions for electron-donating substituents in the para position. The observed substituent effect on the $\sigma$-electron structure of the nitro group, sEDA($\ce{NO2}$), is significantly greater (∼1.6 times) for meta derivatives than for the para ones.

Although, there is no general correlation between these two contributions, pEDA(ring) and sEDA(ring), to the description of the electronic structure of the transmitting ring, it's worth comparing effect of $\ce{NO}$ substituent with the known effect of $\ce{NO2}$. It is clearly the population of the $\pi$-electrons are lower in nitrobenzene when $\ce{NO}$ group substituted in para- or meta-position of it compared to that of second $\ce{NO2}$ group substituted in either position (see $\color{green}{\text{green circle}}$ of the plot). Both $\ce{NO}$ and $\ce{NO2}$ groupd have shown $\pi$-electon withdrawing effects on the ring compared to nitrobenzene itself ($\ce{X = H}$; $\color{blue}{\text{blue circle}}$). Therefore, it is safe to say that the electron density on the phenyl ring decreases in nitrozobenzene compared to that of benzene. This conclusion is supported by the positive Hammett substituent constant ($\sigma$-value) listed for $\ce{NO2}$ substituent in benzoic acid: $\sigma_{p-\ce{NO}} = 0.91$ and $\sigma_{m-\ce{NO}} = 0.91$. By comparison, $\sigma_{p-\ce{NO2}} = 0.78$ and $\sigma_{m-\ce{NO2}} = 0.71$ (The authors referred Ref.3 for these $\sigma$-values).

References:

Halina Szatylowicz, Anna Jezuita, Tadeusz M. Krygowski, “On the relations between aromaticity and substituent effect,”  Structural Chemistry 2019, 30, 1529–1548 (https://doi.org/10.1007/s11224-019-01360-7).
Halina Szatylowicz , Anna Jezuita, Krzysztof Ejsmont, Tadeusz M Krygowski, “Substituent Effect on the σ- And π-Electron Structure of the Nitro Group and the Ring in Meta- And Para-Substituted Nitrobenzenes,”  J. Phys. Chem. A 2017, 121(27), 5196-5203 (https://doi.org/10.1021/acs.jpca.7b03418).
Corwin  Hansch, A. Leo, R. W. Taft, “A survey of Hammett substituent constants and resonance and field parameters,”  Chem. Rev. 1991, 91(2), 165–195 (https://doi.org/10.1021/cr00002a004).


Acid-catalysed Reaction of Ketone with Benzene to Make Benzyl Alcohol

I haven’t ever seen a reaction directly like that. To generate the alcohol you drew on your piece of paper, I would probably use phenyl lithium or phenyl magnesium bromide and add that to acetone; it is easier and faster. (Note that the reaction would stop at the tertiary alcohol and not react further under elimination of water.)

Note that the reaction to synthesise DDT starts from trichloroacetaldehyde. The three chlorine atoms draw electron density away from the carbonyl group, further strengthening the cation. I would suppose that this is not seen often because of the activation a simple aldehyde would need. Let’s face it: benzene is a very bad nucleophile.


Evaporation velocity of ethanol: adding ethanol vapour to a gas mixture

Please note that I have never done this.

But are you sure that you want to rely on approximation/assumptions?

I realize that it needs additional changes to the experimental setup, but is measuring the ethanol concentration out of question? Gas sensors for ethanol are commercially available and a setup, in which they steer mass flow controllers is conceivable.

As a side note, you might be interested to see that ethanol sensors can be build in the lab:

A Novel Flexible Room Temperature Ethanol Gas Sensor Based on SnO2 Doped Poly-Diallyldimethylammonium Chloride.


Regarding acid-base reactions, how does one know when a hydrophobic base substance is fully protonated?

Yes, that is correct.
As you add HCl, the pH will remain more or less constant at around 5, since in solution there is a salt of a strong acid and a weak base.
Then when all the base has been protonated, the pH will fall rapidly.

If you need to know how much insoluble base you have, it must be done through back titration. You add an excess of HCl to the sample and then titrate the remaining free HCl.


You are correct, the reaction order cannot be deduced from the brutto-equation and is usually determined experimentally.

However, here you are presented with the elementary reactions and their rates. Obviously, the slowest step is the bottleneck and controls the overall kinetics, hence the total rate matches the rate of the slowest step. And as this is an elementary reaction, you just apply the law of mass action and use the stoichiometric coefficients for the rate equation.


dsp3 and sp3d hybridization — what is the difference and which applies to square pyramidal?

We can split this into two cases: main group elements and transition elements.

Main group elements do not use their d-orbitals for bonds and thus cannot be hybridised in any d-containing way. Bromine in $\ce{BrF5}$, to cite your example, is unhybridised. The $\mathrm{p}_z$ orbital forms a traditional two-electron-two-centre bond with one of the fluorines. The $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals each form a four-electron-three-centre bond with two additional fluorines for a total count of five fluorines. The bond order of these four-electron-three-centre bonds is $\frac12$.

Bromine’s 3d orbitals are completely populated and effectively belong to the core orbitals. They do not participate in bonding. Bromine’s 4d orbitals are energetically even further removed than the 5s orbital is — remember the aufbau principle. Thus, neither the lower nor the higher d-orbitals can participate in bonding to any notable extent.

This extends well to all main group elements as hinted above. Practically no main group element compounds will include d-orbitals.

For transition metals the case is slightly different since we are in the midst of the d-block where d-orbitals can play a role. However, invoking hybridisation for transition metals puts a big blanket over the bonding situation declaring things equal which really aren’t. A much better description for transition metals would be to use a molecular orbital scheme as I have done in countless answers of mine. Most importantly, the bonding situation can, in my opinion, not be understood without this depiction.

Additionally, many transition metal geometries provide unintuitive ‘hybridisation’ solutions — octahedral being the notable exception which actually hybridises six orbitals (three d-orbitals are nonbonding). For example, a tetrahedral metal complex should be understood as using all orbitals for bonding. Likewise, the square-planar ligand arrangement only has one nonbonding d-orbital.

My suggestion is never to use hybridisation approaches for transition metal complexes.

Formally, one may say that putting the d before s and p implies a lower-shell d-orbital. That would be the transition metal case above. And placing the d behind s and p would use d-orbitals from the same shell. That would be the main group case. Thus, the difference between the two (tl;dr) is:

The first is very strongly discouraged, the second is wrong.


Calculate the wavelength of the radiation released when an electron moves from n= 5 to n=2

The Rydberg formula (for the hydrogen atom) is usually given as

$$\tilde{\nu} = \frac{1}{\lambda} = R \left( \frac{1}{n_\mathrm f^2}-\frac{1}{n_\mathrm i^2} \right)$$

Here, $R$ is the Rydberg constant; take care of the unit ($\mathbf{m^{-1}}$  or $\mathbf{cm^{-1}}$) here! Wikipedia quotes a value of $\pu{1.097 \times 10^7 m-1}$ for a nucleus of infinite mass. In principle, the value for a nucleus of finite mass (as is the case here) is slightly different, but for the purposes of this question, we can ignore that.

$n_\mathrm f$ is the final state, and $n_\mathrm i$ is the initial state.

So, if you substitute $R = \pu{1.097 \times 10^7 m-1}$, $n_\mathrm{f} = 2$, $n_\mathrm i = 5$, you can arrive at $$\tilde\nu = \pu{2.304 \times 10^6 m-1}$$

and the wavelength is $\lambda = 1/\tilde\nu = \pu{4.341 \times 10^-7 m}$, or if you convert to nanometers, $\lambda = \pu{434.1 nm}$. Double-checking against the Wikipedia page on the Balmer series, this value is indeed correct.


molar mass of a substance is the mass of one mole of a substance

This is not quite correct. If you said the speed of a car is the distance it travels in one hour, cars would go at speeds of 120 miles (race car) or 30 miles (in town). However, speed is defined as distance traveled per time. As a result, speed has the dimension of distance divided by time, and a typical unit (at least where I live) is miles per hour.

Molar mass is defined as the mass per amount of substance. The dimensions are mass divided by amount of substance, and typical units are g/mol.

  Why does molar mass have the same value as molecular mass?

If the molar mass is given in g/mol and the molecular mass is given in u (unified atomic mass unit) or Dalton, the number is the same. The reason is that both the mole (at least until May 2019) and the Dalton are based on the mass of the carbon-12 isotope. A mole of a substance is defined as the amount that has the same number of particles as present in 12 grams of carbon-12. The Dalton is define as 1/12th of the mass of a carbon-12 atom.

From these definitions, you will find that the molar mass of the carbon-12 isotope is 12 g/mol, exactly, and the molecular mass of carbon-12 is 12 Dalton, exactly.

  What will change in May 2019?

The mole will be defined as "One mole contains exactly 6.02214076×10^23 elementary entities." The molar mass of carbon-12 will be an experimentally determined quantity.


Effect of pressure on transition temperature

The mathematical answer is based on Clapeyron's equation (written in what follows in form inverted relative to the more usual):
$$\frac{dT}{dp}= \frac{\Delta V}{\Delta S}$$
which must hold on any point on the phase coexistence line.
If $\frac{dT}{dp}=0$ then it follows that $\Delta V=0$.


The percentage of ingredients in a glass of tea

You can't really talk about the 'concentration' of tea, because 'tea' is not a well defined compound. It contains a lot of different chemicals, some of which are dissolved into water when brewing.

You could talk about the concentration of caffeine, epigallocatechin gallate, or manganese in a particular cup of tea, but they depend on the type of tea, the temperature of the water, the infusion time, etc.

The minimum strength of a cup of tea that still allows it to be called tea is a fuzzy question without an answer. Take a pile of sand. Remove grains of sand one by one. When does it stop being a pile? A reasonable limit for the tea would be when a normal human being can't taste the tea anymore.


Does alumina become transparent at high temperatures?

Well, alumina is sapphire so stand to reason that it is transparent at a low temperature too.  I think what is meant by what you heard is that the higher intensity of infrared light is transmitted by the alumina causing thermal loss.

Alumina is transparent to UV, visible and infrared light. To our visible light vision, the scattering and reflection of polycrystalline alumina make it appear as opaque white. In the IR spectrum though it may appear transparent. This is because scattering is proportional to the inverse of the wavelength to the fourth power ($\alpha _{\text{scat}} \propto \lambda ^{-4}$).  Thus a wavelength just twice as long a red light (the longest wavelength of visible light; $\lambda = \pu{750nm}$ ), in this case, $\pu{1.5 \mu m}$, which is near infrared, would scatter one-sixteenth ($6.25 \% $) as much as the red light would. A wavelength twice as long as that wavelength ($\pu{3\mu m}$) would be in the thermal infrared spectrum and be scattered one-two-hundred-fifty-sixth as much as the red light ($0.39\%$) This reduced scattering allows more transmission. At high temperature when infrared radiation is more intense, the energy loss due to transmission becomes more significant, but the alumina was always transparent.


What is the mechanism for the reaction of an organic hydroperoxide with sulfuric acid?

As Marko has commented it is like the Hock rearrangement, which is the migratory step in the cumene process.

It is somewhat similar to the Beckmann rearrangement except that there is an oxygen instead of a nitrogen.

The acid protonates the peroxide, generating a great leaving group on electron-deficient oxygen.
Migration towards electron-deficient oxygen occurs with loss of water to give an oxonium ion.
The migratory aptitude here is determined in the same way as for the Baeyer-Villiger reaction. There is a buildup of positive charge on the migrating carbon during the migratory step, and therefore the group that can best stabilise the charge preferentially migrates. The methoxy-substituted ring is electron-rich and the positive charge is more stabilised on it than on the unsubstituted phenyl group. The methyl group is even worse at stabilising positive charge.
Under the dilute acid conditions this oxonium ion is hydrolysed.


I'm not sure whether calibration for a dynamic assay is any different from one that is stable. But let's think of the situation like this. What you have done is no different from what an operator would do to obtain a calibration curve with an internal standard (IS) correction (by plotting concentration against the ratio of the intensities of analyte/IS) except you have used the absolute intensity of your background signal (no analyte) instead of an internal standard for correction. This is not correct by any standard since the background is a bad choice for correction (the background is NOT stable, and has a terrible RSD due to low counts). What you should do is to plot your calibration curve without normalising it to the background.


Why is magnesium ascorbate, which is a conjugate base of a weak acid, acidic?

$\mathrm{p}K_\mathrm{a}$ relation in the answer of Kav

Wikipedia says pKa's are 4.10 (first), 11.6 (second)

gives the hint a monobasic salt with alkali metals should be very mildly alkalic, as pKa2 is farther from pH 7.
$pH \approx \frac{\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}}{2} = \frac{4.10+11.6}{2}=7.85$
OTOH, it will probably form a complex with Mg, affecting acidity of the 2nd hydrogen. Like
$$\ce{2 H2A(aq) + Mg^2+(aq) -> [MgA2]^2-(aq) + 4 H+(aq)}$$
respectively
$$\ce{Mg(HA)2(s) ->[H2O] [MgA2]^2-(aq) + 2 H+(aq)}$$
See also citrate or EDTA complexes for this effect. AFAIK, before suitable metalochrome indicators came, metal content with excess of Na2H2-EDTA had been sometimes determined by back titration using NaOH.


Why calcium chloride is used to melt ice over sodium chloride

That one mole of $\ce{CaCl2}$ is more effective in melting ice than one mole of $\ce{NaCl}$ is explained by the van 't Hoff factor. The freezing point depression of a solution is calculated by
$$\Delta T=K_fbi$$
where $K_f$ is a cryogenic constant which is specific to the solvent, $b$ is the molal concentration of the solute, and $i$ is the van 't Hoff factor, which indicates number of solute particles. In this case, it is the number of ions produced upon dissociation. Because $K_f$ and $b$ are constant, and $i=3$ for $\ce{CaCl2 \rightarrow Ca^2+ + 2 Cl-}$ and $i = 2$ for $\ce{NaCl \rightarrow Na+ + Cl-}$ the freezing point depression is greater for $\ce{CaCl2}$.

The reason $\ce{CaCl2}$ is widely used as melting agent may have more to do with keeping the concrete intact. Concrete is composed of many calcium-containing species. When water flows over the concrete, the calcium can be leeched out, and the concrete becomes brittle. The higher the calcium content in the water, the less leeching occurs, and the concrete remains structurally sound.


Is it possible for a group to substitute itself?

Yes - a nucleophile can replace itself in a substitution reaction. Under the SN1 mechanism, once the carbocation is formed, it becomes prone to attack from any negatively charged species, including the one that left it in the first place. Likewise, in an SN2 reaction, there is the possibility of a nucleophile replacing itself. Given enough time, I would expect alkyl halides to become racemic in solutions that contain the halide anion; however, substitution reactions remain stereo-selective. This is because when you are preforming a substitution reaction, you normally use a better leaving group than the nucleophile that you are adding. Because of this, the nucleophile in solution will preferentially replace the initial leaving group over its resulting product. This is because the rate laws for both SN1 and SN2 reactions depend on the concentration of the alkyl group - implying that an alkyl group with a better leaving group will react faster than one with a weaker leaving group. In SN1, the rate determining step is the leaving group dissociating from the carbocation, which happens quicker with a better leaving group. In an SN2 reaction, the rate determining step involves both the nucleophile attacking and the leaving group leaving. This - once again - means that the reaction will proceed quicker with a better leaving group. Given enough time, your product should racemize; however, your substitution reaction should be done before this can have a major effect on your yield.


Why does the distance between atoms in a hydrogen bond impact the strength of the bond?

The main attractive force between the nuclei and the electrons is electromagnetic, positive and negative attracting. When two atoms are "pushed" together each one shares its own electrons with the other one's nuclei, and they bond. So why wouldn't they nuclei just get as close to each other as possible, so that they each would feel as many electrons as possible? At the same time, the electrons are repelling each other, and the nuclei repel (same charges repel). The bond length is a function of this "equilibrium" that is reached between the attraction of the nuclei to the electrons, and the repelling of the nuclei  and the repelling of the electrons. Therefore, if they begin to "close the gap" as you say, there will be more repelling than attracting, and they will move farther away from each other, and return to the lowest energy distance, the original bond length.


Understanding Gibbs free energy and enthalpy

The equation for Gibbs Free Energy is:

$$\Delta G = \Delta H - T \Delta S$$

The change in Gibbs Free Energy already includes the amount of energy associated with the change in energy. So to reverse the reaction, the total change in energy is given by $-\Delta G$.

Now the question is how much energy do you NEED to get the reaction to go backwards. The minimum amount of energy for it to go backwards spontaneously is $-\Delta G$ because that puts the product on the same energy level as the reactants so in principle they are energetically equivalent. However, what you should really be concerned about was the activation energy of the forward reaction, $E_a$, because this is the mountain that the reaction needs to climb. There is a diagram below, but it is based on the $\Delta H$ values. You can convert this to a graph with $\Delta G$ as well.


Volume increase when air is enriched with water

The short answer is that the behavior of the system you describe should very well approximate that of an ideal gas at room temperature and 1 atmosphere, and that the addition of $\pu{0.01L}$ of water vapor to $\pu{1.00L}$ of air will give you a total volume very close to $\pu{1.01L}$ (assuming constant temperature and pressure).

The departure from ideal behavior for water vapor strongly depends on relative humidity. While the absolute humidity given in the question is only $1\%$, this would correspond to $100\%$ relative humidity at $\pu{6^oC}$, and greater than $100\%$ relative humidity at lower temperatures. Gas molecule interactions so strong that condensation occurs essentially represents the opposite extreme of the assumptions made for an ideal gas.

So, at the extreme of very low temperatures, essentially all of the water vapor would condense and your calculation would have a high bias of 1% if you were to assume ideal gas behavior.  However, at room temperature (i.e. $\pu{20^oC - 25^oC}$) this will not be the case and the assumption of ideal behavior would introduce negligible error.


A simple way to compare the $\mathrm pK_\mathrm a$ of $\ce{-NH3+}$ and $\ce{-COOH}$ in such a scenario would be to assume the situation when the amino acid becomes a zwitter-ion.
A zwitter-ion is a molecule that has both positive and negative charges on it. In an amino acid, the $\mathrm {pH}$ at which this is seen is known as the isoelectronic point.

Let's consider this molecule to have reached the isoelectronic point. We know that the value would be less than $7$ since there are two acidic groups (meaning that this is an acidic amino acid). When we reach that point, we see that the both the carboxyl groups are de-protonated and the amine group still exists as $\ce{-NH3+}$.
This means that the $\ce{-NH3+}$ group is less acidic than the the acid groups. Therefore, Hd is is less acidic than Ha or Hb.
As you further increase the $\mathrm{pH}$, we see that at a point, the $\ce{-NH3+}$ group becomes $\ce{-NH2}$, before Hc is removed. This means that Hd is more acidic than Hc.
Now, we need to compare the two acid groups. As you may have noticed, The $\ce{-NH3+}$ group is closer to Hb than Ha, since this group is a highly electron withdrawing group [-I effect], the negative charge formed by the de-protonation of Hb would be more effectively withdrawn making it more stable than the conjugate base formed by the de-protonation of Ha.
Therefore the order of stability would be:

Hb > Ha > Hd > Hc


Can anyone explain, why there are coefficients 2 in the equation for rate of change [A]?

$$\ce{A_n ->[$k$] nA}$$
For the reaction given above, rate of formation of $\ce{A}$ is defined as:
$$\frac{1}{n}\frac{\mathrm d\ce{[A]}}{\mathrm dt} = k\ce{[A_n]}$$
In this case, we have three different reactions that involve A. They are:
\begin{align}
\ce{A_n &->[$k_1$] nA} \tag{1}\label{1}\\
\ce{nA &->[$k_1'$] A_n} \tag{2}\label{2}\\
\ce{A + B &->[$k_2$] P} \tag{3}\label{3}\\
\end{align}
Writing the rate equations for $\eqref{1}$, $\eqref{2}$, $\eqref{3}$ and summing them up to get total rate, we get:
\begin{align}
\frac{1}{n}\frac{\mathrm d\ce{[A]1}}{\mathrm dt} &= k_1\ce{[A_n]} \tag{4}\label{4}\\
-\frac{1}{n}\frac{\mathrm d\ce{[A]2}}{\mathrm dt} &= k'_1 [\ce{A}]^n\tag{5}\label{5}\\
-\frac{\mathrm d\ce{[A]3}}{\mathrm dt} &= k_2\ce{[A][B]} \tag{6}\label{6}\\
\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} &= \frac{\mathrm d\ce{[A]1}}{\mathrm dt} + \frac{\mathrm d\ce{[A]2}}{\mathrm dt}+\frac{\mathrm d\ce{[A]3}}{\mathrm dt} \tag{7}\label{7} \\
\end{align}
Substituting values from $\eqref{4}$, $\eqref{5}$, $\eqref{6}$ into $\eqref{7}$, we get:
$$\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} = \color{red}{n}k_1\ce{[A_n]} - \color{red}{n}k'_1\ce{[A]^n} - k_2\ce{[A][B]} $$
Here, $n = 2$, therefore
$$\frac{\mathrm d\ce{[A]_\mathrm{tot}}}{\mathrm dt} = \color{red}{2}k_1\ce{[A_n]} - \color{red}{2}k'_1\ce{[A]^2} - k_2\ce{[A][B]} $$


How does hydrogen bonding affect the boiling points of chemical compounds?

A hydrogen bond is a particularly strong dipole:dipole interaction with some covalent character. If we ignore the covalent character for a moment, we can say the higher the partial positive charge on the hydrogen, the stronger the hydrogen bond (keeping the O-H bond distance constant). If you relate "electron deficient" to positive partial charge, it might help you to make sense of the textbook answer.

You write:

  Both hydrogen, in methanol and water, are connected to oxygen, which has the same electronegativity and takes electrons off the hydrogen making it have a partial positive charge.

Yes, that is the primary reason compounds like water, methanol and - say - hydrogen peroxide and acetic acid are hydrogen bond donors. But the chemistry beyond the OH group also plays a role. To illustrate, just compare the acidity of the proton connected to the oxygen in these four cases. They are very different. In a similar manner, the hydrogen bonding will be different (not as different, though).


Calculating new volume during expansion of an ideal gas

I dont understand where you are getting stuck up since everything is already given to you and you just have to substitute stuff using:

$\Delta U= W+Q$, the basic conservation equation.

$Q$ is the heat supplied. Think about the change in internal energy ($U$) during an isothermal process.


How do I calculate the pH of this buffered solution?

You can think of $\ce{NH4OH}$ as a hydrate: $\ce{NH3_{(aq)}}$; no need for complicated reactions interconverting the two.

Concerning the problem you described, at these concentrations all subtleties are negligible, and you basically only need to calculate the nominal concentrations of $\ce{NH4+}$ and $\ce{NH3}$, and plug them into the Henderson-Hasselbach equation.

You start with $\ce{0.5 M NH3}$ and $\ce{0.5 M NH4Cl}$, and add $\ce{HCl}$ to a total nominal concentration of $\ce{0.15 M}$, without altering the total volume (the question is written ambiguously, but that's what it means, most likely). As $\ce{HCl}$ is a very strong acid, it will convert all $\ce{NH3}$ to $\ce{NH4Cl}$ (actually $\ce{NH4+}$, because we are in solution). Therefore your final situation is: $\ce{(0.5-0.15)=0.35 M NH3}$ and $\ce{(0.5+0.15)=0.65 M NH4+}$. This should give you a pH of about 8.98 (lower than the one you would have without the HCl, consistently with the addition of an acid).

So your main error above was that you dissociated $\ce{HCl}$ (correctly), but then just added the resulting $\ce{H+}$ to the total, forgetting that there was a ton of $\ce{NH3}$ around that would react with it.

Another example is when you add a solution of some particular transition metal salt to a solution of a weak acid (could be iron sulfate + hydrogen sulfide for instance). Despite the fact that the salt solution isn't acidic at all, the pH drops considerably compared to the solution of the weak acid alone, simply because the metal and the anion form a very insoluble salt that precipitates, subtracting weak acid anions from the solution and forcing the previously non-dissociated acid to dissociate.

In general, my advice is stick to what you know, don't go looking for complications: if they did not teach you to do very exact/sophisticated pH calculations, they probably don't expect you to pull them out of a magic hat at the exam. And as my stoichiometry teacher used to say, always write down all species you have in solution.


TL;DR Ab initio HF molecular orbital studies of the potentially antiaromatic $8\pi$-electron systems of distorted and planar 6-membered rings revealed that conformational rigidity (calculated or measured folding energies) is most likely the best extension to a $(4n+2)\pi$ criterion. 1,4-Dithiins experience large energy decreases when permitted to fold, and when no external strains are applied, there is no reason for sulfurs to unevenly donate electronic pairs.

Crystal structure of 1,4-dithiin [1] is indeed suggesting a non-planar molecule with the fold angle between "red" and "blue" planes (both plotted on the mirrored $\ce{-S-C=C-S-\,}$ fragments) of $137.14^\circ$:

Same goes for the structure of (1,4)-Dithiino(2,3-b)-1,4-dithiine [2] with the dihedral angle of $131.03^\circ$:

  The chair conformation of the molecule prevents conjugation, and the C-C bond distances clearly corresponds to normal unconjugated bonds, (simple double bond = 1.337(6) Å) so the molecule is non-aromatic in the solid state.

Theoretical calculations ($\Delta_fH^\text{MNDO}$) suggest that both molecules remain bent not only in crystal structure, but they also are flexible in gas phase and in solution [3].

At this point one might think that geometrical conformation dictates the aromaticity or the lack thereof, which is in my opinion an example of violation of the law of cause and effect. Under certain "strains" 1,4-dithiins can exhibit planar structure, like in case of 1,4-dithiine-tetracarboxylic N,N'-dimethyldi-imide acridine [4], where $\pi$-stacking causes a coplanar arrangement of both molecules:

  The intermolecular distances ... are shorter than the van der Waals distance (3.40 Å). This suggests that $\pi-\pi$ interactions between molecules ... exist in these parts and the molecules form a paired molecular group. The dihedral angle between the least-squares planes of molecules ... is $2.3^\circ$}.
  Distances between paired groups are larger than the van der Waals distances... The stacking of paired molecules is in accordance with the poor electrical
  conductivity of this complex.

  Anti-aromatic 1,2-dithiins display properties opposite to those of 1,4-dithiins, whose dications show aromatic stabilization. Unlike other antiaromatic compounds, the 1,2-dithiin derivatives, with eight $\pi$-electrons ..., appear in natural products synthesized by composite plants. The Dewar resonance energies of nonplanar 1,2-dithiins are close to zero. The aromatic 1,4-dithiin dication is generated by two-electron electrooxidation of the neutral 1,4-dithiin. The aromaticity was established by NMR measurements and theoretical calculations.

Conformational criterion for aromaticity and antiaromaticity has been extensively investigated for a number of $8\pi$-electron systems [5], and for 1,4-dithiin the energy required to force it into planar geometry is less than $\pu{2.7 kcal/mol}$ (max. value of 3-21G folding energy), allowing for a high degree of conformational flexibility. This seems to be one of the most recent methods of quantitative determination of both flexibility and aromaticity, as authors propose:

  Relative to planar reference geometries of maximum symmetry, folding of aromatic and antiaromatic systems results in large endothermic and exothermic changes, respectively. Aromatic compounds are planar and strongly resistant to
  folding, whereas antiaromatic systems exhibit large energy decreases when permitted to relax (fold) from planar geometries to nonplanar, minimum-energy conformations. According to this criterion, systems ... which have
  shallow folding energy curves, are nonaromatic.

Of course, $6\pi$-aromaticity can be achieved via electrooxidation of non-aromatic 1,4-dithiine, which leads to a planar dication which exhibits aromatic character [6]:

Bibliography

Howell, P. A.; Curtis, R. M.; Lipscomb, W. N. Acta Cryst, 1954, 7 (6–7), 498–503. DOI 10.1107/S0365110X54001582.
Varma, K. S.; Sasaki, N.; Clark, R. A.; Underhill, A. E.; Simonsen, O.; Becher, J.; Bøwadt, S. Journal of Heterocyclic Chemistry 1988, 25 (3), 783–787. DOI 10.1002/jhet.5570250314.
Bock, H.; Roth, B.; Lakshmikantham, M. V.; Cava, M. P. Phosphorus and Sulfur and the Related Elements 1984, 21 (1), 67–77. DOI 10.1080/03086648408073129.
Yamaguchi, Y.; Ueda, I. Acta Cryst C, 1984, 40 (1), 113–115. DOI 10.1107/S0108270184003218.
Podlogar, B. L.; Glauser, W. A.; Rodriguez, W. R.; Raber, D. J. J. Org. Chem. 1988, 53 (9), 2127–2129. DOI 10.1021/jo00244a059.
Balaban, A. T.; Oniciu, D. C.; Katritzky, A. R. Chem. Rev. 2004, 104 (5), 2777–2812. DOI 10.1021/cr0306790.


Could an extremely strong reducing/oxidising agent attack nitrogen molecules?

I know there exist industrial methods for that already, but im poking
around to see if maybe there's a way to do it under atmospheric
pressure without tailored molecules like enzymes or metalorgnic
complexes

Though Haber process which was used to industrially produce ammonia was well-known, some researchers have also attempted to split nitrogen molecules using metal complexes:

Chirik et.al (Nature, volume 427, pages 527–530(2004)) attempted to split nitrogen molecules by using a zirconium complex containing two bulky chemical groups called methylated cyclopentadienyl. The complex was the right size to snuggle up to both nitrogen atoms and give an electron to each atom. This double donation take place in an organic solvent at 100 °C and atmospheric pressure, disrupts the triple bond. Further reactions in the same solution then finish the job of splitting the nitrogen molecules and adding hydrogen atoms to make ammonia. (source)
Carbon monoxide and a molecule containing hafnium to break apart nitrogen and create a fertilizer called oxamide. (source)
John Mark P. Martirez and Emily A. Carter at Princeton University propose a gold-molybdenum catalyst that could split the nitrogen triple bond at room temperature using energy from visible light. That dissociation step is the primary limit on the reaction rate in the Haber-Bosch process. (source)


To go from being confused to less confused, here are some facts.

Hydrogen has one valence electron which can participate in reactions so it is a group 1 element.
Hydrogen is not a group 1 alkali metal, nor is it a group 1 compound, so simply stating hydrogen is in group 1  is ambiguous and confusing.  It is a group 1 element.  The word element needs to be included for the statement to make any sense.
Elemental hydrogen has no charge at standard conditions because it exists as dihydrogen gas, $\ce{H2}$.  If hydrogen gas were to be ionized it could hypothetically break into two ions, one with a +1 charge and one with a -1 charge.  If no ions are involved then the concept of ionic charge is irrelevant.
Elemental hydrogen, again as dihydrogen gas, $\ce{H2}$, has a zero oxidation state.  The oxidation state refers to a hypothetical charge of an element within a compound. If no compound is involved then oxidation states are irrelevant.
Elemental hydrogen has an oxidation state of +1 in most compounds and -1 in metal hydrides because of an agreement within the International Union of Pure and Applied Chemistry (IUPAC). -- http://goldbook.iupac.org/O04365.html --  Oxidation states, as many of the concepts in basic chemistry, should be viewed as a tool to gain a rudimentary understanding of how compounds form.


Hydration of acetone is an equilibrium process.
$$\ce{(CH3)2C^16O + H2^18O <=> (CH3)2C^18O + H2^16O }$$
The equilibrium constant of this reaction will depend on how the bond strengths of the $\ce{O-H}$ bond in water and the $\ce{O=C}$ bond in acetone is affected by isotope exchange. The isotope-dependent strength of hydrogen bonds might be a secondary effect, as well.
The ratio of heavy to light acetone will not only be dependent on the equilibrium constant, but also on the ratio of acetone to heavy water initially present.
So none of the statements are sufficient to explain the observation completely. Statements (a) and (b) are correct. Statement (c) is not directly related to the ratio of the species. You also have to consider the isotope effect on the strength of hydrogen bonds with water (as one acetone changes from light to heavy, one water changes from heavy to light).


How to predict the products of a organic chemical reaction

You need to build a working knowledge of known reaction mechanisms that include the functional groups of your reactants. Put another way, (1) identify the functional groups present in your reactants and (2) check known reaction mechanisms that proceed by reaction of these functional groups.

Really, the only way to be good at this is to study and memorize organic reactions. There's a number of resources to help with this (e.g. online flashcards, and I really like Chemistry by Design, an app that you can use on a smartphone).


Is there any database, or reference, for retardation factors (Rf)?

No there is not database for retardation factors, because Rf values are useless outside your lab. It is not a fundamental property like melting point or boiling point. One batch of silica is slightly different from another. It also depends on exact solvent composition and history of the treatment of the stationary phase.


Why is activation enthalpy found and not just rate of reaction?

One of the Factors Influencing reaction rate is given below (Source)

  Temperature: Usually conducting a reaction at a higher temperature delivers more energy into the system and increases the reaction rate by causing more collisions between particles, as explained by collision theory. However, the main reason that temperature increases the rate of reaction is that more of the colliding particles will have the necessary activation energy resulting in more successful collisions (when bonds are formed between reactants). The influence of temperature is described by the Arrhenius equation. As a rule of thumb, reaction rates for many reactions double for every 10 degrees Celsius increase in temperature, though the effect of temperature may be very much larger or smaller than this.

Reaction rate , Activation energy are connected to temperature through Arrhenius equation. We are interested in Activation Energy as it gives an idea what temperature the reaction starts. We want to know $E_{a}$ in order to manipulate reaction rate essentially. Catalysts or Inhibitors work essentially on this principle.

Just looking at rate wont help us in varying the rates to our interest.


If you look at the structure closely, you'll find that there are no Hydrogens on the C with the double bond attached. Adjacent to this C is the methyl group. Due to the n+1 rule, the singlet should be due to that Methyl group connected to the "Hydrogenless" Carbon. Hope this simplified answer helps


Reason of dissociation of water molecule in electrolysis

The dissociation of water to $\ce{OH-}$ and $\ce{H+}$ happens first before any interaction with the cathode. $\ce{H+}$ ions then gain an electron at the cathode to produce hydrogen. At the anode, water is oxidized (loses electrons) to produce oxygen and more $\ce{H+}$ ions.

In pure water there is a very small concentration of $\ce{H+}$ and $\ce{OH-}$ which is why the reaction is very slow and not thermodynamically favorable. It is also the reason why pure water is such a good insulator. But when you add an electrolyte the process becomes much more favourable kinetically and thermodynamically. Further proof that $\ce{H+}$ and $\ce{OH-}$ are generated before any interaction with the cathode.


Wow. Good question. Graphene is not considered a polymer in the traditional sense. While it is used a doping agent for other polymers it does not have the designation of being one itself, the allotropes of carbon, at least in my experience and that of my adviser, are typically in a class of their own. Unique structures such as buckminsterfullerene and diamond differ from graphene only by the angles between the carbon atoms, 3-d shape, (and number of carbons present). All of which are definitely not considered polymers. Also remember that a macromolecule is composed of polymers.
http://www.macrogroup.org.uk/schools/polymer_chemistry.php

https://answers.yahoo.com/question/index?qid=20100906193125AAzb9T0

So the short answer would be no, graphene is not considered a polymer.


Reactivity of meta- and para-bromo-nitrobenzene towards nucleophilic substitution

p- Bromonitrobenzene is certainly more reactive towards Nucleophilic Substitution than m-Bromonitrobenzene. The reasons are as follows,  Firstly, due to $-R$ effect of $\ce{NO2}$ group, the positive charge is created on the carbon attached to bromine in para position, which helps to increase thre electrophilicity of that carbon and that makes easier for the electrophiles to attack the benzene ring. If $\ce{NO_2}$ group would have been in  meta position, the positive charge would have been created on the adjacent carbon, so, the electrophilicity of the bromine attached carbon is not increased so much which in turn will not help it to undergo Nucleophilic attack easily. The image below demonstrates this effect on para position.

Secondly, the Nucleophilic substitution in these systems undergo by addition-elimination mechanism. So, when the nucleophile attacks and forms the intermidiate complex, that complex is much more stabilised due to charge delocalisation in the $\ce{NO2 }$ group also if it is in para position.

But if $\ce{NO2}$ is in meta position, the charge can only delocalise in the ring, but it can not delocalise in the $\ce{NO2}$ group (you can verify by looking at the  image also), so it is not so stabilised as the para isomer intermidiate.  These effects are mainly dominating which makes the p-Bromonitrobenzene more reactive than m-Bromonitrobenzene in Nucleophilic substitution.


Which of these thermodynamic relationships is true for a system at equilibrium?

At equilibrium, $\Delta {G}$ is zero, so according to the equation $\Delta {G} = \Delta {H} - T \Delta {S}$ the first option is correct.


Why does small value of van der Waals gas constant "b" ensure easier liquefication?

The $b$ in the Van der Waals equation of state signifies the volume that is effectively taken out per molecule by intermolecular repulsive interactions. You can imagine that a small $b$ will help the gas to liquefy, since in most ordinary liquids molecules are packed closer together than in a gas.

As written in Atkins Physical Chemistry:


Why is acetylsalicylic acid still more acid than benzoic acid?

The acetyl moiety in acetylsalicylic acid has a $(-I)$-effect, i.e. compared to a proton attached to the aromatic ring (as in the case of benzoic acid), it withdraws electron density out of the aromatic ring.  This provides an additional stabilization of the conjugated base, that is equivalent with rendering acetylsalicylic acid a stronger Brønsted acid than benzoic acid.  Hence, the $pK_a$ of acetylsalicylic acid is lower than the one of benzoic acid.

Regarding the second paragraph, please notice $pK_a$ is a thermodynamic property -- where time does not play a role.  In contrast, the substitution reactivity you likely refer to (Hammett parameters) refers to reaction kinetics and rate constants.  Here, time does play a role.


Under what circumstances can the stereochemistry at nitrogen atom be faithfully retained and identified?

You could try to lock the conformation of structure. For example, derivatives of a  1-azabicyclo[2.2.2]octane system could have a chiral nitrogen, and the fact that the substituents are tied back prevents inversion of the nitrogen. Quinine has this kind of structural moiety.

This example isn't exactly what you're looking for, but it does it by using diastereocontrol to limit the amount of undesired isomer: J. Org. Chem. 2004, 69 (3), 714–718.

As Ivan suggested, you can hold onto the lone pair with something. Here's an example where metal coordination prevents inversion: Inorg. Chem. 2012, 51 (15), 8516–8523.

There are plenty of others. I just did a quick search on the JACS website for stereogenic nitrogen.


What makes a species "fluorescence quencher"?

A fluorescence quencher is any species that causes your fluorophore to stop fluorescing. There are several ways this can happen, but there are two broad categories: chemical reactions and energy transfer. I will give simple examples of each category below. However, which method is operating is going to depend on the fluorophore. Bromide and fluoide will not behave the same way with every fluorophore. Some might be quenched by fluoride and not bromide. Some might be quenched by both. Some might be quenched by neither.

However, in the case of the specific fluorophore in the article, it is quenched by bromide and not be fluoride. The article likely provides an explanation or hypothesis about this behavior, or it cites something that does. Likely this article then is about a selective sensing application. Since bromide turns the sensor off (quenches fluorescence) and fluoride does not, the sensor is selective for bromide over fluoride.

  Quenching by chemical reaction

Any chemical reaction will change the relative energies and energy spacing of the frontier molecular orbitals, which could lead to quenching. This chemical change could be something irreversible like oxidizing or reducing the fluorophore (probably not what is happening here). Other chemical reactions are possible, and it could be that bromide reacts and fluoride does not.

This change could also be a reversible one, like protonation/deprotonation, ligand binding, etc. If the fluorophore in your article is a transition metal complex, than it could be quenched by complexation with bromie anion (which displaces some ligand that has charge-transfer absorption/fluorescence with the metal). Fluoride might not do this.

This behavior could be more complex. Perhaps the fluorophore only fluorescences in the presence of some analyte. Maybe bromide reacts with the analyte (and fluoride does not) or bromide competes with the analyte but does not promote fluorescence.

  Quenching by energy transfer

Fluorescence works as follows. The fluorophore absorbs a photon and is excited from its ground state to its excited state. When the excited state relaxes back to the ground state it emits another photon.

Fluorescence can be quenched if the excited state can be intercepted. An example is as follows. You have a system with your fluorophore and another molecule. Your fluorophore has a smaller HOMO-LUMO gap than this other molecule, so the fluorophore selectively absorbs photons of a certain frequency range and is excited. However, if the other molecule has a lower energy excited state than your fluorophore, then energy transfer can occur between them if they are close. Now your fluorophore is back in its ground state and the other molecule is in its excited state. If this molecule does not fluoresce, then fluorescence is quenched. This is probably not happening in your example. Note this phenomenon can be given various names like "resonance energy transfer" or "Forster energy transfer."

This type of quenching can get more complex as well. The energy acceptor could go and do chemistry or it could be so eager to be reduced that it not only steals the energy from the fluorophore, but it grabs an electron as well. This is not likely what is going on with bromide and fluoride.

The other type of behavior that can go on here is analyte-induced aggregation. Aggregation of the fluorophore causes the electronic properties to smear out over the aggregate, which likely means no more fluorescence (and perhaps no more absorbance) at the frequencies you care about. This phenomena could be happening in your example. Bromide could induce aggregation while fluoride does not.


Can a low concentration of an acid disolve solids just as well?

The rate of dissolution depends on $\mathrm{pH}$ and $\mathrm{pH}$ dependent equilibrium reaction chain :

$$\begin{align}
\ce{2 CH3COOH + CaCO3 v &<=> (CH3COO)2Ca + Ca(HCO3)2}\\
\ce{2 CH3COOH + Ca(HCO3)2 &<=> (CH3COO)2Ca + 2 H2CO3}\\
\ce{H2CO3 &<=>   H2O + CO2(aq)}\\
\ce{CO2(aq) &<=>   CO2(g)}\\
\end{align}$$

what can be written in the equivalent ionic form:

$$\begin{align}
\ce{CaCO3 v &<=> Ca^2+ + CO3^2-}\\
\ce{CH3COOH &<=> CH3COO-  + H+}\\
\ce{H+ + CO3^2- &<=> HCO3-}\\
\ce{HCO3-  + H+ &<=> H2CO3}\\
\ce{H2CO3 &<=>   H2O + CO2(aq)}\\
\ce{CO2(aq) &<=>   CO2(g)}\\
\end{align}$$

6% vinegar (1M) acetic acid has $\mathrm{pH=p}K_{\mathrm{a}}/2=2.37$.

Vinegar with 1/11 acid  spent has pH 3.75, with activity of $\ce{H2O+}$  24 times lower, so does is lower the reaction speed.

Vinegar with the acid half spent has pH 4.75, with activity of $\ce{H2O+}$ 240 times lower.

Vinegar with 10/11 acid  spent has pH 5.75, with activity of $\ce{H2O+}$  2400 times lower.

If pH raises enough, equilibrium concentration of $\ce{CO2(aq)}$ does not cross the gas solubility limit anymore and reaction approaches equilibrium, as production of gaseous carbon dioxide ceases..

If initial concentration crosses some minimal threshold, $\mathrm{pH}$ raises to level not able to produce gaseous carbon dioxide before water gets saturated by $\ce{CO2}$.

Some carbonate gets dissolved as bicarbonate, but with low initial acetic acid concentration, this dissolution is slow.

See also carbonic acid Wikipedia article.

Another effect is diffusion limitation. Without stirring, the acid near walls may get spent, while the acid abundance in free volume is intact, without access to limescale. Rate of acid diffusion to walls limits rate of dissolution.

Personally I would suggest rather citric acid, that have much higher acidity.

And/or, instead of filling the whole volume by diluted solution , rather using wall washing by concentrated solution.


What are some ways to measure sugar concentration of fruits?

The solute will alter the density of the solvent, which is used for example to determine the concentration of sulfuric acids by a hydrometer and in particular by help of a saccharometer.  Depending on your equipment you will need at least about 20 mL of analyte.
A popular alternative among winemakers is to rely on the change of the refractive index of a solution (refractometry) instead.  With portable devices as on the picture, you basically determine the critical angle of total reflection of the juice of few grapes placed on a prism.  The analysis is quick (seconds) and needs little (mL or less), the reading typically yields refractive index and content of sugars present as expressed for example in Oechsle, Baumé, Brix, or mass%.

(source)

(source)
Note, however, that these techniques are not specific to sugar(s).  The presence of other compounds soluted equally may affect the refractive index, as well as the wavelength of light used and temperature during the recording.  To prevent adulteration of wine, for example, some more advanced refractometers already determine the refractive index at multiple wavelengths simultaneously, too.


How to calculate atoms of *each* element in a compound

Think of the mole as a SI prefix, like 'kilo' or 'mega'. If you have 5k $\ce{H2SO4}$ molecules, you will have 10k hydrogen atoms, not 5k. Just as 'kilo' means '1000', a mole means $6.02 \cdot 10^{23}$ of something.

Therefore, the first approach gives the correct answer, but I don't understand what you mean by 1gr/mol. Grams are out of the picture here.


Flory–Schulz type distribution with stoichiometric imbalance

I came up with a way to solve my own problem as I was writing my question, but I feel as though it is a bit hand-wavy.  I would appreciate any input regarding the validity of this solution.

The following is a workaround to force the Flory distribution to have a maximum value that is consistent with the average degree of polymerization due to stoichiometric imbalance via Carothers' equation.  Carothers' equation models how stoichiometric functional group imbalances limit the average degree of polymerization for the step-growth mechanism.  This average degree of polymerization $(\text{DP}_\text{average})$ can be calculated very simply:

$$\begin{align}
\text{DP}_\text{average} &= \frac{r+1}{r+1-2rp} \\
\end{align}$$

When there are no stoichiometric imbalances, and the Flory distribution is valid:

$$\text{DP}_\text{average} = \frac{1}{1-p}\\
f(\text{DP}_\text{average}) = x \cdot p^{x-1} \cdot (p-1)^2$$

Let $p_\text{eff}$ be the effective extent of reaction, which is related to $r$ and $p$ via:

$$\frac{1}{1-p_\text{eff}} = \text{DP}_\text{average} = \frac{1+r}{1+r-2rp}\\
p_\text{eff} = 1 - \frac{1+r-2rp}{1+r}$$

This $p_\text{eff}$ can be used to calculate the ideal Flory distribution according to

$$F(\text{DP}_\text{average}) = x \cdot p_\text{eff}^{x-1} \cdot (p_\text{eff}-1)^2$$

The function $F(\text{DP}_\text{average})$ will force the molecular weight distribution to be properly centered about the  $\text{DP}_\text{average}$ that is consistent with Carothers' equation.  I'm not sure if this is chemically sensible, or accurately describes a real-ish polymer system.  Comments/alternative answers would be appreciated.


Stereoelectronic effects in carbamate carbonyl oxygen attack on halide

Path A resembles a 5-enolexo-exo-tet reaction which is allowed by Baldwin rules while path B represents a 5-enolendo-exo-tet reaction which is disfavored. The orbital interactions of path B suffer severe geometrical constraints (even in the conformation where the C-Br bond is pointing towards the viewer when looking at the attached image). Path A is presented in the attached image.


Is the chemical structure of an amide bond (-CONH) or (-CONH2)?

Amides are derivative of carboxylic acids. It consists of 2 parts, and the bond between carbonyl carbon and nitrogen of ammonia is called is amide bond.

The bond structure of 1° amides is as shown in the figure.

What you have shown in the question is bond structure of 2° amide, which can be obtained by removing 1 hydrogen from 1° amide.

Now if you observe structure of nylon, you will see there are n numbers of 2° amides.

  And also, I see online that lots of bond structures (like NH2 and COOH) are prefixed with a dash, so they'd appear like -NH2 and -COOH. Is this just a personal preference, or is there a more elaborate reason behind this?

The dash means it is Functional Group like -R means hydrocarbyl functional group, -COOH means carboxylic acid functional group, etc.


Alcohol (ethanol) distillation: mystery congener

People turn apple cider all the way to apple cider vinegar, so during the process, there will be a time when the amount and ratio of ethanol and acetic acid are optimally balanced for making ethyl acetate - but it doesn't seem to happen. However, given enough time (and oxygen and bacteria), maybe different things happen. Interesting that the expired cider didn't have the nasty taste, but the concentrate did. I wonder if you actually made the ethyl acetate by your heating and distillation procedure.

Ethyl acetate presence gives nail polish remover/ solvent flavor/aroma. This is not related to volatile acidity. It's known to be produced by Pichia (and probably other microbes) in the presence of oxygen.
Source: https://www.homebrewtalk.com/threads/need-a-little-help-with-cider-flaws.666728/

The CRC Handbook shows a ternary azeotrope of water (9.0%), ethanol (8.4%) and ethyl acetate (82.6%) that boils at 70.23 °C vs the water-ethanol azeotrope that boils at 78.17 °C (96% ethanol). The boiling points are sufficiently close that a one-plate column might not do a significant separation, but the heads still should taste worse.
Suggestion: if the apple cider is acidic from a little acetic acid, neutralize to pH 7 or 8 before distillation ($\ce{NaHCO3}$). This might reduce the possibility of ethyl acetate formation. Another experiment would be to take fresh apple cider and see if you can get good tasting alcohol from it.
TLC of volatile solvents is difficult, but you might be able to detect different spreading patterns for the heads, the nasty heart, and some other heart (which tastes OK).


These are called non-stoichiometric compounds or berthollides. Because the ratio of the fractional component is not fixed, but can vary over a small but continuous range, it would be pointless to try to express the formula as least common multiples (LCM).

For example, iron(II) sulfide can have a ratio from about $\ce{Fe_{$0.8$}S}$ to $\ce{FeS}$. So in LCM terms, the formula varies continually from $\ce{Fe4S5}$ through $\ce{FeS}$, including $\ce{Fe9S10}$, $\ce{Fe19S20}$ and so forth, all with similar physical and chemical properties.


If a chemical reaction is repeated multiple times, will it produce the same result?

The answer depends on whether you are looking at the level of a single atom or a large population of atoms. Because of quantum effects, there is a lot of randomness in the behavior of single atoms. If there are multiple possible outcomes, we can fairly accurately determine the probability of each outcome, but which outcome occurs is not predetermined.
With a large population of atoms (such as we have in any normal laboratory experiment like the ones your teacher is talking about), the large sample size means that we get the same result every time if we control every relevant variable (which is actually quite hard to do).


PubChem gives dipole moment ($\mu$) of 1,4-dibromobenzene (p-dibromobenzene) as $\pu{1.43 D}$ in gas phase and $\pu{1.87 D}$ in liquid phase at $\pu{20 ^\circ C}$. It has given 1987 version of Handbook of Organic Chemistry (J. A. Dean, Ed.) as the reference. PubChem also gave $7.77$ as dielectric constant at $\pu{10 ^\circ C}$ and $6.7$ at $\pu{40 ^\circ C}$ with the same reference. However, I afraid to tell you that all of these values are incorrect. The 1999 version of Handbook of Organic Chemistry (same reference in different edition) listed followings for 1,4-dibromobenzene and its isomers (also included 1,4-diiodobenzene and its isomers for comparison):
$$
\begin{array}{c|ccc}
\hline
\text{Compound} & \text{Dipole moment, }\mu & \text{Dielectric constant, }\epsilon  & \text{reference page}\\
\hline
\color{blue}{\text{1,4-Dibrmobenzene}} & \color{blue}{\pu{0.0 D}} \ (-) & 2.57 \ (\pu{95 ^\circ C}) & \text{p. 5.111}\\
\text{1,3-Dibrmobenzene} & \pu{1.5 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 4.21 \ (\pu{20 ^\circ C}) & \text{p. 5.111}\\
\text{1,2-Dibrmobenzene} & \pu{2.13 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 7.86 \ (\pu{20 ^\circ C}) & \text{p. 5.111}\\
\color{red}{\text{1,4-Diiodobenzene}} & \color{red}{\pu{0.19 D}  \ (\pu{20 ^\circ C}\text{ in benzene})} & 2.88 \ (\pu{120 ^\circ C}) & \text{p. 5.113}\\
\text{1,3-Diiodobenzene} & \pu{1.22 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 4.11 \ (\pu{50 ^\circ C}) & \text{p. 5.113}\\
\text{1,2-Diiodobenzene} & \pu{1.70 D} \ (\pu{20 ^\circ C}\text{ in benzene}) & 5.41 \ (\pu{50 ^\circ C}) & \text{p. 5.113}\\
\hline
\end{array}
$$
Accordingly, 1,4-diiodobenzene actually has a dipole moment, but, every other 1,4-dihalobenzenes do not have a dipole moment. The values given in PubChem must be corrected during 15th edition of the book.
Reference:
John A. Dean, Editor, Lange's Handbook of Organic Chemistry, Fifteenth Edition; McGraw-Hill, Inc.: New York, NY, 1999, p. 5.111-5.113 (ISBN: 0-07-016384-7).