Autosampler - is a piece of equipment that aspirates the sample from a vial and introduces it into a chromatographer or a detector. It has nothing to do with peaks or their identification - that comes later.
After the sample is injected into the detector and the measurements are taken you get back data describing physical properties of the sample. If you inject directly into the detector you can plot it as signal intensity vs. physical property, here's an example of spectra from Mass Spectrometer:

You may need to find peaks on such graph in order I don't know.. to identify how many impurities there are.
But often before introducing the sample directly to the detector chemists would run chromatography - it'd separate different components of the mixture and make them elute at different time. Now we'd measure physical properties not once - but do it at some rate. If we measure (and sum up or average) the signal at different time we may get peaks when the compound is eluted from chromatographer:

Chemists would then have to define the boundaries of the peaks and integrate their area in order to find how many substances there are in the mixture, in some cases calculate how much of each there are. Software can automate some of this work - identify the peak boundaries for you. This may be what your professor is asking for. Or maybe he wants you to write algorithms determining what substance is represented by this or that peak.
Note that this problem is very challenging and not universally solved - finding peaks can be problematic because:

different chromatograms will have different amount of high-frequency noise
there could be a baseline drift (low-frequency noise)
peaks often are skewed
compounds can elute very closely to each other, so you get merged peaks (rider peaks, shoulders, round peaks)

Due to these reasons you may not always be able to find the peaks reliably e.g.

So for autodetection you may want to go over several algorithms. And the person who uses them may have to understand and tune the parameters for it to work successfully on a particular chromatogram.


Which chemicals could be used to dissolve Polybutylene Terephthalate (PBT), but not dissolve brass?

PBT is a thermoplastic and its melting point is only 223 °C. Depending on its composition, brass doesn't melt until 900 °C or so. High enough heating may change the structural properties of the brass, but perhaps there is a temperature to which you can soften the plastic enough to detach the pins without damaging them.

Unfortunately, brass is easily attacked by many chemicals and PBT is resistant to many solvents, but something like hot DMSO or N-methyl-2-pyrrolidone may swell the polymer enough to separate the parts.


Wording of the Question...

The question asks: Which formula can be used to calculate the exact hydronium concentration present in sodium hydrogen carbonate solution?

What does "exact" mean?

The word exact gives me pause. In general there isn't usually an exact answer, but rather only an answer that is within the required experimental error. The values of $K_\mathrm{a1}$, $K_\mathrm{b1}$, and $K_\mathrm{b2}$ are given to three significant figures. The value for $K_\mathrm{a2}$ is only given to two significant figures, but

$$K_\mathrm{a2} = \dfrac{K_\mathrm{w}}{K_\mathrm{b2}} = \dfrac{1.00\times10^{-14}}{2.13\times10^{-4}} = 4.69483\times10^{-11}\tag{SF1}$$

So it seems reasonable to assume that $K_\mathrm{a2} = 4.70\times10^{-11}$ and that exact means to within 1 part per thousand (ppt).

Since the pH of pure water is taken to be 7.000, this means that for any pH between 4.000 and 10.000 the autoionization of water will have to be taken into account to get results within 1 ppt. (Remember that the characteristic of the pH doesn't count in significant figures, only the mantissa does.)

Realistically that is terribly precise. At 25 $^\circ \mathrm{C}$ $K_w \approx 13.995$ and the change is about 0.033 per $^\circ \mathrm{C}$. So a 1 ppt change is only a 0.03 $^\circ \mathrm{C}$ change in temperature. For such calculations I'd probably expect the typical error would be on the order of ± 5% of reality. But the problem poses 1 ppt precision and so that is how the calculations "should" be done. If the problem poser wanted less precision them the equilibrium constants should have been specified with fewer significant figures.

Setup

Assumptions

Ignoring relationship between $\ce{CO2}$ gas phase and dissolved $\ce{CO2}$ in the liquid phase. Mass balance would require relative volumes of gas and liquid phases.
Ignoring relationship between dissolved $\ce{CO2}$ and carbonic acid. Accept  that $K_{\mathrm{a1}}$ is the "combined" equilibrium constant which takes that relationship into account.
The equilibriums will be calculated using concentrations instead of the more appropriate activities. So above 1 millimolar concentration of sodium bicarbonate the calculated results are suspect.

Starting with the basic relationships.

$$K_{\mathrm{a1}} = \dfrac{\ce{[H+][HCO3^-]}}{\ce{[H2CO3]}}\tag{S1}$$

$$K_{\mathrm{a2}} = \dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^-]}}\tag{S2}$$

$$K_\mathrm{w} = \ce{[H^-][OH^-]}\tag{S3}$$

Letting $\mathrm{[C_T]}$ be the total concentration of all the various carbonate species (which we know from is nominal molarity of the sodium bicarbonate).

$$\mathrm{[C_T]} = \ce{[H2CO3] + [HCO3^-] + [CO3^{2-}]}\tag{S4}$$

But this leaves us with five unknowns ($\ce{[H+], [OH^-], [H2CO3], [HCO3^-]}$ and $\ce{[CO3^{2-}]}$) but only four equations (S1), (S2), (S3) and (S4). So another equation is needed relating the five unknowns to completely numerically solve the system.

Fraction of Species

The fraction of each species can be calculated for a particular pH value using equations (F1), (F2), and (F3). Basically you substitute values into (S4) as a function of $\ce{[H+]}$ and whatever species ($\ce{[H2CO3], [HCO3^-], [CO3^{2-}]}$) that you're interest in. The equations for the fractions of the species are exact.

$$\dfrac{\ce{[H_2CO_3]}}{\ce{[C_T]}} = \dfrac{\ce{[H^+]^2}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F1}$$

$$\dfrac{\ce{[HCO3^-]}}{\ce{[C_T]}} = \dfrac{\ce{K_\mathrm{a1}[H^+]}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F2}$$

$$\dfrac{\ce{[CO3^{2-}]}}{\ce{[C_T]}} = \dfrac{\ce{K_\mathrm{a1}K_\mathrm{a2}}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F3}$$

Think of dissolving $\ce{H2CO3}$ ($\ce{C_T}$) and then adding between 0 and 2 equivalents of $\ce{NaOH}$. The equations relate the three variables pH, $\ce{[C_T]}$ and the $x$ equivalents of $\ce{NaOH}$. Knowing any two fixes the third.

Another interesting fact is found by differentiating equation (F2) for $\ce{[H^+]}$ which shows that the maximum of $\ce{[HCO3^-]/[C_T]}$ (or any diprotic acid...) occurs when

$$\ce{[H^+] = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}}\tag{F4}$$

It is the pH as which the bicarbonate solution has its maximum buffering capacity. Also at that pH $\ce{[H2CO3] = [CO3^{2-}]}$.

For the bicarbonate system the maximum concentration of $\ce{HCO3^-}$ is 97.99% which occurs at a pH of 8.340. At that pH $\ce{[H2CO3] = [CO3^{2-}] = 1.01\%}$. So for an accuracy of 1 ppt the hydrolysis of water and the multiple species cannot be neglected.

If the pH is above 7.648, then the concentration of the bicarbonate species will be at least 95%.

\begin{array}{|c|c|c|c|c|}\hline
\ce{[H+]} & \mathrm{pH} & \ce{F(H2CO3)} & \ce{F(HCO3^-)}&\ce{F(CO3^{2-})}\\ \hline
1.00E+00 & 0.000E+00 & 1.000E+00 & 4.450E-07 & 2.091E-17 \\ \hline
8.00E-01 & 9.691E-02 & 1.000E+00 & 5.562E-07 & 3.268E-17 \\ \hline
6.00E-01 & 2.218E-01 & 1.000E+00 & 7.417E-07 & 5.810E-17 \\ \hline
4.00E-01 & 3.979E-01 & 1.000E+00 & 1.112E-06 & 1.307E-16 \\ \hline
2.00E-01 & 6.990E-01 & 1.000E+00 & 2.225E-06 & 5.229E-16 \\ \hline
1.00E-01 & 1.000E+00 & 1.000E+00 & 4.450E-06 & 2.091E-15 \\ \hline
8.00E-02 & 1.097E+00 & 1.000E+00 & 5.562E-06 & 3.268E-15 \\ \hline
6.00E-02 & 1.222E+00 & 1.000E+00 & 7.417E-06 & 5.810E-15 \\ \hline
4.00E-02 & 1.398E+00 & 1.000E+00 & 1.112E-05 & 1.307E-14 \\ \hline
2.00E-02 & 1.699E+00 & 1.000E+00 & 2.225E-05 & 5.229E-14 \\ \hline
1.00E-02 & 2.000E+00 & 1.000E+00 & 4.450E-05 & 2.091E-13 \\ \hline
8.00E-03 & 2.097E+00 & 9.999E-01 & 5.562E-05 & 3.268E-13 \\ \hline
6.00E-03 & 2.222E+00 & 9.999E-01 & 7.416E-05 & 5.809E-13 \\ \hline
4.00E-03 & 2.398E+00 & 9.999E-01 & 1.112E-04 & 1.307E-12 \\ \hline
2.00E-03 & 2.699E+00 & 9.998E-01 & 2.225E-04 & 5.228E-12 \\ \hline
1.00E-03 & 3.000E+00 & 9.996E-01 & 4.448E-04 & 2.091E-11 \\ \hline
8.00E-04 & 3.097E+00 & 9.994E-01 & 5.559E-04 & 3.266E-11 \\ \hline
6.00E-04 & 3.222E+00 & 9.993E-01 & 7.411E-04 & 5.805E-11 \\ \hline
4.00E-04 & 3.398E+00 & 9.989E-01 & 1.111E-03 & 1.306E-10 \\ \hline
2.00E-04 & 3.699E+00 & 9.978E-01 & 2.220E-03 & 5.217E-10 \\ \hline
1.00E-04 & 4.000E+00 & 9.956E-01 & 4.430E-03 & 2.082E-09 \\ \hline
8.00E-05 & 4.097E+00 & 9.945E-01 & 5.532E-03 & 3.250E-09 \\ \hline
6.00E-05 & 4.222E+00 & 9.926E-01 & 7.362E-03 & 5.767E-09 \\ \hline
4.00E-05 & 4.398E+00 & 9.890E-01 & 1.100E-02 & 1.293E-08 \\ \hline
2.00E-05 & 4.699E+00 & 9.782E-01 & 2.177E-02 & 5.115E-08 \\ \hline
1.00E-05 & 5.000E+00 & 9.574E-01 & 4.260E-02 & 2.002E-07 \\ \hline
8.00E-06 & 5.097E+00 & 9.473E-01 & 5.269E-02 & 3.096E-07 \\ \hline
6.00E-06 & 5.222E+00 & 9.310E-01 & 6.905E-02 & 5.409E-07 \\ \hline
4.00E-06 & 5.398E+00 & 8.999E-01 & 1.001E-01 & 1.176E-06 \\ \hline
2.00E-06 & 5.699E+00 & 8.180E-01 & 1.820E-01 & 4.277E-06 \\ \hline
1.00E-06 & 6.000E+00 & 6.920E-01 & 3.080E-01 & 1.447E-05 \\ \hline
8.00E-07 & 6.097E+00 & 6.426E-01 & 3.574E-01 & 2.100E-05 \\ \hline
6.00E-07 & 6.222E+00 & 5.741E-01 & 4.258E-01 & 3.336E-05 \\ \hline
4.00E-07 & 6.398E+00 & 4.733E-01 & 5.266E-01 & 6.187E-05 \\ \hline
2.00E-07 & 6.699E+00 & 3.100E-01 & 6.898E-01 & 1.621E-04 \\ \hline
1.00E-07 & 7.000E+00 & 1.834E-01 & 8.162E-01 & 3.836E-04 \\ \hline
8.00E-08 & 7.097E+00 & 1.523E-01 & 8.472E-01 & 4.977E-04 \\ \hline
6.00E-08 & 7.222E+00 & 1.187E-01 & 8.806E-01 & 6.898E-04 \\ \hline
4.00E-08 & 7.398E+00 & 8.239E-02 & 9.165E-01 & 1.077E-03 \\ \hline
2.25E-08 & 7.648E+00 & 4.803E-02 & 9.500E-01 & 1.984E-03 \\ \hline
2.00E-08 & 7.699E+00 & 4.291E-02 & 9.548E-01 & 2.244E-03 \\ \hline
1.00E-08 & 8.000E+00 & 2.188E-02 & 9.735E-01 & 4.576E-03 \\ \hline
8.00E-09 & 8.097E+00 & 1.756E-02 & 9.767E-01 & 5.738E-03 \\ \hline
6.00E-09 & 8.222E+00 & 1.320E-02 & 9.791E-01 & 7.670E-03 \\ \hline
4.57E-09 & 8.340E+00 & 1.006E-02 & 9.799E-01 & 1.008E-02 \\ \hline
4.00E-09 & 8.398E+00 & 8.806E-03 & 9.797E-01 & 1.151E-02 \\ \hline
2.00E-09 & 8.699E+00 & 4.372E-03 & 9.728E-01 & 2.286E-02 \\ \hline
1.00E-09 & 9.000E+00 & 2.142E-03 & 9.531E-01 & 4.479E-02 \\ \hline
8.00E-10 & 9.097E+00 & 1.695E-03 & 9.429E-01 & 5.540E-02 \\ \hline
6.00E-10 & 9.222E+00 & 1.249E-03 & 9.262E-01 & 7.255E-02 \\ \hline
4.00E-10 & 9.398E+00 & 8.037E-04 & 8.941E-01 & 1.051E-01 \\ \hline
2.00E-10 & 9.699E+00 & 3.638E-04 & 8.094E-01 & 1.902E-01 \\ \hline
1.00E-10 & 1.000E+01 & 1.528E-04 & 6.802E-01 & 3.197E-01 \\ \hline
8.00E-11 & 1.010E+01 & 1.132E-04 & 6.298E-01 & 3.700E-01 \\ \hline
6.00E-11 & 1.022E+01 & 7.560E-05 & 5.607E-01 & 4.392E-01 \\ \hline
4.00E-11 & 1.040E+01 & 4.133E-05 & 4.598E-01 & 5.402E-01 \\ \hline
2.00E-11 & 1.070E+01 & 1.342E-05 & 2.985E-01 & 7.015E-01 \\ \hline
1.00E-11 & 1.100E+01 & 3.942E-06 & 1.754E-01 & 8.246E-01 \\ \hline
\end{array}

The fourth equation

The fourth equation comes from the charge balance in the solution. The solution must be electrically neutral so the total charge on cations and anions must be equal.

$$\ce{[Na^+] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C1}$$

But in a solution of $\ce{[NaHCO3]}$ the $\ce{[Na^+] = [C_T]}$ so:

$$\ce{[C_T] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C2}$$

Which yields and exact solution as a quartic equation. But rather than solve only that particular quartic equation, let's consider the more general case. Assume we start with $\ce{H2CO3}$ in water and add amounts of $\ce{NaOH} according to the equation:

$$\ce{H2CO3 + xNaOH -> Na_xH_{2-x}CO3 + xH2O}\tag{C3}$$

where $x$ is a real number bounded by $ 0 \le x \le 2$. The special cases of $x = 0,1,2$ correspond to adding $\ce{H2CO3, NaHCO3}$ or $\ce{Na2CO3}$ to pure water. Then it is intuitively obvious to the casual observer that (a la analytical chemist Dr. Peter Carr, University of Ga.) the general equation is:

$$\begin{multline}
  [\ce{H+}]^4
  + (K_\mathrm{a1} + xC_T)[\ce{H+}]^3
  + (K_\mathrm{a1} K_\mathrm{a2} + (x - 1) C_T K_\mathrm{a1}
    - K_\mathrm{w})[\ce{H+}]^2\\
  + K_\mathrm{a1} ((x - 2)C_T K_\mathrm{a2} - K_\mathrm{w})[\ce{H+}]
  - K_\mathrm{a1} K_\mathrm{a2} K_\mathrm{w}
  \tag{C4}
\end{multline}$$

It is not hard for a computer to solve the quartic equation numerically, but to do so by hand is extremely tedious. Hence books never(?) required such a solution in problem sets.

DERIVATION OF EQUATION 1 $\ce{[H^+]} = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}$

$$K_\mathrm{a1}K_\mathrm{a2} = \dfrac{\ce{[H+][HCO3^-]}}{\ce{[H2CO3]}}\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^-]}} = \dfrac{\ce{[H+]^2[CO3^{2-}]}}{\ce{[H2CO3]}}\tag{EQ1-1}$$

rearrange (EQ1-1)

$$\ce{[H+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[H2CO3]}}{\ce{[CO3^{2-}]}}}\tag{EQ1-2}$$

assume $\ce{[H2CO3] = [CO3^{2-}]}$

Then

$$\ce{[H+]} = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}\tag{EQ1-3}$$

DERIVATION OF EQUATION 2 $\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T]}}}$

Sodium is a spectator ion. Thus it is be possible to model the system as a monoprotic Brønsted–Lowry acid which has $\ce{H^+}$ as the acid and $\ce{HCO3^-}$ as the conjugate base.

This derivation has two problems. First it ignores the autoionization of water. Second the equation models the system as a monoprotic base, but the carbonate system has three species - carbonic acid, bicarbonate, and carbonate, and all three are likely to be present in significant quantities in alkaline solutions.

Start with the reaction of the conjugate base

$$\ce{HCO3^- + H2O <=> H2CO3 + OH^-}\tag{EQ2-1}$$

and the equilibrium expression

$$\ce{K_\mathrm{b1}} = \dfrac{\ce{[H2CO3][OH^-]}}{\ce{[HCO3^-]}}\tag{EQ2-2}$$

Assume that little of $\ce{HCO3^-}$ accepts a proton, thus $\ce{[HCO3^-] = [C_T]}$ and $\ce{[H2CO3] = [OH^-]}$. Now substituting:

$$\ce{K_\mathrm{b1}} = \dfrac{\ce{[OH^-]^2}}{\ce{[C_T]}}\tag{EQ2-3}$$

but $\ce{K_\mathrm{b1}} = K_w / K_{a1}$ and $\ce{[OH^-] = K_w / [H^+]}$. Now making more substitutions

$$\dfrac{K_\mathrm{w}}{K_\mathrm{a1}} = \dfrac{(\frac{K_\mathrm{w}}{\ce{[H^+]}})^2}{[C_T]}\tag{EQ2-4}$$

rearranging and solving for $\ce{[H^+]}$ gives the desired result.

$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T]}}}\tag{EQ2-5}$$

The desired result could have been obtained more expeditiously by starting with equation (2) and letting $\ce{[H2CO3] = [OH^-] = K_w/[H^+]}$, but that route doesn't seem as clear chemically and seems more like math magic.

DERIVATION OF EQUATION 3 $\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}C_T + K_{a1}K_w}{C_T+K_{a1}}}$

Starting with equation (C2) for the charge balance of a solution of $\ce{NaHCO3}$ where $\ce{[C_T] = [Na^+]}$

$$\ce{[C_T] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C2}$$

subtract equation (S4) which is for mass balance of all the carbonate species

$$\mathrm{[C_T]} = \ce{[H2CO3] + [HCO3^-] + [CO3^{2-}]}\tag{S4}$$

yielding

$$\ce{[H^+] = [CO3^{2-}] - [H2CO3] + [OH^-]}\tag{EQ3-1}$$

rearranging

$$\ce{[OH^-] = [H2CO3] - [CO3^{2-}] + [H^+]}\tag{EQ3-2}$$

substituting $K_\mathrm{w} / \ce{[H^+] = [OH^-]}$ and rearrange

$$\dfrac{K_\mathrm{w}}{\ce{[H^+]}} \ce{+ [CO3^{2-}] = [H2CO3] + [H^+]}\tag{EQ3-3}$$

Solve equation (S1) for $\ce{[H2CO3]}$ and solve equation (S2) for $\ce{[CO3^{2-}]}$, and then substitute those results into equation (EQ3-3)

$$\dfrac{K_\mathrm{w}}{\ce{[H^+]}} + \dfrac{K_\mathrm{a2}\ce{[HCO3^-]}}{\ce{[H^+]}} = \dfrac{\ce{[H^+][HCO3^-]}}{K_\mathrm{a1}} + \ce{[H^+]}
\tag{EQ3-4}$$

Multiplying both sides by $\ce{[H^+]}$, then $K_\mathrm{a1}$, and finally collecting terms

$$K_\mathrm{a1}K_\mathrm{w} + K_\mathrm{a1}K_\mathrm{a2}\ce{[HCO3^-]} = \ce{[H^+]^2([HCO3^-] + K_\mathrm{a1})}
\tag{EQ3-5}$$

rearranging and solving for $\ce{[H^+]}$ yields the exact equation for $\ce{[H^+]}$ in terms of the particular species $\ce{[HCO3^-]}$ when $\ce{NaHCO3}$ is dissolved in pure water.

$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[HCO3^-]} + \mathrm{K_{a1}K_w}}{\ce{[HCO3^-] + \mathrm{K_{a1}}}}}\tag{EQ3-6}$$

As indicated previously the above equation is exact in terms of the relationship between $\ce{[H^+]}$ and the carbonate species $\ce{HCO3^-}$. Over the whole pH range $\ce{[C_T] \ne [HCO3^-]}$, however for the range $ 7.648 \lt \mathrm{pH} \lt 9.00$ the species $\ce{HCO3^-}$ is at least 95% of $\ce{[C_T]}$. So we can "reasonably" substitute $\ce{[C_T] = [NaHCO3]}$ and get the desired equation even though the equation is not exact.

$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[C_T]} + K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T] + K_\mathrm{a1}}}}\tag{EQ3-7}$$

!! CHECKS !!

Why aren't students taught to do checks any more!?!

The column $\ce{[C_T]}$ is the nominal molarity of the sodium bicarbonate solution. It is also the molarity of the spectator cation $\ce{[Na^+]}$.

The column $\Sigma \space \text{cations}$ should equal the column $\Sigma \space \text{anions}$.

The column % Diff is a measure of the inconsistency of the calculations of the concentration of anions and cations and is calculated by

$$\text{% Diff} = 200*\dfrac{\Sigma\text{ cations} - \Sigma\text{ anions}}{\Sigma\text{ cations} + \Sigma\text{ anions}}$$

Again, these calculations use concentrations instead of the more appropriate activities. Concentrations are probably fine for these solutions below 1 millimolar. Above that and activities should really be used.

EQUATION 1 $\ce{[H^+]} = \sqrt{K_{a1}K_{a2}}$

\begin{array}{|c|c|c|c|c|c|c|}\hline
\ce{[C_T]} & calc \space [H^+] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff\\ \hline
1.00E-01 & 4.5733E-09 &  8.340 &  1.0000E-01&  2.1866E-06 &9.7986E-02 &2.0140E-03 &1.0000E-01 & 0.0 \\ \hline
1.00E-02 &4.5733E-09 &  8.340 &1.0000E-02 &2.1866E-06 &9.7986E-03 &2.0140E-04 &1.0002E-02 & 0.0 \\ \hline
1.00E-03 &4.5733E-09 &  8.340 &1.0000E-03 &2.1866E-06 &9.7986E-04 &2.0140E-05 &1.0022E-03 & -0.2 \\ \hline
1.00E-04 &4.5733E-09 &  8.340 &1.0000E-04 &2.1866E-06 &9.7986E-05 &2.0140E-06 &1.0219E-04 & -2.2\\ \hline
1.00E-05 &4.5733E-09 &  8.340 &1.0005E-05 &2.1866E-06 &9.7986E-06 &2.0140E-07 &1.2187E-05 & -19.7\\ \hline
1.00E-06 &4.5733E-09 &  8.340 &1.0046E-06 &2.1866E-06 &9.7986E-07 &2.0140E-08 &3.1866E-06 & -104 \\ \hline
1.00E-07 &4.5733E-09 &  8.340 &1.0457E-07 &2.1866E-06 &9.7986E-08 &2.0140E-09 &2.2866E-06 & -183 \\ \hline
\end{array}

So assuming that the pH is constant only works for concentrated solutions (greater than  1 millimolar) for which activities should really be used...

EQUATION 2 $\ce{[H^+]} = \sqrt{\dfrac{K_{a1}K_w}{\ce{[C_T]}}}$

\begin{array}{|c|c|c|c|c|c|c|}\hline
\ce{[C_T]} & calc \space [H^+] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff\\ \hline
1.00E-01 &2.1095E-10 & 9.676 &1.0000E-01 &4.7405E-05 &8.1748E-02 &3.6427E-02 &1.1822E-01 & -16.7 \\ \hline
1.00E-02 &6.6708E-10 & 9.176 &1.0000E-02 &1.4991E-05 &9.3287E-03 &1.3145E-03 &1.0658E-02 & -6.4\\ \hline
1.00E-03 &2.1095E-09 & 8.676 &1.0000E-03 &4.7405E-06 &9.7369E-04 &4.3388E-05 &1.0218E-03 & -2.2\\ \hline
1.00E-04 &6.6708E-09 & 8.176 &1.0001E-04 &1.4991E-06 &9.7844E-05 &1.3787E-06 &1.0072E-04 & -0.7 \\ \hline
1.00E-05 &2.1095E-08 & 7.676 &1.0021E-05 &4.7405E-07 &9.5271E-06 &4.2453E-08 &1.0044E-05 & -0.2 \\ \hline
1.00E-06 &6.6708E-08 & 7.176 &1.0667E-06 &1.4991E-07 &8.6910E-07 &1.2247E-09 &1.0202E-06 & +4.5\\ \hline
1.00E-07 &2.1095E-07 & 6.676 &3.1095E-07 &4.7405E-08 &6.7830E-08 &3.0225E-11 &1.1527E-07 & +91.8 \\ \hline
\end{array}

If 1 ppt is required, then Equation 2 doesn't work. For a +/- 5% error, the equation works in the range from $1\times10^{-2}$ to $1\times10^{-6}$ molar.

EQUATION 3 $\ce{[H^+]} = \sqrt{\dfrac{K_{a1}K_{a2}C_T + K_{a1}K_w}{C_T+K_{a1}}}$

\begin{array}{|c|c|c|c|c|c|c|}\hline
\ce{[C_T]} & calc \space [H] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff \\ \hline
1.00E-01 &4.5781E-09 & 8.339 &1.0000E-01 &2.1843E-06 &9.7986E-02 &2.0119E-03 &1.0000E-01 & 0.0 \\ \hline
1.00E-02 &4.6216E-09 & 8.335 &1.0000E-02 &2.1638E-06 &9.7986E-03 &1.9930E-04 &1.0000E-02 & 0.0 \\ \hline
1.00E-03 &5.0352E-09 & 8.298 &1.0000E-03 &1.9860E-06 &9.7977E-04 &1.8291E-05 &1.0000E-03 & 0.0 \\ \hline
1.00E-04 &8.0700E-09 & 8.093 &1.0001E-04 &1.2392E-06 &9.7660E-05 &1.1376E-06 &1.0004E-04 & 0.0 \\ \hline
1.00E-05 &2.1120E-08 & 7.675 &1.0021E-05 &4.7348E-07 &9.5267E-06 &4.2400E-08 &1.0043E-05 & -0.2\\ \hline
1.00E-06 &5.5624E-08 & 7.255 &1.0556E-06 &1.7978E-07 &8.8822E-07 &1.5010E-09 &1.0695E-06 & -1.3\\ \hline
1.00E-07 &9.0382E-08 & 7.044 &1.9038E-07 &1.1064E-07 &8.3082E-08 &8.6408E-11 &1.9381E-07 & -1.8 \\ \hline
1.00E-08 &9.8897E-08 & 7.005 &1.0890E-07 &1.0111E-07 &8.1785E-09 &7.7735E-12 &1.0930E-07 & -0.4\\ \hline
\end{array}

If 1 ppt is required, then Equation 3 works for concentrations grater than  $1\times10^{-5}$ molar. For a +/- 5% error, the equation works in the range from $0.1$ to $1\times10^{-8}$ molar.

Summary of $\ce{[H^+]}$ values

the table below is a summary of the results for the three derived equations, and the exact solution using equation (C4) by Wolfram Alpha.

\begin{array}{|c|c|c|c|c|}\hline
\ce{[C_T]} & Eq(1)& Eq(2) & Eq(3) & exact\\ \hline
1.00E-01  & 4.5733E-09 & 2.1095E-10 & 4.5781E-09 & 4.5782E-9 \\ \hline
1.00E-02  & 4.5733E-09 & 6.6708E-10 & 4.6216E-09 & 4.6226E-9 \\ \hline
1.00E-03  & 4.5733E-09 & 2.1095E-09 & 5.0352E-09 & 5.0443E-9 \\ \hline
1.00E-04  & 4.5733E-09 & 6.6708E-09 & 8.0700E-09 & 8.1354E-9 \\ \hline
1.00E-05  & 4.5733E-09 & 2.1095E-08 & 2.1120E-08 & 2.1602E-8 \\ \hline
1.00E-06  & 4.5733E-09 & 6.6708E-08  & 5.5624E-08 & 5.7983E-8 \\ \hline
1.00E-07  & 4.5733E-09 & 2.1095E-07  & 9.0382E-08 & 9.1835E-8 \\ \hline
1.00E-08  & -- & -- & 9.8897E-08 & 9.9096E-8 \\ \hline
\end{array}


Bond stability of C-C in ethane, ethene, and ethyne

The truth is: in reality, sigma bonds, bond orders and hybridization do not exist (not even bonds or orbitals do). They are just concepts to explain the reactivity, stability and geometry of a molecule (and in the case of orbitals to build approximate many-particle wave functions). There is thus no way to compare a sigma-bond strength without choosing some model which introduces bias.
The least-biased way to get there is probably starting from the experimental dissociation energy already mentioned. These can be divided by the bond order (which, again, is not a thing, but quite well established):

ethane: 350 kJ/mol -> 350 kJ/mol per bond
ethene: 610 kJ/mol -> 305 kJ/mol per bond
ethyne: 835 kJ/mol -> 278 kJ/mol per bond

In general: for any chosen approach, the sigma bond in ethane will be the strongest even though the bond distance of ethyne is the smallest. The additional electrons of the pi bond(s) destabilize the bond.


Compare reactivity of carbocations and stability of carbanions, radicals, and carbocations

Order of stability: 3 > 2 > 1 > 4
I would say that the third structure is the most stable (least reactive) as an octet-complete resonance structure of it exists which is much more stable that the initial carbocation. This is basically what you said, the carbocation is stabilized due to the electron donation tendency of the nitrogen atom in this case.

Then structure 2 then 1 since 2 is tertiary while 1 is secondary. More the alkyl groups, more the positive inductive effect stabilizing the carbocation.
I would say that the fourth structure is the least stable. If you try to draw the resonance structure in which oxygen ends up donating electrons stabilizing the carbocation, you'll notice that the oxygen ends up with a positive charge which is even more unstable than the initial carbocation. Resonance does not help here.

(the structure above should have a positive charge on the oxygen; I wasn't able to draw that using the software available)
Do note that resonance structures don't actually exist and there are better ways to explain more or less the same thing. I've stuck with resonance since you've mentioned that in the question.


Mixing a Tagged Sample of Lead Iodide with another Iodide Salt

You are missing a basic aspect of how a precipitate in solution behaves. The precipitate isn't just statically immersed in the solution but rather some is dissolving and some of the ions (or molecules) in solution are precipitating out. So for a salt like $\ce{PbI2}$ there will be an exchange of ions which are in the precipitate with those which are in solution.


The only case I know of where activated carbon causes reduction at room temperature would be filtration of ozone, $\ce{O3}$. Even fluorine will not oxidize plain carbon:

Carbon is stable in a fluorine atmosphere up to about $\pu{400 ^\circ C}$ .

Oh, well perhaps dioxygen difluoride, $\ce{FOOF}$, could also oxidize carbon (or could be reduced by carbon, should you prefer a passive voice). Read Derek Lowe's Things I Won’t Work With: Dioxygen Difluoride, if you'd care to test that. Please let me know so that I could arrange to  be elsewhere.


Calculate ion concentrations in mixed buffers

You can ignore K+, Na+ and Cl-. You have to deal with H2O, H3O+, OH-, Ap3-, HAp2-, H2Ap-, H3Ap, Ac3-, HAc2-, H2Ac-, H3Ac. Here H3Ap = H3PO4 and H3Ac = citric acid in neutral state. You have 11 unknowns.

(+1)    [$\Delta$H+] = 3*[citric3-]+2*[Hcitric2-]+1*[H2citric-]+[(PO4)3-] - [HPO4-]-2*[H3PO4]
(+1) [H(from_water)+ + $\Delta$H+][OH(from_water)-]=10^-14 (dissociation of H2O)
(+1) [H(from_water)+]  = [OH(from_water)-]
(+3) Three dissociation constants of H3PO4 are known
(+3) Three dissociation of citric acid are known
(+1) Concentration of [Ac3- + HAc2- + H2Ac- + H3Ac] is known
(+1) Concentration of [Ap3- + HAp2- + H2Ap- + H3Ap]is known

You have 11 equations.

OH(from_water)- corresponds to total OH-
H3O+ corresponds to H(from_water)+ + $\Delta$H+]

//(+1) Water concentration is assumed 1 in solutions

Forget about (PO4)3- and H3PO4. They will have negligible impact on overall pH.

You didn't tell the ratios of Na2HPO4 and citric acid and these ratios are critical. Given pKa of citric (3.15, 4.77, 5.19) and phosphoric (2.15, 7.20, 12.35) acids you will most likely deal with either of these regions:
2.15-3.15 H3Citric + H2Phosphoric-
3.15-4.77 H3Citric+H2Citric-
4.77-5.19 H2Citric- - HCitric2-
5.19-7.20 H2PO4- - (HPO4)2-

If you know which interval you are working with (check ratio of the reagents, aka stoichiometry) then you can pick the relevant equilibrium and set all other concentrations to zero for initial evaluation. 10^-15 M of ions will have zero impact on overall pH.

I regrouped some equations and expanded the one you asked about.

Can you solve it for a general case. Yes, you can. But it is like accounting for air friction, Special and General Theories of Relativity and quantum effects. It is harder, but it does cover more cases. In this case you will be dealing with x^11+a*x^10+...=0 instead of a square equation. You can solve it using Wolframalpha. But at the end of a day you will find that in a given region the impact of majority of members is negligible. Think of it this way. You can approximate the system with a spline (5 polinomials of second degree) or one polinomial of 11th degree.


How can a substance be diluted to nothing?

You correctly point out that the number of molecules in a solution is finite and constant, however the volumetric concentration (that is, how many molecules per litre) changes upon dilution. If, for instance, you take one liter of a 1 mol/L solution of ethanol in water ($\approx{6.02\times{10}^{23}}$ ethanol molecules per liter) and add 9 liters of distilled water, you now have a 1 mole of ethanol in 10 liters, giving a concentration of 0.1 mol/L. If you were to draw out 1 liter of this and count the number of ethanol molecules in that liter, you would find approximately $\approx{6.02\times{10}^{22}}$ - a tenfold reduction. The rest of the ethanol is retained in the other 9 liters of solution.

Equivalently, if you were to measure out 100 mL of 1 mol/L ethanol solution and then make it up to 1 liter by adding distilled water, you would also get a 0.1 mol/L solution. This is the kind of thing taken to absurd extremes in homoeopathy, where tiny quantities of solution are repeatedly diluted, with some shaking. Obviously, they don't literally add billions of liters of water to their solutions, but rather pipette out a tiny amount of solution and discard the remainder, adding water to make up the desired volume. This process is repeated many times.

The finite number of molecules present in a solution leads to the concept of the dilution limit, which is where you have carried out sufficient dilutions that statistically speaking, your chances of finding even a single molecule of the dissolved ingredient become vanishingly small and that by extrapolation, you would have to drink approximately a zillion liters of the preparation to get a single molecule of it.

Where did all those other molecules go? They didn't simply disappear - they were poured down the drain during the dilution process, which curiously is the same place money goes when spent on homoeopathy.

Bonus: homoeopathic plutonium! (courtesy of Theodore Gray's periodic table)


Standard state of a substance - definition

The IUPAC definition of standard state is:

  State of a system chosen as standard for reference by convention. Three standard states are recognized: For a gas phase it is the (hypothetical) state of the pure substance in the gaseous phase at the standard pressure p = p°, assuming ideal behaviour. For a pure phase, or a mixture, or a solvent in the liquid or solid state it is the state of the pure substance in the liquid or solid phase at the standard pressure p = p°. For a solute in solution it is the (hypothetical) state of solute at the standard molality m°, standard pressure p° or standard concentration c° and exhibiting infinitely dilute solution behaviour. For a pure substance the concept of standard state applies to the substance in a well defined state of aggregation at a well defined but arbitrarily chosen standard pressure.

Specifying that the standard state is a specific temperature and pressure, would mean that the standard state is not even remotely realizable for many gases.  For example, how could you have pure water vapor at standard temperature and pressure?


EZ-water is not a breakthrough. It is not new, nor is it valid. This appears to be one of the many claims about the healthful benefits of drinking "ionized water".
The original "article" linked, contains a number of misleading statements.

Water molecules make up 99% of your body. This is true, but misleading. According to various sources, the human body is between 57% and 75% water by mass. Since water has a much lower molar mass than most biomolecules, 99% of the molecules in your body may well be water.
The journal listed in which the research is published is Water, which is not a high impact journal, despite its name. Water had an impact factor of 0.973 in 2012. That means that articles published in 2011 in Water were cited less than 1 time each by anyone in 2012. Compare with Cell, which had in impact factor of 31.957 in the same period - each article in Cell is cited an average of nearly 32 times.
Water has a "fourth phase" and no single boiling point or melting point. This is true! Any student of chemistry should be able to tell you this. Take a look at the phase diagram of water below. Water apparently has at least 18 phases (15 solid phases, plus liquid, vapor, and supercritical fluid). You can also see that the solid-liquid and liquid-vapor phase boundaries are not single points. They are curves varying in temperature and pressure. Those curves represent the conditions in which the two phases have the same vapor pressure. This is not a new discovery or a startling secret. This behavior has been known for many decades (if not centuries).

Later on the "article" tells us that EZ-water is not $\ce{H2O}$ (note that they cannot be bothered to format their chemical formulas) but $\ce{H3O2}$, which is alkaline with a negative charge. Hmmm... If it has a different formula and different properties, it must be a different compound. Note that this "formula" is essentially $\ce{H3O2 = H2O + OH}$, and hints at what happens inside these water ionizers.
EZ-water has a negative charge. It is possible for objects to build up charge, but it is not possible for them to remain permanently charged. Overall, the universe is balanced in charge. Consider lightning as a consequence of attempting to maintain separation of charge. This "negative charge" comes from a misunderstanding of pH. The water definitely has a higher pH, which means more $\ce{OH-}$ ions than $\ce{H+}$ ions, but something (probably $\ce{Na+}$) needs to replace the missing $\ce{H+}$.
I can go on, but you get the point.

A water ionizer electrolyzes water, which requires electricity (not light as suggested by Mercola). I had the good fortune to be involved in the testing and dismantling of one of these ionizers recently. The device claimed to take ordinary tap water and convert it to a "super-cleaning solution that contained only negatively charged oxygen particles" or something like that. From experiments:

The ionizer increases the pH of the water.
The ionizer does not work with deionized or distilled water!
The ionizer does not produce measurable oxygen gas (a smoldering splint will not relight in the vicinity)
The device contains an electrolysis cell and an ion exchange resin

Here is what happens:

Water is electrolyzed.
Doing so requires ions in the water already to conduct the electricity, which is why the device must use tap water (and not purified water). At the cathode, water is reduced to hydrogen and hydroxide. At the anode, it is oxidized to oxygen and protons.
\begin{align}
\ce{2H2O +2e- &-> H2 + 2OH-}\\
\ce{2H2O &-> O2 +4H+ + 4e-}
\end{align}

Protons are exchanged for sodium ions
To prevent the recombination of hydroxide and protons to form water and neutralize the solution, there is an ion exchange membrane between the anode and the rest of the solution.
$$\ce{H+ + OH- -> H2O}$$
The ion exchange resin collects the protons and releases sodium ions into solution to maintain charge neutrality, which must happen or you would have a lightning gun and not a mystic device to produce super water.

Ultimately, drinking this stuff would probably have no effect on your health. But we are no doctors, so don’t listen to us either. It might help or it might hurt, depending on your condition, it also might not matter.
First, your stomach secretes acid, so any extra base would be neutralized immediately. Let's say an ionizer gets the water's pH to 8 (slightly basic, and more basic than your blood). This means that the concentration of protons is $\ce{[H+]}=10^{-\text{M}pH}=10^{-8} \text{ M}$ (remember that pH is a logarithmic scale. The concentration of hydroxide is $10^{-14} / 10^{-8} = 10^{-6} \text{ M}$. The pH in your stomach ranges from 1.5 - 3.5, which represent proton concentrations between $10^{-3.5}= 3.16\times 10^{-4}$ and $10^{-1.5}=3.16\times 10^{-2}$. If we use the higher pH stomach acid to neutralize an equal volume of EZ-water, the final pH would be:
\begin{align}
[\ce{H+}]_f &= \frac{3.16\times 10^{-4} - 1\times 10^{-6}}{2}=\frac{3.15\times 10^{-4}}{2}=1.07_5 \times 10^{-4}\\
\text{pH} &=-\log[\ce{H+}] =-\log(1.07_5 \times 10^{-4})=3.97
\end{align}
Since your stomach acid would likely have a greater volume than the ionized water, the change in pH would be even smaller. This calculation assumes your stomach acid is not a buffer, which it is.
Second, the acid-base homeostasis of your blood prevents the pH of your body from changing. Yes, your blood pH is 7.4 or so (which is slightly basic), but that does not mean your blood is negatively charged. Outside of range of 7.38 - 7.42 all of your blood proteins (like hemoglobin) start to denature, which will kill you.
Well, to be more precise, you will suffocate before that happens as hemoglobin/myoglobin oxygen binding capacity is pH sensitive. And if anything, we can tolerate acidic blood more robustly than basic blood pH (down to 6.8 or 7, but only up to about 7.6) before death. You might want to look up acidosis.


Placing oxygen molecule into electric field

Yes, an oxygen molecule (much like any other molecule) can be broken by electric field alone.
No, it won't be broken into separate oxygen atoms, precisely because the atoms are neutral and the bond between them non-polar, like you said. Instead, it would be broken into $\ce{O2+}$ and an electron.
This is not important anyway, since it would require electric field much stronger than those present in air ionizers.

What happens in air under voltage, then? Well, air always contains some ions from whatever sources (cosmic rays, background radiation, etc). As you apply voltage, your positive and negative ions start speeding up in the opposite directions. Once the voltage is high enough, they accelerate to the point that they might break neutral molecules by bumping into them. In doing so, they produce more ions, and those do the same, so pretty soon there is a breakdown going on. You see a blinding spark, the fuses blow up, the voltage shuts off, and the ions start to recombine into molecules. As luck would have it, sometimes they combine into $\ce{O3}$ - that's an important source of ozone both in nature and in the lab.

In fact, the situation is much more complicated. Gas discharge does not have to be that dramatic. It comes in various forms, some of them pretty controllable.


Does aminosugars give positive response in alkaloid tests?

Most alkaloids are tertiary amines. Almost all tests for alkaloids, reagents of which are  are solutions of the salts of heavy metals, are precipitation tests. For example, all of tests with Dragendorff's reagent, Mayer's reagent, Hager's reagent, and Wagner's reagent give color precipitates: The heavy metal atom in the reagent with the nitrogen in the alkaloid would form ion pairs. These ion pairs form an insoluble precipitate. The color of the precipitate depends on reagent, plant species, etc. (Ref.1). Most amino sugars are either primary amines or acetamides, thus, I doubt they'd give precipitates with sought reagents. However, some reagents give false-positive tests for non-nitrogenous extracts indicating the presence of alkaloids. The best example is Dragendorff's spray reagent (Ref.2).

Mean time, some reagents are giving negative results even with alkaloid extracts, e.g., Mayer's reagent (Ref.3). Therefore, my answer is purely opinion based.

References:

Felix G. Coe, Gregory J. Anderson, "Screening of medicinal plants used by the Garífuna of Eastern Nicaragua for bioactive compounds," Journal of Ethnopharmacology 1996, 53(1), 29-50 (https://doi.org/10.1016/0378-8741(96)01424-9).
Abdel‐Azim M. Habib, "False‐positive alkaloid reactions," Journal of Pharmaceutical Sciences 1980, 69(1), 37-43 (https://doi.org/10.1002/jps.2600690111).
J. W. Field, M. Kandiah, "A note on the use of mayer's reagent for the detection of quinine in alkaline urine," Transactions of the Royal Society of Tropical Medicine and Hygiene 1935, 28(4), 385-390 (https://doi.org/10.1016/S0035-9203(35)90133-X).


How fast does MgSO4 absorb water from the air?

This is a great question, and unfortunately, I don't have an answer for you.

My first instinct would be to question how well you heated the material and under what conditions. Insufficient heating would mean that you might still have water remaining in the material.

Anhydrous, powdered magnesium sulfate will absorb water relatively quickly at ambient conditions, though it will depend on the temperature of the solid, how anhydrous it is and the humidity of the air. Something more to consider is that the crystalline form will absorb water at a slower rate. Were you dealing with powder or crystals?

Unfortunately, it isn't possible to judge the kinetics without more information. Perhaps you can provide your procedure?


Excited states and electron addition/removal energies within Hartree-Fock, CI, etc

There are actually quite a few methods post-HF that you can do to calculate excitation energies. The simplest one you will want to look at is Configuration Interaction Singles (CIS). To do this, you express your Hamiltonian in the basis of singly-excited HF determinants and diagonalize it. It doesn't involve correlation and is roughly the same cost as a HF calculation. The matrix elements are given in Szabo and Ostlund,  Modern Quantum Chemistry pg 236.

You are already familiar that for DFT, the most common way to get excitation energies is through a time-dependent DFT calculation (TDDFT). You can do a time-dependent Hartree-Fock (TDHF) as well, though you may see this in the literature as the Random Phase Approximation (RPA). They are essentially the same thing. TDHF and TDDFT are pretty much identical save for the fact that TDDFT is based off Kohn-Sham density, and TDHF is based off of Hartree-Fock density. The equations are almost the same, and it isn't that much more expensive to compute.

If you want to include electron correlation in wavefunction-based methods to a greater degree, look for EOM-CCSD like methods. These are based off a HF, followed by a coupled-cluster ground state calculation, followed by an extension to get excited state energies.

My graduate research uses and develops a lot of these methods. In fact, I wrote about these methods not too long ago --- you can read more about it here: http://bit.ly/1anA9dZ


How to determine the concentration after a dilution with Beer's law?

Beers law is

$$ E= \epsilon_{\lambda}\ \cdot c\ \cdot d$$

Where $c$ is the concentration of the solution. Then, you've found your answer once you've found $c$.

As Jerepierre points out, if you were to account for units in the answer, the question is actually missing some information. The pathlength $d$ is unaccounted for (but we're assuming its $1$), however if it was say $\mathrm{1\ cm}$, then the units for $\epsilon$ would need to be $\mathrm{AU\cdot mg^{-1} \cdot cm^{-1}}$. Which would mean that the units for your concentration would be in $\mathrm{mg}$. But, for concentration we would need to know antibody per some amount of volume. So taking units into account shows that something is missing.

What you're adding at the end is a step you would take to find the original concentration of the undiluted solution. Given that they diluted the concentration 5 times, but we don't know to what factor so we would have trouble solving for that.


How does iodide replace bromide when the latter is more basic?

As you suggest "$\ce{Br−}$  is a stronger base than $\ce{I−}$ and keeping in mind, stronger the base, lesser is the leavability, $\ce{I−}$ should not replace $\ce{Br−}$ "

I actually agree with the logic and according to me, infact, the reverse reaction, i.e.

$$\ce{CH3CH2CH2I + Br- (presence of DMSO) -> CH3CH2CH2Br + I-}$$

should be much more favourable. Because we already know, that in the aprotic solvents, the nucleophilicity of halides is  $\ce{F− > Cl− > Br− > I−}$, so $\ce{I−}$ should be a better leaving group and $\ce{Br−}$ should be a better nucleophile .


Which of mono-, di-, triphenylmethane is the most acidic?

You should consider the stability of carbanion that is left behind, after removing $\ce{H+}$.   One could make arguments for resonance stabilization of each of the carbanions. You could "push" the minus charge around the phenyl rings, and there is a general rule that the more "good" resonance structures that one can draw, the more stable the molecule.

It turns out that the tertiary carbanion in 2) is flanked by three phenyl groups (thus many good resonance structures compared to 1) and 3) ).


Oxygen/nitrogen ratios vary seasonally and geographically even at sea level on the order of 200 ppm.  Oxygen concentration has also been slowly decreasing in correlation with carbon dioxide increase.

Try using the MSIS-E-90 Atmosphere Model which allow location, date, time and height input and provides composition output.


Why does Fluorine have a high affinity for Calcium?

Actually it's not the question of affinity but insolubility of CaF2. As this salt has solubility product of ~4×10−11 which is mostly caused by nice packing and strong ionic interaction (see enter link description here)


Why does road salt reveal this bus's frame?

No. the best two explanations don't need any chemistry at all.

If you look closely at the panelling on a bus like this you will find that there is physical  evidence on the outer skin about where the structural frame is. The skin immediately over the frame tends to stick out a little. These physical features are probably enough to ensure that more dirt sticks to those parts of the skin. You can see a similar effect above warm radiators when small bumps on the wall above tend to pick up a disproportionate amount of the dust in the moving air caused by the convection currents from the warm radiator, becoming darker as a result.

Another possible contributor would be the fact that the vibrations in the panel skin of the bus will also be higher at the centre of each panel farthest away from the frame. This might contribute something as well.


Secondary amines react with nitrous acid to give N-nitroso compounds reference here

N-Methylaniline will give N-Methyl-N-nitrosoaniline as shown below


Nomenclature of saturated aliphatic hetrocyclic amines with an alkyl group on the nitrogen

According to Subsection P-22.2.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) of the functional parent compound is the retained name ‘pyrrolidine’, the nitrogen atom being numbered ‘1’.

Therefore, the name of the compound given in the question is ‘1-propylpyrrolidine’.


tl;dr: You are likely to see more soot, $\ce{CO}$, $\ce{H2}$, less of $\ce{CO2}$ and same amount of $\ce{H2O}$.

There are different type of combustion that you can talk about. For example when you mix fuel and oxidizer before burning them, it is referred to as premixed system. For example a gas cooking range, gas turbine engines, IC engines, etc. On the other hand if you do mix the fuel and oxidizer you get what's called a diffusion system. For example a candle flame, flame thrower, apartment fires, etc. Typically premixed systems will produce a blue flame, whereas diffusion systems will produce yellow-orangish flame.

It is important to distinguish between a premixed and diffusion flame system because they will behave differently. A lighter produces a diffusion flame. For this answer we will focus on diffusion flame.

An important characteristic of a diffusion flame is that it burns where there is just the right amount of fuel and oxidizer. That means irrespective of whether there is fuel entering oxidizer or oxidizer entering fuel, the flame will only be present where there is just the right amount of fuel and oxidizer. So for example for a methane-oxygen system at the flame we get, $$\ce{CH4 + 2O2 -> CO2 + 2H2O}$$

So in both the cases we would expect to see same amount of $\ce{CO2}$ and $\ce{H2O}$ begin produced when same amount of fuel is burnt. That doesn't sound interesting at all. Plus our ideal situation ignored the effect of heat produced and its effect on the surrounding.
What happens when we take this in account?

Soot

Well in most methane systems it is very difficult to produce soot, so let us consider something like propane $(\ce{C3H8})$ or higher order hydrocarbon (anything with more carbon atoms), which is more likely to soot. Soot is produced when unburned $\ce{C}$-atoms combine to form larger molecules, which combine to form particles. How soot is produced is still an active research area. But we do know that most soot is produced at relatively low temperatures (compared to the flame temperature) in fuel rich regions.

In a candle/lighter flame the soot is produced near the wick, but most of it is oxidized as it reaches the flame. Only some close to the center escapes from the tip because there is not enough oxidizer there to burn it all. This is the main reason diffusion flame appear yellow.

Now imagine your lighter which starts off with oxidizer in the center, there won't be any soot particle inside the flame. So the flame will appear bluish. But the heat produced from the flame may be enough to break (pyrolyse) the fuel in the surrounding region. This will cause soot to be produced. Also because there is no reason for the soot to approach the flame it will not get oxidized easily and you will be left with a lot more soot.

Unburned fuel

Next consider other unburned combustion products. When you burn methane in oxygen you produce more than just carbon dioxide and water. The unbalanced reaction can written as:
$$\ce{CH4 + O2 -> CO2 + H2O + CO + O2 + H2 + C + ...}$$

The unburned $\ce{C}$ forms the soot. A side effect is production of $\ce{H2}$. So now not only will you see more soot, there will be excess hydrogen too. Also in case of higher hydrocarbon fuels you might see some pyrolyzed fuel which is yet to form soot or get oxidized fully.

In a premixed flame system when you have lot of fuel lying around you are more likely to produce $\ce{CO}$ rather than $\ce{CO2}$. A simple reason being you first need to produce $\ce{CO}$ before you get $\ce{CO2}$. A side effect of this is that the temperature will be lower. This is because most heat is released when $\ce{CO}$ is oxidized to $\ce{CO2}$.
So what happens in a diffusion flame system?

As mentioned earlier in diffusion flame systems, the flame is around where there is just the right amount of fuel and oxidizer. The products formed at the flame diffuse out (into the fuel and oxidizer) and since there is scarcity of oxygen, $\ce{CO2}$ will get reduced to $\ce{CO}$ i.e. you will see increase in $\ce{CO}$ production and decrease in amount of $\ce{CO2}$.

Side note: Saturn's moon Titan has an atmosphere which primarily consists of nitrogen (98.4%). The next abundant gas is methane (1.4%) followed by hydrogen (0.1-0.2%). So it's not a bizarre world after all.


How many electrons does each shell hold?

This is a great question that allows us to learn from the periodic table. The previous answer asked if you have learned about orbitals, and if you haven't this is probably why you are having difficulty.
If you look at the periodic table, you will see that the elements are arranged in two ways, vertically (group) and horizontally (period), there is good reason for this. Electrons can only occupy space based on mathematical probability; each of these probability functions is what we refer to as orbitals. Each orbital can only hold a maximum number of electrons ( the equation for total electrons is 4l+2 electrons where l is a quantum number starting at l=0 for s, l=1 for p etc.) in the valence shell (for 1, it can hold 2 Starting on the left side of the periodic table with Hydrogen we know that it is in group 1, which means that it has 1 electron in its valence orbital. Element number 2 (He) is a bit of a special case because although it only has 2 electrons in its valence orbital, it is full because the orbital number is 1. What I mean by this is that if you look at period 2 elements, starting with Li, you see that it falls under Hydrogen, and has 1 valence electron. This is because the 1s orbital is full with 2 electrons, and the additional electron goes into the next lowest energy orbital which is called the 2s orbital. Be has 4 electrons, 2 of them in the valence orbital, and 2 of them in the 1s orbital. When we move to B(oron) the 1s and 2s orbitals are full, and we start to fill the 2p orbital. The p orbital can hold a total of 6 (4l+2 where l = 1) electrons and so the boron atom has the following configuration (1s)2(2s)2(2p)1. What you see is that the highest energy orbital (in this case 2 (s + p) is the valence orbital and so B(oron) has 3 valence electrons, but also has 2 electrons in the much lower energy 1s orbital. Think of these orbitals like layers of an onion, where the further from the center, the greater the energy of the orbital. There are rules to how the orbitals are filled and the rules are dictated by quantum mechanics but generally speaking the orbitals are filled as follows 1s 2s 2p 3s 3p 4s 3d 4p 4d using what is know as the Madelung rule. s orbitals can hold 2 electrons, p orbitals can hold 6 electrons, d orbitals can hold 10 electrons. As you work your way from low atomic number on the period table to high you will see why it is organized the way it is. The break between H and He is because of electrons, when you move to period 2 you start filling valence shell 2, when you move to period 3 you start filling valence shell 3 and so on.  I hope this helps, and let me know if you need clarification on anything!


Extraction of calcium carbonate from liquid

The stuff you're seeing is solid calcium carbonate that most likely did not dissolve, or some other insoluble calcium compound that formed by reaction of calcium with whatever is in the apple juice.

You might want to try using sodium carbonate instead of calcium carbonate. Sodium compounds are usually more soluble than calcium compounds, and it's a more common food additive than calcium carbonate.

That said, I wouldn't recommend adding chemicals to your food without knowing exactly what you're doing. Some seemingly safe chemicals may have hazardous impurities and do not qualify as food-grade. If you're doing this because of health problems then you should consult with a doctor on better ways of action than pouring carbonates in your drinks.


Will the solubility of a solute increase over the time a solute dissolve?

Ivan's comment really nailed the problem, but I'll throw in a bit more.

In the question the term solubility would be ok in an English class, but the term is incorrectly used within chemistry. The more correct tense would be solvation. Likewise the diagrams should have solvation rather than solubility.

  In an investigation, two substances X and Y of the same mass are dissolved in the same volume of water. It is found that X dissolves faster than Y. Which of the following graphs correctly shows the solubility solvation of the substances against time in this investigation?


Heat capacity (specific heat) varies inversely with atomic mass, the Dulong–Petit law. Al is about 27 amu, and Fe is about 56 amu, so as you noted, it would be expected that aluminum stores more heat. An extreme example is lead solder, which has such low specific heat that a calloused plumber's hand can wipe a solder joint with little discomfort.

As for thermal conductivity, that property is associated with the rigidity of crystal lattices, where sound travels as phonons, as in diamond, and with electrical conductivity, where "freely moving valence electrons transfer not only electric current but also heat energy." Aluminum and copper are among the best metallic conductors, so are used for cooking-pan bottoms to spread heat evenly. Since stainless steel has many additions to Fe, such as Ni and Cr, these inclusions provide discontinuities at grain boundaries that further impede heat transfer. Dewar flasks are made of stainless steel or glass, rather than Al, because they are poor conductors of heat.


Avogadro Conformer Properties with Linux Version

I can't reproduce the problem.

I'm using Ubuntu 16.04 and installed Avogadro from the repository. The version is 1.1.1, just as described on Launchpad.

This version does have the menu entry in question:
Ansicht > Eigenschaften > Konformereneigenschaften

After a conformation search, it does open a table with the found conformers and their energies.


Simply the glass degrades because it dissolves in the water.

First the how. There are two factors which impact this process which is the dissolution of sodium and the dissolution of silica. The dissolution of sodium occurs first as evidenced by a plot of pH vs time for a glass powder in water and proceed via electrophilic attack:
$$\ce{Si-O-Na_{(glass)} + H2O_{(l)} \longrightarrow Si-O-H_{(glass)} + OH-_{(aq)} + Na+_{(aq)} }$$

The second step is the dissolution of the silica which proceeds via nucleophilic attack:
$$\ce{(-O-)2Si(-O-H)2_{(glass)} + H2O_{(l)} <-> Si(OH)4_{(aq)} }$$

The solubility of the silica is, of course, low, but measurable and significant. This process at high temperature increases the solubility of the silica (hydrothermal) and is how natural quartz crystals were formed (hydrothermal vents).

Second you say that degradation occurs slower in seawater than distilled water and that, in fact, is correct. At first, one might suspect that the dissolved sodium reduces the dissolution due to a common ion effect or reduced chemical potential but is actually not the case. The operative word here is seawater.  Seawater already has a large amount of dissolved silica in it from the rocks and sand that lines coastlines and ocean beds such that it is saturated. Since the second reaction is reversible the increased concentration prevents the silica from being leached appreciably faster than it can be redeposited thus reducing the corrosion of the fiber


How is plane polarized light rotated by chiral compounds?

Circularly polarized light is like a helix that twists through space. The two components are mirror images of each other.
Now, every molecule interacts with both the left-handed twisting light and the right handed twisting light. The interactions differ. Every molecule, in different orientations, interact differently with the left-handed and the right-handed circularly polarized light.

Animation Reference: https://en.wikipedia.org/wiki/Circular_polarization
Now, if the molecules in solution have any mirror symmetry themselves, then if averaged over all the molecule interactions the left-handed and right-handed circularly polarized light interact in the same way, and so the polarization does not change.
But this does not hold for chiral molecules: for a particular interaction between a molecule in a certain orientation and the left-handed circularly polarized light there is no corresponding identical interaction with the right-handed circularly polarized light. There is no such molecule of that mirrored shape around.
This results in a net difference in interactions of the molecules with the left-handed and right-handed circularly polarized light, which can be described as a difference in refractive index for the two light waves. And this difference in refractive index can be detected as a change in the direction of the polarization for the sum of the two light waves.

Now, if we want to consider this using the concept of the speed of light varying in media, light goes slower in matter than in vacuum. This slowing down of light in matter is measured using the refractive index n. A higher refractive index means that the light goes slower and so keeps the light longer in the medium. Now if the medium is chiral, it gives the light two different speeds, one for the light that rotates its polarization clockwise and the other one for rotating polarization counter-clockwise. Any polarized light has only two parts (clockwise and counter-clockwise).
The two parts are combined and so the light shows a direction of polarization. When these two parts of light pass through chiral matter, one goes faster and the other slower. The result is that the polarization of light is rotated.

Response to @ Michael Seifert comment;
The way of splitting the light really depends on the matter, rather not decided by the light itself. . If matter has two directions (like a calcite crystal), light is split linearly (parallel and perpendicular). If matter is chiral, (like a solution of bio-substance) light is split circularly (clockwise and counter-clockwise). If matter is both, then all four polarizations are possible, which makes the out-going light hard to nail down.

References:
https://en.wikibooks.org/wiki/Organic_Chemistry/Chirality/Optical_activity#Why_Polarized_Light_Is_Affected
https://en.wikipedia.org/wiki/Optical_rotation
https://www.quora.com/How-do-chiral-molecules-rotate-the-plane-of-polarised-light


As molarity is defined as the ratio of molar amount of the dissolved solute and solution volume:
$$c=\frac nV$$
molarity increases when solvent ( like water ) evaporates.
Molarity can be considered constant only in context of later refilling the evaporated solvent and mixing the solution. Then molarity returns to the original value. It assumes the solute ( like sodium chloride ) content is constant, not undergoing any reaction.


Why And How are Unbranched chains of hydrocarbons easily biodegraded?

Here is a study1 showing this is the case, that generally speaking, branched hydrocarbons are more difficult to degrade, especially those exhibiting anteiso-terminal branching (branching  on the third carbon). Both the study and a book on oil spill forensics2 hypothesize that steric hindrance is the  mechanism responsible. They state, respectively:

  If alkyl branches are located near the terminus, it seems likely that decreased biodegradability could be the result of steric inhibition of terminal oxidizing enzymes

  Complex branching hinders both the initial oxidation and the subsequent lipid catabolism apparently because tertiary and quarternary carbon atoms interfere by steric hindrance with the oxidation enzymes.

However, this is not always the case as Boethling3 explains when discussing design objectives for molecule biodegradability:

  Similarly there is a commonly held view that any branching, for example, even a single methyl group on an otherwise linear alkyl group, is to be avoided. This is a gross oversimplification. All that can be said with any confidence is that quaternary carbon is usually to be avoided, as is extensive methyl chain branching, which has no strict definition.

1 Schaeffer, T. L., Cantwell, S. G., Brown, J. L., Watt, D. S., & Fall, R. R. (1979). Microbial Growth on Hydrocarbons: Terminal Branching Inhibits Biodegradation. Applied and Environmental Microbiology, 38(4), 742-746. Retrieved from https://aem.asm.org/

2 Wang, Z., & Stout, S. (2006). Oil Spill Environmental Forensics: Fingerprinting and Source Identification (1st ed.). Burlington, MA: Academic Press.

3 Boethling, R. S., Sommer, E., & DiFiore, D. (2007). Designing Small Molecules for Biodegradability. Chemical Reviews, 107(6), 2207–2227. https://doi.org/10.1021/cr050952t


How can the bulky octadecyltriethoxysilane molecule promote single-crystal growth?

The function of the octadecyltriethoxysilane monolayer in this case is not to promote epitaxial growth but to reduce the surface energy of the dielectric layer (SiO2), on top of which the semiconductor (pentacene/rubrene/etc.) is about to be deposited. The reduced surface energy allows the crystal to grow "comfortably".

Organosilanes are typical compounds for modifying oxide layers (Pernstich et al., 2004):


Energy a system at room temperature can supply?

Echoing Jon Custer's comment:

  R = 8.314 J/K/mol

Multiply by 298 K, and you have a rough-and-ready rule-of-thumb number:

The "thermal energy" at room temperature is about 2.5 kJ / mol.


Where does the charge on an electrode, responsible for the formation of a double layer once submerged in an electrolyte, come from?

This can be answered on more than one level. Where did the word "electrode" come from, to begin with? Supposedly it is due to the fact that you apply some electric potential to that metal needle (stuck into the brain of some creature or other electrolytic medium), and that's why the electrode gets charged.

Now what if we apply no potential? Well, for example, if two pieces of different metals are in contact, they would form a galvanic couple and develop some potential on their own, resulting in galvanic corrosion.

But wait, what if there is just one metal in the electrolyte? It turns out that some contact potential would develop even then. As the link says:

  Electrode potential appears at the interface between an electrode and electrolyte due to the transfer of charged species across the interface, specific adsorption of ions at the interface, and specific adsorption/orientation of polar molecules, including those of the solvent.

Think of a metal in a distilled water (which, BTW, is a poor electrolyte). Inevitably, a tiny portion of metal would dissolve, for such is the nature of chemical equilibrium: it is almost never shifted all the way to one side. Now, that metal would be in the form of cations, and the extra electrons would stay in the bulk, giving it a negative potential (not so tiny, because the electric forces are quite powerful). And if we consider other possible effects, especially if the water is not distilled but contains some ionic species, then the resulting potential on the metal can be pretty much anything (and not necessarily negative).

All in all, a random piece of metal in a random electrolyte will most likely have some potential difference with the electrolyte and hence form a double layer.


Is calcium hydroxide or potassium hydroxide more efficient at removing carbon dioxide?

$\ce{KOH}$ (and $\ce{NaOH}$) absorb $\ce{CO2}$ more quickly and completely than $\ce{Ca(OH)2}$, but they are much more dangerous to work with: caustic to skin, deliquescent (forming a highly concentrated solution that can pour out of the apparatus), etc. For that reason, "lime water" is preferred for classroom use to demonstrate the presence of $\ce{CO2}$.


I have 100 mg of a proteinase K lyophilized powder and I need to make it to a working concentration of 25 mg/mL

Simply adding 4 mL of solution to 100 mg of enzyme is not an accurate way of preparing this solution. I am going to describe what to do from an analytical chemist's point of view. If you don't need this level of precision, you could do this following the same steps without volumetric glassware.

The first thing you need to do is decide what volume of your 25 mg/mL solution you need. Ideally, you will have a volumetric flask corresponding to this volume. For now, let's go with the 4 mL that you mentioned.

Next you will need to make a stock solution of your buffer. This will include all of the components except for the enzyme:

Decide what volume of buffer to make. You will need to make at least enough for your working solution, and again you will want to use a volume for which you have a volumetric flask. Also, you want to be weighing out reasonable amounts of your reagents (because weighing out <10 mg of something can really suck). Let's assume you're making 1 L. You're reagents are cheap anyway and the larger scale will make everything easier.
Calculate how much $\ce{TrisHCl}$ and $\ce{CaCl2}$ you will need to obtain the desired concentrations in 1 L of solvent.
Weight out the $\ce{TrisHCl}$ and $\ce{CaCl2}$ and add it to your volumetric flask.
Add 300 mL glycerol to your flask using a graduated cylinder. 30% glycerol probably refers to v/v.
Fill nearly to volume with water. You need everything to be in solution so you can adjust the pH, but you also need to leave some room to work with to do the actual adjustment. I'd shoot for approximately 80% full.
Adjust the pH by adding concentrated $\ce{HCl}$ dropwise.
Fill to volume with water and thoroughly mix the solution.

Now that you have your buffer solution, you just need to weigh out the appropriate amount of your enzyme and dissolve it in your buffer. From the example above, you would weigh out 100 mg of the enzyme (don't just assume you have 100 mg, even if the bottle says so) and dissolve it in your buffer using a 4 mL vol. flask.*

*
If you don't have a 4 mL vol. flask and you actually need all 100 mg of your enzyme in this solution, you will need to mess around with dilutions. Start by by creating a 50 mg/mL (2 mL vol. flask) or 100 mg/mL (1 mL vol. flask) solution.


Can I make a pH 7 buffer from phosphoric acid and sodium hydroxide?

You should be able to do this with little trouble.  The only caveats:

Make a stock solution of phosphoric acid and also of NaOH.  When you add them to each other to make your buffer, heating could be a problem, much moreso than if you were using the mono- and di-basic sodium phosphate salts.  So you may want to cool on ice when mixing these compounds.
You can calculate approximately how much NaOH and phosphoric acid to mix in order to get a pH 7 result, but it's best to use a pH meter to verify that you've reached pH 7.


Why is 295 nm a good wavelength to check absorbance when testing for aspirin?

I think this was basically an educational experiment. Your analytical method assumes that the hydrolysis was complete (this is the key requirement of a successful calibration curve). Your experiment did not include a test to assess the completion of hydrolysis. Acetylsalicylic acid begins to hydrolyze in alkaline solution to salicylic acid and basically it is an indirect method of detection.
You want to analyze acetylsalicylic acid but you are detecting salicylic acid. Guess where does this method fail? It will fail if you had a mixture of acetylsalicylic acid and salicylic acid in the beginning and you were asked to quantify both.
As EdV explained to you, being 1 nm off from the lambda max is not a big deal in UV-Vis spectra of solutions. Even the monochromator may be slightly off!


Regioselectivity in Friedel–Crafts acylation of thiophene

Friedel-Crafts acylation reactions are well documented on thiophenes, and go with high regioselectivity for the 2-position (or 5-position, which is equivalent in the case of unsubstituted thiophene) rather than the 3-position (or 4-position). Joule and Mills' Heterocyclic Chemistry (5th ed., p 327) gives the following two examples:

The regioselectivity can be explained through consideration of the intermediates involved when the thiophene attacks the electrophilic carbenium species.

One is able to see that the intermediate species produced by attack at the 2-position is able to be drawn in three resonance forms, whilst the intermediate produced by attack at the 3-position is only able to be drawn in two resonance forms. As a general rule of thumb, the more resonance forms one can draw, the more 'stable' and intermediate, and hence the energy required to access it is lower, hence in the case of thiophene we get preferential electrophilic substitution at the 2/5-position.

(When looking at the intermediates, we can also consider the type of conjugation involved. Attack at the 2-position yields a linearly conjugated intermediate, which is lower in energy than the cross conjugated intermediate from attack at the 3-position).


Including H⁺ from the dissociation of the acid in a titration pH calculation?

There are numerous approximations made when pH calculations like these are made.

In a problem like this the approximations that would usually be made are

pH = -log[H+] (as opposed to pH being -log (H+ activity) which is the true definition).
NaOH is an infinitely strong base, with NaOH completely dissociated. (really, [Na+][OH-]/[NaOH] is about 4)
The dissciation of water  can be ignored (really (Kw =[H+][OH-]= 10^-14).
There exists a constant Ka for HF that is [H+][F-]/[HF], (whereas a true equilibrium constant would be a function of activities, rather than concentrations, of these species, and of water.  The Ka expressed as concentrations is dependent upon ionic strength for example in accordance with the Davies equation approximation).
Particular to this example, it is being assumed that the only fluorine species are HF and F- (really there is also HF2- and another equilbrium constant for it).

So you can take away one or more of these approximating assumptions to try to more accurately calculate pH, but it makes the problem very difficult.

  However, don't you have to consider the dissociation of HF first, and then consider the reaction between those produced  H+  and the added OH-? Because whenever an acid reacts with a base isn't it truly the hydrogen ion formed from the deprotonated acid reacting with the hydroxide ion from the base? How can you consider reactions of unprotonated acids/bases directly?

At the end of the titration (or at any other point during the titration along as the solution is homogeneously mixed) the system will be in equilibrium.  The correct approach is to consider all the species that are present and all the equilbrium constants amoung the species.

Here, the species would be H2O, H+, OH-, HF, F-, Na+, and NaOH (and if I want to be picky HF2-).

The equilbrium constants are Kw, Ka of HF, Kb(NaOH), (and Ka of HF2-).

You need the same number of independent equations as unknowns to solve the problem.

[F-] + [HF] = (0.1/0.35)0.3M

[Na+] + [NaOH] = (0.1/0.35)0.3M

[H+] + [Na+] = [F-] + [OH-]

[H20] = 55M (or, if that is too much of assumption [H2O] + [OH-] + [NaOH] = constant by conservation of oxygen atoms)

plus the 3 equilibrium constant equations gives you 7 equations and 7 unknowns.   That would be a comprehensive consideration of the problem dropping assumptions number 2 and 3.


I worked on a project using this reaction fairly recently. There are a significant number of intermediate species involved in the reaction with a significant number of autocatalytic steps.

In order to obey the rules of free energy, the reaction may not oscillate about its equilibrium as a pendulum does- there is always a monotonic decrease in reactant concentration and an increase in product concentration, but the multitude of intermediates oscillate in concentration as the reaction switches its reliance on certain sections of the mechanism (dependent upon the instantaneous concentration of stuff in the mixture). The indicator used dictates the colour changes- I used ferroin, and so when the concentration of Ce(4+) (the catalyst used) is high the mixture is red, and when the concentration of bromate is high the mixture is blue.

It should be noted that the reaction will always come to a stop unless you use a continuously stirred flow reactor in which products are removed and reactants added as the reaction proceeds.

There are several mechanisms such as the FKN, MBM, GTF and Oregonator. The Oregonator is by far the most simple but lacks chemical accuracy, the former 3 display real chemical conversions yet still fail to perfectly model the reaction. The mechanisms can be modelled using coupled differential kinetic equations, which I did during my project in Mathematica.

The simplest general reaction scheme for a chemical oscillator is given by the Brusselator:

A → X
2X + Y → 3X
B + X → Y + D
X → E

In this scheme, A represents the reactions and E the products. X is autocatalytic (it catalyses its own production).

Another interesting example comes into play in population dynamics, the Lotka-Volterra equations (also known as predator-prey equations). They can be used to quantify the changes in populations in a predator-prey system such as foxes and rabbits. The reaction itself has also been suggested as a very primitive form of chemical order- interesting stuff.

Disclaimer: Sorry if this seems a little off-topic; the details of the specific BZ mechanisms are far less interesting than the concept of the oscillators themselves. If you want to look into the actual mechanisms though, start with the FKN model, it's fairly easy to digest and even model.


Oxidizing 2-Methylpropan-2-ol using potassium dichromate. Why is heating required?

Tertiary alcohols (such as 2-methyl-2-propanol, in your specific case) cannot be directly oxidized by $\ce{Cr(VI)}$ salts like $\ce{K2Cr2O7}$ because they have no $\alpha$-hydrogen. On the other hand, a primary alcohol (such as 1-butanol) will be readily oxidized, even under mild conditions and with reagents much gentler than $\ce{K2Cr2O7}$.

With heating and sufficiently acidic conditions, a tertiary alcohol can be eliminated to form an alkene, which can then be oxidatively cleaved to yield two molecules with carbonyl groups. In the case of 2-methyl-2-propanol, one would presumably obtain acetone and either formaldehyde or formic acid, depending on how far the oxidation is permitted to go. (Though, under conditions vigorous enough to effect the oxidative cleavage of the alkene, it's highly unlikely the aldehyde could be isolated. I would also note that carbon dioxide is a product of oxidative cleavage of terminal alkenes when $\ce{KMnO4}$ is the oxidizer, but whether this occurs with $\ce{Cr(VI)}$ salts is unknown to me.) This is really the only reaction I can imagine to have plausibly occurred, if any.

To the best of my knowledge, $\ce{K2Cr2O7}$ is not very efficient for this purpose, and the procedure itself strikes me as pretty odd if the oxidative cleavage described above is the intended reaction. It may be that no reaction was intended to happen, since oxidation reactions of this type can be used as chemical tests to differentiate alcohols by degree of substitution.


Recall the definition of σ-bonds. They are caused by the head-on overlapping of the orbitals.

$s$ orbitals always, and only form σ-bonds while the $p$ orbitals in the direction of bond axis (which is $p_x$), also form σ-bonds.

The other two $p$ bonds cannot form σ-bonds as they are perpendicular to the bond axis, and cannot overlap head-on. In such cases, π-bonds are formed, by the sideways overlapping of the orbitals.

The $p$ bonds not perpendicular to the bond axis involve in π-bond formation ($p_y$ and $p_z$).


How do I calculate the cost of a single Aluminum atom?

You can compute the area of aluminum foil in the roll as:

$$30.5\;\rm{cm}\cdot 3050\;\rm{cm} = 93025\;\rm{cm}^{2}$$

You have measured the mass per square centimeter as:

$${1.35\;\rm{g}\over 18.6\;\rm{cm}\cdot 12.5\;\rm{cm}} = {1.35\;\rm{g}\over 232.5\;\rm{cm}^{2}}$$

You have been given the cost of the roll (and thus the cost per area):

$${\$5.53\over 93025\;\rm{cm}^{2}}$$

Using those numbers, the molar mass for aluminum, and Avogadro's constant (here, it's just the number of atoms so I designate it as such):

$$\rm{cost\;per\;atom} = \left({\$5.53\over 93025\;\rm{cm}^{2}}\right)
\left({232.5\;\rm{cm}^{2}\over 1.35\;\rm{g}}\right)
\left({26.98\;\rm{g}\over 1\;\rm{mol}}\right)
\left({1\;\rm{mol}\over 6.022\cdot 10^{23}\;\rm{atoms}}\right)
$$
$$\rm{cost\;per\;atom} = \$4.59\cdot 10^{-25}$$


Flood Control - electrolysis - limit to rate of conversion?

Floodwaters are immense in quantity.  Let's estimate how much energy would be required to convert that quantity of water to $\ce{H2}$ and $\ce{O2}$.

The Mississipi river has a (non-flood) flow rate of about 593,000 cubic feet per second, or 17000 cubic meters per second.  That's $1.7 \times 10^7$ liters per second, or $1.7 \times 10^{10}$ grams per second, or roughly $10^9$ moles per second.
Faraday's constant is the conversion factor between moles and Coulombs.  It's value is about 96000 ampere-seconds per mol.  Let's approximate it as $10^5$ ampere-seconds per mole.
It takes four moles of electrons per mole of water to complete the electrolysis.
Multiply the numbers together, and you get that to electrolyze all the waters of the Mississippi river, you'd need $4 \times 10^{14}$ amperes of current.  That's about the amount of current present in a billion bolts of lightning.  Yikes!
Exactly how much voltage would be required depends on your design of electrolytic cell you use for this giga-bolt scale.  Let's say it's 5 volts.  That means the power you need for the electrolysis is $2\times10^{15}$ watts.  That's 2000 terawatts.  But the world's current total energy consumption is about 12 terawatts...


Why does acetone have a greater dipole moment than dimethyl ether?

The ether is fairly symmetrical though there is a bond angle of ~111°, "exposing" the oxygen on the side (and its electron cloud).  Acetone, however, has an oxygen off to the side, with its electron cloud offset ~120 picometers (pm), making it more polar.

Remember that oxygen is a strong electron acceptor, so that there will be a "lump" of electron probability surrounding it.

[Sorry about the hodgepodge illustration!]


My teacher said there is one structural isomer

I think (hope) you are talking about stereo isomers (i.e. same Lewis structure, different substance).
There are two ways of counting. Either you can ask "How many substances are stereo isomers of cis-1,2-dimethylcyclohexane?" Or you can ask "How many stereo isomers exist of 1,2-dimethylcyclohexane?". The correct answers are off by one.
This is like asking "how many siblings do you have" vs. "how many siblings are there in your family (including you)".

How many structural isomers does 1,2-dimethyl-cyclohexane have?

The molecular formula for this substance is $\ce{C8H16}$. There are lots of substances with that formula but different Lewis structure.


Is this a feasible synthetic route for 1-(phenylethynyl)cyclopentan-1-ol?

The route definitely looks feasible and it is more or less what I would have done to arrive at your desired product. A few notes:

Make sure you actually have sodium amide $\ce{NaNH2}$ and not sodium in ammonia $\ce{Na/NH3}$. As pointed out in the question comments, they react very differently; the latter generates a trans-alkene.
I’m not sure I would have chosen the same base. But the choice of base is likely arbitrary anyway. n-Butyllithium would definitely also do the trick but might be overkill.
Typically in a lab setting, reactions with strong bases (as is the case here) are carefully quenched with something aquaeous. So you can be sure that there will be an aquaeous proton source ($\ce{H2O}$ or $\ce{H+/H2O}$, depending on your markers’ preferences) in the end. There should be no need to additionally specify conditions. It would make no sense to attempt to isolate the alcoholate anyway.


What is the order of acidic strength among phenol, catechol, resorcinol and hydroquinone?

A quick search on says that resorcinol is more acidic that catechol, followed by phenol.

Hydroquinone - strong +M effect and weak -I effect

Phenol - the normal resonance you'd expect in all phenols

Resorcinol - -I effect

Catechol - +M effect, stronger -I effect and hydrogen bonding

Taking these factors into account, it is easy to see why hydroquinone is least acidic. Hydrogen bonding and the vicinity of the $\ce{-OH}$ to the anion (for the stronger -I effect) make cactechol more aciic than the phenol despite its +M contribution. Resorcinol has only -I effect and no counteracting electron-donating effects, which makes it the most acidic.


What is the "curse of dimensionality" in molecular dynamics?

It's certainly more of a "back of the envelope" calculation than a rigorous proof. The main point is, that for a certain system, the cost of obtaining a set of trajectories with a given statistical accuracy is fixed with ab initio MD, as only the dynamically relevant parts of the PES are calculated.

With what is here called "classical MD" (but also applies to "quantum dynamics" and "wave packet" methods), the whole PES is computed in advance, and this cost is system dependent (scales with $3N - 6$ for nonlinear systems); the individual trajectories over this PES are then calculated comparably quickly.

The ratio of the two scalings is $\frac{10^{3N-6}}{10^{M+n}}$, increasing with system size (making ab initio MD favourable) and decreasing with required statistical accuracy (making classical MD favourable).


Reactions involving trimethylaluminium Al2Me6

For a): Many organometallic compounds are easily hydrolyzed to the respective hydrocarbons and metal (hydr)oxides. In this case, the products are $\ce{CH4}$ and more likely $\ce{Al2O3}$, because the formation of $\ce{Al(OH)3}$ requires more moles of water.

Reactions b) and c) should be transmetallations where the methyl groups are transferred from aluminium to the other metal. The excess of $\ce{Al2Me6}$ helps to shift the equilibrium of the reaction to the product side. $\ce{SiMe4}$ and $\ce{SnMe4}$ would be the correct products which you have identified, but aluminium is also converted into its respective oxide/chloride.


Why does coumarin predominantly exist in keto form?

The answer is as simple as it will be unhelpful to your original question: There is no tautomer possible.

Coumarin does not have a keto group, and there are no enolisable protons either. Coumarin consists of a benzene ring, connected to a heterocycle with the functional group lactone. The molecule is flat and the conjugation extends throughout it.
What you have drawn are two resonance structures. Both of them are valid configurations (there are more), but neither actually exist; see What is resonance, and are resonance structures real? In both of your drawn structures the nuclei positions (in the clamped nuclei approximation) do not move.

The concept which applies here is completely different from keto-enol-tautomrie, where there are two different, observable molecules.

The question which you are probably asking is: Which of the drawn structures is the major contributor to the (so-called) resonance hybrid.

I have run a quick calculation of the DF-B97D3(BJ)/def2-SVP//GFN2-xTB level of theory, which suggests that the conjugation is skewed towards the lactone group. More investigations would be necessary, but the π orbitals should give you a general idea.


Chemoselectively dihydroxylate a terminal alkene in the presence of an internal alkene?

The Lemieux-Johnson oxidation (details here) using catalytic $\ce{OsO4}$ or $\ce{RuO4}$ turned over with $\ce{NaIO4}$ will do the transformation you want in one step. These conditions should be selective for the exo double bond.


Why doesn't 4 hydrogen and 1 carbon form a ionic bond? What is the limit of electrons that can be given away?

I can understand your reasoning. But we need to consider many factors. When a compound is to be formed it should be stable enough. Or else the compound breaks down. Here the electronegativity of Carbon is 2.55 and Hydrogen is 2.2 (According to  Pauling's scale). So the electronegativity difference between these two atoms is 0.35. In the table given below...,

We can see that the nature of bonds between atoms of different electronegativity. As the difference is 0.35, the bond is Non-polar. And also for removing electrons from carbon are...,
Ionisation energy number & Enthalpy(kJ/mol)
1st------------------------->1086.5;
2nd------------------------->2352.6;
3rd------------------------->4620.5;
4th------------------------->6222.7.

So total energy required is 14282.3 KJ/mol. But electron affinity of hydrogen is 72.8 KJ/mol.So four hydrogens give 291.2 energy. So totally we need 13991.1 KJ/mol of energy to form methane.

While for removing successive hydrogen atoms from methane the bond-dissociation energies are 104 kcal/mol (435 kJ/mol) for D(CH3–H), 106 kcal/mol (444 kJ/mol) for D(CH2–H), 106 kcal/mol (444 kJ/mol) for D(CH–H) and finally 81 kcal/mol (339 kJ/mol) for D(C–H). The bond energy is, thus, 99 kcal/mol or 414 kJ/mol (the average of the bond-dissociation energies). Here the formation of methane is exothermic and is spontaneous while in the above situation it highly endothermic. All these factors give us why C-H bond is covalent but not ionic. Also you can do the same for either two bonds or three bonds anything. So we can say that $\ce{CH4}$  is covalent and cannot be ionic. And two bonds also cannot be ionic.

But in October 2016, astronomers reported that the very basic chemical ingredients of life—the carbon-hydrogen molecule (CH, or methylidyne radical), the carbon-hydrogen positive ion (CH+) and the carbon ion (C+)—are the result, in large part, of ultraviolet light from stars, rather than in other ways, such as the result of turbulent events related to supernovae and young stars, as thought earlier. So we can say these ions are in existence in our universe.

I hope it helps you.
I'm new to Stackoverflow so sorry if I've made any mistake in giving your desired answer
.


Is this the right formula to use to solve the above problem

Yes, this is it, if the initial concentrations of the reagents are the same (which is not mentioned explicitly in the problem statement, but probably is implied, because otherwise the problem would be unsolvable.)

Indeed, as A and B are spent in equal amounts, their concentrations are bound to be the same not only initially, but at all times: $[\ce{A}]=[\ce{B}]=c(t)$. Therefore $\dot c=-k\cdot[\ce{A}]\cdot[\ce{B}]=-k\cdot c^2$, which you seem to already know how to solve.

Update. Come to think of it, you don't actually need to solve for $c(t)$. You know that the rate is proportional to the first power of $[\ce{A}]$, and also to the first power of $[\ce{B}]$, and that both $[\ce{A}]$ and $[\ce{B}]$ have decreased by 50%; what would happen to the rate?

The concept of half life does not apply here, since the decay is not exponential.

Also, welcome to Chem.SE, and pay attention to the formatting.


The Carnot cycle uses an idealized heat engine to illustrate basic thermodynamic principles. Net work is done either by or on the system, depending on the direction in which it is run. As part of this idealization, the piston is massless and frictionless. The piston merely serves as a moveable impermeable boundary between the system and the surroundings—i.e., as an interface that allows the system to do work on the surroundings, and visa versa. The piston exerts no force of its own, so no work is done by or on the piston; the piston merely serves as a conduit for force exerted by the system or surroundings, and is not part of either.
To more specifically address your question, since the piston is massless, its gravitational potential energy is zero, independent of position.
Also, as this is a cyclic process, the system is returned to its original state at the end of each cycle, and thus the piston is returned to its original height. [Hence, as Chet Miller points out in his comment, even if the piston had mass, its PE would be unchanged at the end of each cycle.]


Why the magnitude of dipole moment influences on FTIR % transmittance?

Light has both oscillating electric and magnetic fields.  In the common UV-Visible-IR spectroscopy experiments it is the interaction of the molecule with the  electric field that is being observed. A molecule is very small compared to the size of a beam of light, so it is fair to view the molecule as being immersed in or surrounded by the light.  In other words, it is fair to view the molecule as being immersed in a uniform electric field.

The strength of the interaction ($I_s$) of the light's electric field ($E$) with the molecule is dependent upon both the strength of the light's electric field and the vector description of the molecule's electric field.  The molecule's electric field is described by the spatial arrangement of the molecule's electrons.  The spatial arrangement of the molecule's electrons is described by the dipole moment ($\mu$), a vector sum.  Hence,
$$I_s = E \cdot \mu $$

The absorption intensity of a spectral transition (the area under the absorption curve - usually it correlates with the peak height) is given by the same equation except that rather than use the molecular dipole moment $\mu$, one uses what is called the transition dipole moment.  Rather than consider the ground state distribution of electrons around the molecule, one must use the rearranged distribution of electrons around the molecule that occurs when the molecule absorbs light.

If the transition dipole moment is always similar to the ground state dipole moment, then what the video says is correct.  This may be true for larger molecules when we are looking at an absorption that excites a $\ce{C-H}$ or $\ce{C-C}$ bond, or something similar.  In such cases, the small perturbation in electron density around the specific bond probably doesn't radically change the overall electron distribution around the entire molecule.  In smaller molecules like $\ce{CO}$ it seems possible to me that light absorption could make a significant change to the overall electron distribution and that consequently the ground state dipole moment might differ substantially from the transition dipole moment.


Why put zinc solution in a galvanic cell?

It's tempting to look at the whole cell here and ask: hmm what's going on? But try taking a reductionist approach and look at each electrode system one at a time, in isolation.

Let's consider a basic model.If we immerse an electrode of $\ce{Cu}$ metal into a solution of $\ce{Cu^2+}$, the following can be thought to take place:

$$\ce{Cu(s) <=> Cu^2+(aq) + 2e-}$$

Where the ions in the solution are bombarding the electrode surface, and are being taken up by gaining two electrons from the metal, whilst the metal is giving up ions from its surface, keeping hold of two electrons. At equilibrium, this forward and backward bombardment occurs at the same rate.

You might say to me: okay but we can't actually probe this system with a voltmeter to say that this is going on, because this would require a circuit to be made with an external system, and hence would perturb any effect we wish to observe. I'd then say to you: you're correct but bear with my argument, just noting what I've said.

By a similar token, in our other isolated beaker we have:

$$\ce{Zn(s) <=> Zn^2+(aq) + 2e-}$$

Now, the whole idea behind an electrochemical cell is to extract chemical energy by systems which are not at equilibrium. If a system is at equilibrium then no energy will be available. The equation which gives the free energy available from an electrochemical cell is (derivation not given):

$$\Delta G=−RT\ln K_\mathrm{eq}=−nFE_\text{cell}$$

Where $R$ is the ideal gas constant, $T$ is the parameter of temperature, $F$ is the Faraday constant (the amount of charge carried by a mole of electrons) and $E_\text{cell}$ is your available electrochemical energy.

So for these two individual single-electrode systems at equilibrium, can we extract chemical energy? Well no, because Gibbs at equilibrium is zero, and hence so is $E_\text{cell}$.

What then do we do?

Complete the circuit between the two, and if the potential between two half-reactions is sufficient to warrant the moving of electrons, a dis-equilibrium will be created until the reaction has completed (which is indeed what we get for a system of zinc/copper). Once the electron transfer between zinc and copped has been achieved i.e. equilibrium has been achieved, then we can no longer extract any useful chemical energy. Our cell is dead!

So in summary: the dual metal electrode/same-metal ion solution is needed for your initial half-reactions, which you then perturb by completing your circuit. This perturbation of the individual half-reactions results in our being able to extract useful chemical energy by shifting the equilibrium until a new equilibrium is established.


Different possible Lewis structures of peroxymonosulfate anion

So shouldn't the one that satisfies the octet rule be favored over the one with an extended octet?

That's only true when you are comparing resonance structures. Resonance structures only involve movement of electrons and not atoms.

It's not true for different Lewis structures, which are the ones you drew. There are lots of compounds which are perfectly stable but do not (in principle) satisfy the octet rule.

I would have mentioned the extreme instability of your multiple $\ce{O-O}$ single bonds in your alternative proposed structure, but the other answer beat me to it.


Relation between speed of sound c and compressibility

With your result

\begin{equation}
\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}
\end{equation}

you're nearly there. All that is missing is a little mathematical trick. Simply make the substitution $f = \frac{1}{\rho}$ and use the chain rule to differentiate $f$ by $\rho$ and you see that:

\begin{equation}
  \frac{\partial f}{\partial \rho} = - \frac{1}{\rho^2} \quad \Rightarrow \quad \mathrm{d} f = - \frac{1}{\rho^2} \mathrm{d} \rho \quad \Rightarrow \quad \mathrm{d} \frac{1}{\rho} = - \frac{1}{\rho^2} \mathrm{d} \rho \ .
\end{equation}

Using this and the definition of $c^2$ you get to the desired result:

\begin{equation}
\beta_S = -\rho \underbrace{\frac{\partial \frac{1}{\rho}}{\partial p}}_{= - \frac{\frac{1}{\rho^2} \partial \rho}{\partial p}} =
\underbrace{(-\rho) (- \frac{1}{\rho^2})}_{= \frac{1}{\rho} } \underbrace{\frac{\partial \rho}{\partial p}}_{= \frac{1}{c^2}} \quad \Rightarrow \quad c^2 = \frac{1}{\rho \beta_S} \ .
\end{equation}