text:
How is the standard electrode potential measured without allowing the electrolyte concentrations to change?
Let's say we need to measure the standard electrode potential of a zinc electrode dipped in an aqueous solution of $\ce{ZnSO4}$. To do this, we use the standard hydrogen electrode(SHE)-the reference electrode-as the anode and the zinc electrode as the cathode, and measure the emf of the cell.
The electrolyte corresponding to the SHE is a solution whose hydronium ion concentration is 1M. To construct the zinc half cell, we dip a strip of zinc in a 1M solution of $\ce{ZnSO4}$(for the STANDARD value). We connect the two electrolytes(of the zinc electrode and the SHE) with a salt bridge. Now, we measure the emf of this cell using a high resistance voltmeter, to find the potential difference between the electrodes without allowing any significant current through the cell.
When the zinc strip is dipped in the solution, it initially has no excess electrons on it. It is just a neutral piece of zinc metal. Now an equilibrium will be established between the solid zinc and zinc ions in the solution, finally leading to an excess of electrons on the electrode and a specific concentration of zinc ions in the solution. This, and a similar process at the SHE would cause an electrostatic potential difference between the two electrodes, since the relative positions of the two equilibria would be different. The voltmeter measures this potential difference to be the emf of the cell.
Here's my question: In setting up the two equilibria, won't the concentrations of the electrolytes change? Initially the zinc electrode had zero charge. But later it acquires a negative charge. This means that a certain number of zinc ions had to join the solution, leaving behind the electrons on the metal. So the concentration of zinc ions in the solution would now have to be more than 1M. So how would the STANDARD electrode potential be measured? How would the electrode potential be measured at ANY arbitrary concentration of zinc ions? It seems to me that the electrode potential can only be measured when the resulting concentration of zinc ions is equal to the equilibrium concentration. Does such a concentration even exist? Similarly, how is the hydronium concentration of the electrolyte corresponding to the SHE maintained at 1M?
According to me, the concentrations have to change for any net charge to develop on either electrode, hence establishing a potential difference. How does a net charge develop on each electrode without the ion concentration in the solution changing?

Your are really asking the wrong question. The real point is that with a modern high impedance voltmeter virtually no current has to flow to measure the voltage. So there is virtually no change in concentration at the cell's electrodes compared to the bulk solution. If you do draw a high current through the cell then the ions have a lower concentration at the electrode surface because diffusion limits how fast the ions can travel from the bulk of the solution to the electrode surface. So for high currents the EMF of a galvanic cell will drop.

A net charge does not need to develop for you to measure a potential difference. A potential difference by definition is the energy it takes for you to take a unit charge from point A to point B.
By the very act of putting a zinc strip into a solution of $\ce{ZnSO_4}$ you established an electrochemical system in which there is an energy difference associated with moving two electrons out of $\ce{Zn}$ to make $\ce{Zn^2+}$.
Think of how thermocouples work. Two dissimilar metals are put into contact and there is a measureable potential difference that is a function of temperature. You don't need to flow any current to establish a potential difference, nor do you need any charge imbalance.
You may be confusing a zero potential with the potential of zero charge. The potential of zero charge $E_{pzc}$ is the electrochemical potential where there is no net charge on the electrode surface. This potential has no connection to the equilibrium potential of a given reaction.

A voltimeter has very high resistance and consumes a very small amount of current, so if you make the measurement quickly, the concentrations are essentially constant for the duration of the measurement.

But there is no net charge on either electrode initially

So there won't be any potential difference. For charges to develop, the concentrations have to change.

If you're in a large enough system, the concentration change is so small, it would nearly impossible to measure it.


text:
Why do the physical properties of an egg shell change when the egg shell is exposed to vinegar for a week?
When an egg is kept in vinegar for one week, its hard calcium carbonate shell changes into a soft rubbery membrane. As vinegar is weak acetic acid, how does vinegar change calcium carbonate into a soft rubbery membrane? What is the reason for this?

It doesn't make calcium carbonate rubbery, it removes the calcium carbonate.  Egg shells are not purely calcium carbonate they are more like a composite with a continuous matrix of calcium carbonate and a smaller continuous matrix of protien.   When you put the egg in vinegar, you etch away this calcium carbonate matrix leaving the formerly less noticeable protein matrix as a gel filled with vinegar and creates a gel.

Also known as never use vinegar to clean your marble floors...

@JonCuster Unless you want to re-polish them yes.

@A.K. I guess your answer does not provide proper explanation regarding the protein part turning into rubber like mass.

As far as calcium carbonate is considered, it must be getting dissolved in the vinegar solution and left over is the protein matrix.
The concept of denaturation of proteins may be helpful to explain your so called phenomenon of formation of soft rubbery membrane.
A process that changes the physical and biological properties of proteins i.e.,  native state of proteins without effecting the chemical composition of the protein is called denaturation.
It can be caused by change in pH, temperature, presence of salts or certain other chemical substances.
Change in pH has the greatest disruptive effect on the hydrogen bonding and salt bridges in proteins. This denaturation results from the rearrangement of the secondary and tertiary structure of the proteins. The globular proteins present in egg shell change to a rubber like insoluble mass.
This is what happens to the egg albumin when kept in a vinegar solution.

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text:
Why magnesium has the lowest melting and boiling point in its group?

It is written in the third point ( shown in the picture) that presence of d orbitals results in stronger metallic bond.
1) We check the strength of metallic bonding to compare the melting and boiling point in case of metals. According to the general trend, melting and boiling point decreases down in the group( in s block) because the strength of metallic bonding decreases.

2) Presence of d orbitals decrease the size of atom thus increasing the electrostatic attraction between the delocalized electrons in the metal lattice and the cation and as a result, increasing metallic bonding. But magnesium IS IN A DIFFERENT PERIOD.
Can the ineffective shielding by d electrons be so much effective in decreasing the size of cation?
3) We didn't use any of this d orbital argument in comparing melting and boiling point of alkali metals. It decreased down the group following the expected trend.
How should  we explain this fact ?

Explanation based on d-orbitals is rather questionable. Do a search on SE for "magnesium melting point", this question has been asked numerous times.

@Nilay You have >3,000 rep, so you can already vote to close as a dupe by yourself. (Click on "close" next to "edit".)

I read some of the answers posted earlier on this topic, including the one that Nilay Ghosh mentioned but I don't really find them convincing. What should I do if I don't find the earlier answers convincing?

http://chemistry.stackexchange.com/questions/27143/why-does-magnesium-have-the-lowest-melting-point-of-all-earth-alkalis

@Blue Put a bounty on the existing question.

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text:
Error in calculation of rate of reaction through graphs
I'm asked to find the Reaction Rate when t = 120 seconds (which involves differentiation) for an unknown Chemical Equation. The rate of a chemical reaction is defined as:
$$R =\frac{dC}{dt}\ \text{(Differentiation of concentration with respect to time)}$$
I'm given a set of 11 points $(x=t\ (\text{time}),\ y=C (\text{concentration}))$ which I need to plot in order to find a general trend for them.
After plotting all points and observing the resulting graph, I tried performing multiple regression analyses in order to find a function $C(t)$ that best fits the plotted points.
For each of the Functions I obtained from each individual Regression...

I derived each one
Evaluated each resulting derivative for $t = 120$, in order to obtain
the reaction rate for each case.

Finally, I compared each result (Reaction Rate) with the Literature Value, and calculated an error percentage for every case.
These were my results:
$$\begin{array}{|c|c|}
\hline
\text{regression type} & \text{error percentage} \\ \hline
\text{exponential} & 26.92\% \\
\text{quadratic} & 22.89\% \\
\text{cubic} & 19.1\% \\
\text{quartic} & 1.39\% \\ \hline
\end{array}$$
My question is:
Many sources (including my book) claim that a High-Degree Polynomial Regression (Cubed Regression and above) is not appropriate for describing a Physical Phenomenon, even when the Regression Fit is perfect (R=1).
However, based on the table above, the most accurate answer I got was obtained through a Quartic Regression, while the rest had considerably higher error percentages.
Am I missing something or is it just an exception? Any clarification would be much appreciated!

Why are you trying to d a polynomial fit? That doesn't make any sense... For example, for a second order reaction, the integrated rate law would be $\frac{1}{\ce{[A]}} = \frac{1}{\ce{[A]_{0}}} + kt$. This isn't a polynomial...

Please check my updated post

How many parameters were in each fit?  My guess is exponential had two params, quadratic three, cubic four, and quartic five.  Is that right?  If so, can you guess why it gives a better fit?

Am I missing something?

Yes. You are missing the rate law and everything that goes along with it. It is only easy to fit zeroth order raw rate data. To produce linear fits, you need to consider the possibility that your reaction has a first or second order rate.
In addition to rate being defined as the (microscopic) change in concentration respect to time, you can also derive a rate law representing this change since the rate of the reaction is going to be proportional to things like the number of collisions between particles (which increases as there are more particles, i.e. higher concentration). So, we get a rate law that is something like the following:
$$\mathrm{rate}=kC^x$$
where $k$ is the rate constant, which is dependent on things like temperature and the activation barrier for the reaction, and $x$ defines the order of reaction, which is dependent on the mechanism and can be loosely interpreted as being related to the number of particles the must collide in the fundamental step(s). The value of $x$ is determined empirically (one such way is the experiment you did), though common values for simple reactions are integers, with $x=0,1,2$ being most common.
With both equations for rate in hand, we can do some to obtain integrated rate laws, which take the form $C(t)$: just the sort of equations that you are looking for. Note the minus signs in my equations. Rate is usually expressed based on consumption of reactants.

$x=0$, zeroth order

$$\begin{align}-\dfrac{dC}{dt}&=kC^0 = k\\
-dC &= kdt \\
-\int_{C_0}^{C} dC &= \int_0^t kdt \\
C - C_0 &= -kt \\
C &= -kt + C_0
\end{align}$$
A reaction with a zeroth order rate will produce a graph of $C(t)$ that is linear.

$x=1$, first order

$$\begin{align}-\dfrac{dC}{dt}&=kC^1 = kC\\
-\dfrac{dC}{C} &= kdt \\
-\int_{C_0}^{C} \dfrac{1}{C} dC &= \int_0^t kdt \\
\ln C - \ln C_0 &= -kt\\
\ln C &= -kt + ln C_0 \\
C &= C_0 e^{-kt}
\end{align}$$
A reaction with a first order rate will produce a graph of $C(t)$ that shows exponential decay. However, my derivation provides a linear equation: $\ln C = -kt + \ln C_0$. If you determine the natural log of each concentration and plot against $t$, you will get a line (if the reaction is first order).

$x=2$, second order

$$\begin{align}-\dfrac{dC}{dt}&=kC^2\\
-\dfrac{dC}{C^2} &= kdt \\
-\int_{C_0}^{C} \dfrac{1}{C^2}dC &= \int_0^t kdt \\
\dfrac{1}{C} - \dfrac{1}{C_0} &= -kt \\
\dfrac{1}{C} &= -kt + \dfrac{1}{C_0} \\
C &= -\dfrac{C_0}{1-C_0 kt}
\end{align}$$
A reaction with a second order rate will produce a graph of $C(t)$ that shows power decay. However, my derivation provides a linear equation: $\dfrac{1}{C} = -kt + \dfrac{1}{C_0}$. If you determine the inverse of each concentration and plot against $t$, you will get a line (if the reaction is second order).

This is the kind of Insight I was hoping for. My book does eventually talk about the Collision Theory and Reaction Orders; however, so far it has only mentioned the definition for Rate of Reaction shown in my original post and a few techniques on determining Reaction Rates Experimentally. The Exercise at hand was presented before all the topics you mentioned, so I suppose it's more of a generic example.

A quick question... In the Second Order C(t) function, wouldn't both negative signs actually be positive?

@Sam202 - do you mean that the final equation could be written $C = \dfrac{C_0}{C_0 kt -1}$?

I meant this way: C=Co/(Cokt + 1) because the integral of (1/C^2) gets rid of the minus sign outside the integral?

If you have $n$ points, you can always find an order $n-1$ polynomial that will fit your data. There will be no error in your measured set, but it is also very unlikely to have a physical meaning or describe reality.
Consider you take $k$ measurements. Regression analysis yields a polynomial
$$p(x) = c_{k-1}x^{k-1} + \ldots + c_1x + c_0.$$
The curve runs through every single value, so you assume it is a perfect fit. However, then you perform additional $m$ experiments. You now see that none of these are a good match for your function $p(x)$. To combat this, you find a new polynomial, this time of degree $k + m - 1$. Voilà! Error: $0$%.
Has the physical reality changed? Unlikely. So the actual underlying trend or trends are the same. We wish for our model to have predictive power (often with theoretical backing); something that seldom happens when we only seek the 'perfect match'.

As a sidenote, every experinent has an uncertainty in measurement. So even if the actual trend is linear, the higher degree polynomial might lead us to oscillatory nonsense.

So as far as theoretical problems and examples are involved, I can use the Ideal Polynomial Degree Regression for further calculations...  but as far as Real Experiments are concerned, I need to employ the k + m -1 degree Polynomial Regression for each successive experiment?

@Sam202 No, that was an example of why using high-order regression is not a good idea most of the time. To recap, the so-called ideal fit is not ideal at all; if you try to predict something with such a function, it will most likely give results with no meaning.

Hmm that's curious because I repeated the whole procedure for 3 different problems (asking for the same thing) and no matter what, the most accurate value I get for the Reaction Rate is always the one corresponding to the Quartic Regression Function. I'm puzzled.

Statistical evaluation
Other answers have done a good job of explaining the chemistry.  I wanted to add a purely statistical explanation of why your comparison is misleading.
It is misleading to compare fits solely by the error percentage.  The reason is that each model uses a different number of parameters.

Exponential fit

$$y = a \exp(b x)$$
This has two parameters, $a$ and $b$.

Quadratic fit

$$y = a x^2 + bx + c$$
This has three parameters, $a$, $b$, and $c$.

Cubic fit

$$y = a x^3 + bx^2 + cx + d$$
This has four parameters, $a$, $b$, $c$, and $d$.

Quartic fit

$$y = a x^4 + bx^3 + cx^2 + dx + f$$
This has five parameters, $a$, $b$, $c$, $d$, and $f$.
What happens as you add more parameters
John von Neumann, a well-known mathematician, famously said:
With four parameters I can fit an elephant, and with five I can make him wiggle his trunk.
Adding more parameters will always make a fit better, whether or not the parameters are physically meaningful.
An information criterion
Statisticians have devised various metrics to compare models with different numbers of parameters.  Some of them are the Aikake information criterion and the Bayesian information criterion.  The core idea of these metrics is the same -- penalize models with more parameters.  If a simple model with two parameters fits the data almost as well as a complicated one with five parameters, the simple model is likely a much better description of the data.
Those two information criteria are evaluated from the likelihood, a fancy statistical number which turns out to be closely related to the error that you have already calculated.  Let's take the example of the AIC:
$$AIC = 2k - 2\ln{L}$$
Here, $k$ the number of parameters and $L$ is the likelihood.  Lower AIC scores are better.  Unlike just comparing relative error, comparing AICs between models with different numbers of parameters is OK, because the AIC takes the extra parameters into account.
It's easy to compute these information criteria using modern statistical packages like R.  For example, here's some R code:
require(tidyverse)
require(broom)

conc <- c(0.2, 0.153, 0.124, 0.104, 0.09, 0.079, 0.070, 0.063, 0.058, 0.053, 0.049)
time <- c(0, 20, 40, 60, 80, 100, 120, 140, 160, 180, 200)

df <- tibble(conc=conc, time=time)

model_list <- list('quadratic' = conc ~ poly(time, 2),
                   'cubic' = conc ~ poly(time, 3),
                   'quartic' = conc ~ poly(time, 4),
                   'exponential' = log(conc) ~ time,
                   'second_order' = I(1 / conc) ~ time)

num_pts <- length(conc)
num_models <- length(model_list)

fit.results <-
    df %>%
    map(model_list, lm, .) %>%
    map_df(glance)

fit.results

That gives the following result:
$$
\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline
model & r.squared & adj.r.squared & sigma       & statistic   & p.value      & df & logLik   & AIC        & BIC        & deviance     & df.residual \\ \hline
quadratic & 0.9813919 & 0.9767399     & 0.007232647 & 210.9601    & 1.198971e-07 & 3  & 40.36382 & -72.72765  & -71.13607  & 4.184895e-04 & 8           \\ \hline
cubic & 0.9977167 & 0.9967381     & 0.002708474 & 1019.5748   & 1.323163e-09 & 4  & 51.90266 & -93.80533  & -91.81585  & 5.135082e-05 & 7           \\ \hline
quartic & 0.9997019 & 0.9995032     & 0.001057037 & 5030.5170   & 1.059279e-10 & 5  & 63.10056 & -114.20113 & -111.81376 & 6.703963e-06 & 6           \\ \hline
first-order (exponential) & 0.9634713 & 0.9594125     & 0.091628408 & 237.3814    & 8.937773e-08 & 2  & 11.78552 & -17.57104  & -16.37735  & 7.556189e-02 & 9           \\ \hline
second-order & 0.9999197 & 0.9999107     & 0.048309187 & 112032.5265 & 9.653383e-20 & 2  & 18.82683 & -31.65367  & -30.45998  & 2.100400e-02 & 9           \\ \hline
\end{array}
$$
This result shows you the AIC (and BIC) for each model.  You are right that the higher-order models fit better, but not by as much as you would think from looking at the relative error.  The quartic model as an AIC of -114, and the quadratic model has an AIC of -72.  So the quartic model is better (in a statistical, not necessarily chemical) sense, but only by twofold.  You have to penalize the extra parameters you gave it.
Chemical evaluation
The only two models which make chemical sense are the last two: an exponential fit (i.e. first-order reaction) and a inverse fit (i.e. second-order reaction).  Between those two, the second-order fit is much better than the exponential, with an AIC of -31 compared to -17 for the exponential fit.
Graph
There's no substitute for graphing your fits.
fit.data <- df %>%
    map(model_list, lm, .) %>%
    map_df(~augment_columns(data=df, x = .)) %>%
    mutate(model = rep(names(model_list), each = num_pts)) %>%
    mutate(.fitted = ifelse(model == 'second_order', 1/.fitted, .fitted)) %>%
    mutate(.fitted = ifelse(model == 'exponential', exp(.fitted), .fitted))

options(repr.plot.height=4, repr.plot.width=5)
fit.data %>%
    ggplot(aes(x=time, y=conc)) +
        geom_point() +
        geom_line(aes(x=time, y=.fitted, color=model)) +
        theme_bw()

ggsave('SEChem_65921.png', height=4, width=5)

Results in:

It shows the deficiency of the first-order (exponential) fit, as well as the deficiency with the quadratic fit.  It shows that the other models fit pretty well, but keep in mind that only some of these make chemical sense.
Conclusion
Curve-fitting and parameter estimation to fit chemical reaction data requires both statistical and chemical expertise.  You have to (a) make sure you don't give too much credit models with lots of parameters, and (b) make sure your fits make chemical sense.  For your data, a second-order fit seems best, because it (a) gives a good AIC, (b) has few parameters, and (c) makes chemical sense.


text:
Why is brine solution more effective than granular salt in melting ice on a surface?
Why is a brine solution consisting of a small fraction of salt in water, more effective in melting ice on a surface, than just pure salt?

greater surface area in contact with the ice.

This is because the brine solution is already in solution, whereas granular salt has to dissolve first. If ice is present instead of water, this is difficult, as there is no liquid water for the salt to dissolve in.
Secondly, salt dissolution is an endothermic process, so it will further reduce the temperature of the surroundings. For example, if you have an ice and water solution and add salt crystals you can bring the temperature of the ice/water combination to a much lower temperature (even below $\pu{-10^\circ C}$!).

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text:
Why is the melting point of tert-butyl alcohol 140 °C higher than that of sec-butyl alcohol?
This is one of the most drastic differences in a physical property I've ever seen for two such similar molecules, and in a simplistic sense anyway the difference lies in the opposite direction from what one might expect.  tert-butyl alcohol is completely sterically hindered from participating in H-O hydrogen bonding so it would seem to be limited to van-der-Waals interactions. sec-butyl alcohol interactions should be dominated by van-der-Waals forces, but should also have some ability to participate in hydrogen bonding.  Furthermore, sec-buty alcohol has a slightly larger van-der-Waals surface.
These are such common reagents and the unusually high melting point of tert-butyl alcohol is so well known that there is a surprising dearth of information on theoretical or experimental evidence available (in my searching anyway) explaining the reason for this $\pu{140^oC}$ disparity in melting points.
Does anyone know of a concise, coherent explanation for this observation?

Melting point is not just about intermolecular forces. The ability of a compound to form a regular solid can play a major role as well. Symmetry is often a deciding factor in otherwise similar molecules.

The most important point is that all C-C and C-O bond angles are fixed with respect to each other. Only the X-H bonds can move (OH and methly groups can rotate), and that is also possible in the crystalline state.

What gives you the idea that there can be no hydrogen bonding? Check the crystal structure of tBuOH and you'll find that you're wrong.

Two things to consider are that: tert-butyl alcohol has higher symmetry that sec-butanol and also sec-butanol is chiral.

Big kudos to all the symmetry and bond angle responses, I think that's right on track.  I have an idea along those lines that I've always thought was a bit nebulous which is why I've asked for a better explanation.  And Karl, yes I should have left out the word "completely" and just said "sterically hindered from participating in H-O hydrogen bonding".  It is there.  Still, that alone doesn't begin to explain the nearly 2-fold difference in absolute melting point temperature relative to sec-butyl alcohol.

My thoughts are along the lines of the high degree of symmetry as some have pointed out.  Specifically, how does this affect the entropy term of the Gibbs free energy of melting for these compounds?  The absolute entropy of both compounds in the solid phase should have relatively small and similar values.  In the liquid phase, the high symmetry of tert-butanol should result in a smaller change in entropy for the phase change as compared that of sec-butanol.  This is my nebulous theory eluded to above.  My hope is that someone can either better state this or poo-poo it with a better answer.

Almost forgot: big thanks to Loong for the beautifying edits.  I'm a noob who wasn't aware of how to format things properly...much thanks for the education there.

If you check the crystal structure, you will surely find that each OH group has two neighbouring OH groups to which it forms nice hydrogen bonds. And the gauche conformation of n-butane is ~4kJ/mol above anti, kT at 300K are 2.4 kJ/mol. Methinks that should be quite enough to inhibit crystallisation.

Another good point @Karl, though did you mean n-butane or n-butanol?  The melting points of n-butanol and sec-butanol are not notably different from each other, to the degree that each of them are from that of tert-butanol.  I'm having trouble finding a good image / discussion of the crystal structure of tert-butanol; could you provide a reference, preferably a web link, as my access to a good library is limited these days?  Much thanks for you efforts on my question.

I mean butane, but it doesn't  matter. t-butanole is the only one which has no gauche conformational isomer. The energy difference for all of them is not very different from than in n-butane. No sorry, i have no crystalline structure. But the high melting point of t-butanol is a dead-cert proof that there is strong hydrogen bonding.

I am not aware of a solid scientific publication on this matter, but the following points should elucidate what is happening:

The high melting point sort of proves that there is strong hydrogen bonding in tBuOH. I also don't see how the tBu-group could be so bulky that each OH group cannot form two hydrogen bonds.
n-, sec- and iso-butanol all have pairs of methyl- and/or OH-groups that can be in gauche or anti position with respect to each other. The  energy difference is in the range of 4 kJ/mol, which is not much higher than the thermal energy $k_BT$ at 300K. Those alcohols exists more or less as a mixture of conformational isomers, which precludes crystallisation. In tBuOH, there is only one conformation.
Rotation of the methyl groups has a low excitation barrier, but it can also rotate in the solid phase (there is some tunneling involved), so this does not preclude crystallisation.

(The energy of ~4kJ/mol is from n-butane, but I expect no large difference. The gauche conformation of the OH group is probably even lower, energetically.)

If you want to ponder this look up the Boiling Points and Melting Points of the isomeric pentanes and butyl alcohols; their shapes are similar; an oxygen atom with two unshared pairs substituting for a carbon with two C-H bonds. [Should you want to complicate things a bit throw in the 4-carbon ethers or ketones too.] Their Van der Waals forces are reasonably similar and the alcohols add hydrogen bonding. A careful perusal will give a reasonable idea of what is going on.
At the phase equilibrium Delta G = 0 = Delta H - [T Delta S]; T = Delta H/Delta S.  This means an analysis of intermolecular forces, really heats of fusion and vaporization, and the probable entropies of melting and vaporization will give the trends in the transition temperatures.
For simple molecules linear molecules have more surface area and less conformational interactions so Van der Waals forces are greater and H-bonding is less restricted and MPs and Bps should be raised. This is moderated by a greater entropy change from solid to liquid since many rotational modes are freed up. Secondary and iso molecules should have similar entropy changes and lower Heats so should have lower MP and BP. The very symmetrical tertiary isomers have very different heats of melting because of lattice arrangements, although a lower entropy change should also contribute, and lower vaporization energies because of less surface area and possible inhibition of H-bonding for steric reasons. The entropy changes seem to be lower. Symmetrical molecules have higher MPs and lower BPs.
These are general trends that I have surmised. As molecules become more complicated the analysis must become more detailed. This is shown in liquid crystals, additional liquid phases; polymers, glass transition phase changes; and of course, proteins, enzymes and other biomolecules that have exquisite foldings and conformations.

Will someone please give me the reason for the down vote.

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text:
Can the enthalpy change of a reaction be changed, and if so, will this affect the value of the equilibrium constant?
In the case of an arbitrary equilibrium reaction, is it possible to change the absolute enthalpy of the final products, thus the reaction enthalpy change? For example, say pH affects the charge distribution of the product molecule, changing the attractive force between it and the solvent (feel free to come up with a better example).
Given that this does in fact affect $\Delta H$, would the Gibbs free energy change of the reaction $\Delta G=\Delta H-T\Delta S$ be altered as well, or does the entropy change somehow adjust automatically?
If $\Delta G$, changes, I would assume $K$ changes as well through $\Delta G = - RT$ $\mathrm{ln} K$

Short answer: No and yes.
Enthalpy is a state function. That means that it depends only in the current "state" - the identity of the substances, their concentrations, pressures, etc. That means that the absolute enthalpy of some substance is independent of how you produced it. However, that substance has different enthalpy in different states: gas versus liquid, pure versus in solution. If you change the state of the product(s), you with change their enthalpy. Same goes with any of the other thermodynamic potentials: entropy, internal energy, free energy.
And that of course will alter the equilibrium situation.

The equilibrium situation only, or the constant as well?


text:
Degradation of sulfuryl fluoride SO2F2
Recently, the houses next to us were fumigated with "Vikane", which apparently is $\ce{SO2F2}$. To my surprise, the gas was just pumped out in the atmosphere without further ado at the end of the procedure. I was told by the pest control people that this gas is "degraded by light" and that there is no hazard whatsoever for humans living close to the house. However, I've also read that sulfuryl fluoride is a potent greenhouse gas staying in the atmosphere for up to 40 years which it probably wouldn't do if it were deteriorated easily by light. Which is it??

I think this pretty much comes down to three things:

The estimates of the atmospheric half-life of $\ce{SO2F2}$ has
changed over time.
The current estimate of 30-40 years isn't a particularly long time
relative to some other players (i.e. methane's atmospheric lifetime
is around 6-8 years and that of carbon dioxide is around 50 years,
though there is poor agreement on that one).
The emissions relative to the other greenhouse gas players is small
enough to offset the fact that it is "potent" in this aspect.

From sulfuryl fluoride Wikipedia page:

Based on the first high frequency, high precision, in situ atmospheric
and archived air measurements, sulfuryl fluoride has an atmospheric
lifetime of 30–40 years, much longer than the 5 years earlier
estimated.

That addresses my first item.  From the same page regarding my second and third items:

Sulfuryl fluoride has been reported to be a greenhouse gas which is
about 4000-5000 times more efficient in trapping infrared radiation
(per kg) than carbon dioxide (per kg). It is important to
note, however, that amounts of sulfuryl fluoride released into the
atmosphere (about 2000 metric tons per yr) are far lower than the
amounts of $\ce{CO2}$ released by hydrocarbon-burning vehicles, industry, and
other processes (about 30 billion metric tons per year).

That means that annual carbon dioxide emissions will trap 3000 times as much heat as will sulfuryl fluoride.  So the pest control people may not have worded it quite right, but the bottom line is that sulfuryl fluoride emissions have a negligible effect on the planet's heat budget.


text:
Preventing build-up of Diethyl Ether Peroxides
I've never stored Diethyl Ether before but am thinking of producing minor amounts myself and storing said Ether inside of glass/PTFE bottles.
I am aware of possible pressure build-up due to vapors and storage temperature, but what always puts me off is the formation of explosive peroxides.
Now as the above Wikipedia article mentions:

Diethyl ether hydroperoxide decomposes in the presence of sodium hydroxide and $\ce{Fe^2+}$ -containing salts.

This is obviously stating that I can stop the peroxide formation simply by adding in some $\ce{NaOH}$ which I have an abundance on hand.  However that above information is regarding hydroperoxides. The less nasty of the two peroxides.
Ethylidene peroxides are the explosive ones that detonates by simply touching it. This is what scares me. There's no information regarding the stopping of formation on the page, and frankly my google searches for "Safe storage of Diethyl Ether" and "Stop Diethyl Ether Peroxide formation" seem to come up with "put some iron wire in it" or Potassium Hydroxide.
So my question is: what is the most reliable, safe, and long-term solution to prevent building up of both of Diethyl Ether's peroxides, especially the Ethylidene peroxide, with the least amount of contamination possible.

Where does the wikipedia article say etyhlidene peroxide is not reduced with Fe2+?

@Karl Well for me, the real question was, where does the article say it IS reduced? Nowhere.

You state that iron salts like Fe2O3 should help. This is incorrect, since this contains basically Fe3+ only, which cannot reduce the peroxides (that's the manner in which the peroxides are rendered safe). Moreover, Fe2O3 is basically insoluble. In my experience, the canonical reagent for the reduction is FeSO4.

@TAR86 Thankyou for the correction, far out I'm making some errors today! I did see somewhere that Ferrous Sulphate was the way to go but then again, I made this question for a reason. I'll edit it out.

There's no information regarding the stopping of formation [of ethylidene peroxide] on the page...

The required information is given implicitly. The wikipedia page states

Diethyl ether peroxide, also known as ethylidene peroxide, [...] is a polymerization product of diethyl ether hydroperoxide

With other words:
If you prevent the formation of diethylether hydroperoxides by

storing diethylether in a brown bottle over sodium hydroxide

you will automatically prevent the formation of subsequent products, such as the ethylidene peroxide.
Note that there is a different approach, which is recommended in terms of workplace safety:
Substitute diethylether with methyl tert-butyl ether (MTBE), which does not form peroxides.

Ah, I didn't see the polymerization piece.... My bad. :P

Why the hydroxide?

@Karl If I remember correctly, the hydroxide deprotonates the hydroperoxide and renders it insoluble.

What if we just used Ethanol?

@frayment What do you mean? Substitute diethylether with ethanol?

@KlausWarzecha Sorry for the vagueness. Instead of adding in Sodium Hydroxide, will Ethanol act as a deprotonating substance or atleast have the same effect? In other words, is it the same reason I see people adding Ethanol that your telling me to add Sodium Hydroxide?

@frayment No, that wouldn't work. The hydroxide will convert the hydroperoxide to an insoluble precipitate. Moreover, you might run into layer separation problems with water if your diethyl ether does contain (too much) ethanol. In summary, I'd suggest to use MTBE nowadays, if possible.

Sure, the hydroperoxide, if already present, precipitates. Ok, that means  you don't have it in your work if you take ether from that storage bottle, which is of course good. Just have to be careful with the dry NaOH/NaOOEther mixture when the bottle gets empty.

@Karl True. That's why the use use of test strips, which you mentioned in your answer, is crucial.

Do they show anything when the hydroperoxide has precipitated? I guess it's more of an alternative: test strips or NaOH.

@Karl Honestly, I don't know. But better safe than sorry.

The ethylidene peroxide is the polymerisation (polycondensation actually) product of the hydroperoxide. Firstly that only occurs when the solution gets concentrated, i.e. when the ether and residual water evaporates. Secondly, it is also destroyed by Fe(II) just as the hydroperoxide, and thirdly, you must store ether away from light anyway, and then you should never get peroxides. Have your peroxide test strips at hand and check from time to time, and IF they give a positive result, add the reducing agent.
https://en.wikipedia.org/wiki/Diethyl_ether_peroxide

We are medical lab doing occasional water:ether extractions. We have always had the practice of adding deionized water to diethyl ether, and drawing ether from the top of this stock. This apparently removed some minor impurities in the diethyl ether stock able to dissolve in water. I was taught this is a safer way to work with ether, but have never seen this mentioned in documentation from any supplier, such Sigma or any publication or educational text. We only have small amounts and try to not keep it around very long, so I have never been able to determine what the water addition does regarding peroxide formation over longer storage (years).


text:
le chateliers principle
One of le chateliers principle is effect of concentration change :

When the concentration of any of the reactants or products in a reaction at equilibrium is changed , the composition of equilibrium mixture changes so as to minimize the effect of concentration of changes.

So to get more products we need to decrease concentration of products. For decreasing concentration of products can we do it by adding something(example water) because it will also decrease"concentration". But I have not seen like this in my book. So can we?

Adding solvent will decrease the concentration of the products and the starting materials.


text:
Advantages of Travelling wave IMS and Drift tube IMS
I am learning Ion Mobility spectrometry. I am sorry for a very fundamental question. After reading different articles, I really did not understand what is the advantage of Travelling wave IMS over Drift tube IMS (or vice versa). Can anyone explain very fundamentally, what are the adavantages/disadvantages of TW IMS over Drift Tube IMS (and vice versa)?

In a nutshell, since other sources probably would give much more accurate and complete answers:
Advantages of DT-IMS over TWIMS:

Direct access to absolute collision cross sections. This is a major advantage considering the difficulty to provide accurate CCS calibration scales for TWIMS measurement as can be seen in the recent litterature.
Separative resolution is usually reported as higher, although recent work by Dick Smith's group at PNNL seems to show that it is mostly related to separation device lengths which are different.

Advantages of TWIMS over DT-IMS:

Reduced ion losses and increased sensitivity. The stacked-ion ring basis leads to reduced drift towards the sides. This advantage has been reduced by the addition in several DT-IMS setups of ion funnel electrodes which focus the ion trajectories upon reaching the end of the drift tube.
No need for large potentials and thus large voltage generators and all the related problems. To reach a given electric field, the potential difference does not need to be applied to the whole drift tube.


text:
What happens chemically to 5'-radiolabeled aqueous thyroxine (thyroid prohormone) when its I-131 undergoes beta+gamma decay?
Related to this question about $\ce{^{14}C}$ beta decay in $\ce{CO2}$, what happens when the $\ce{^{131}I}$ in radiolabeled thyroxine (thyroid prohormone, or "T$_4$") undergoes beta-plus-gamma decay to form $\ce{^{131}Xe}$?
In order to narrow the scope of the question to something tractable, assume that despite its low aqueous solubility the thyroxine is in dilute solution in deionized water, and that it bears a single $\ce{^{131}I}$ at the $5'$ position (click image to enlarge):

As well, as noted above, assume that the $\ce{^{131}I}$ undergoes its most common mechanism of tandem electron & gamma decay:
$$
\ce{^{131}_{53}I -> ^{131}_{54}Xe + \beta^-~+ \gamma}
$$
I assume the resulting chemical species remaining will start out as a (very transient) xenon-bearing cation:

What happens next?

It is highly unlikely that your organic xenon species will be formed as there will be a recoil of the $^{131}Xe$ atom which is on the order of ~50 eV.  This is minor in terms of the energy released in beta decay so its often ignored, but it will be larger than the bond energy of the C-I bond (<10 eV) so when the iodine decays the resulting daughter will have enough kinetic energy to recoil out of the bond.

Is it supposed to be beta+ or beta- decay? Because you are talking about beta+ but write a beta- decay...

@deusexmachina beta-minus decay. If I understand you correctly, I think I phrased the text poorly. See my edit.

So, what does happen to the rest of the molecule?

When the beta decay occurs it is likely that the C-I (or now C-Xe) bond will break. . The electron emitted in the beta decay will fly away from the molecule, the positively charged xenon atom will now pick up an electron from the aqeuous solution.
The Szilárd-Chalmers effect needs to be considered, it is typcial that when a radioactive decay occurs that the bonds to the atom at which they decay is will be broken. The recoil energy from the beta emission should be able to break the C-I bond.
I suspect that the bond will break to form a carbon centred aryl radical. This is a very high energy organic radical which will be likely to react with dissolved oxygen to form a Ar-O-O. radical. This in turn will abstract a hydrogen from another organic molecule to form Ar-O-O-H.
When this decomposes the O-O bond will break thus forming ArO. and HO. The ArO. will abstract a hydrogen from another organic solute to form a phenol. I know that 8-hydroxy-20-deoxyguanosine (8-OHdG) in urine is associated with exposure to radiation in humans. See S. Kato et. al. JOURNAL OF PEDIATRICS, 2015, volume 167, pages 1369 onwards. Also see the work of Martin Tondel et. al., ARCHIVES OF ENVIRONMENTAL CONTAMINATION AND TOXICOLOGY, 2005, volume 48, pages 515-519.
This work on this DNA marker for radiation damage does to my mind support the hypothesis that the autoradiolysis of the thyroxine will create hydroxy compounds. When S. Cefova et. al. irradated emulsions of nitrobenzene and water they did observe the formation of nitrophenols which lends further weight to the idea that the product will be a phenolic compound.

My guess would be the following:
The electron generated by the decay would most likely go either towards the O or the H. In either case, the oxygen (due to its electronegativity) will atract the electron forming an ion, and the hidrogen will be displaced.
The oxygen will form a double bond and stabilize. The hydrogen will attack the xenon and take it's place, as the molecule will be more stable that way. The xeonon will just float off.
That's what seems logical to me that will hapen in an aqueous solution.
What will happen in the human body might be different, as different molecules will most likely be in the vicinity of the thyroxine molecule.
Below is a crude drawing in paint in order to help you visualize.

The velocity of the emitted $\beta$ particle (or the electron) would be too large for the attraction towards $\ce{O}$ that you mention


text:
E2 step in dehydration of primary alcohol using concentrated sulfuric acid
The familiar two-step $\mathrm{E2}$ mechanism of the dehydration reaction of ethanol using concentrated sulfuric acid is given below (forgive me for not using diagrams):

$\ce{H-CH2-CH2-OH + H2SO4 -> H-CH2-CH2-OH2+ + HSO4-}$
$\ce{HSO4- + H-CH2-CH2-OH2+ -> H2SO4 + CH2=CH2 + OH2}$

I have certain doubts about the viability of the second step, as $\ce{HSO4-}$ is a weak base.
Maybe $\ce{H2O}$ is used as the base in the second reaction instead? If so, is the catalyst regenerated at all?

What actually happens, as always, depends on the exact conditions and concentrations which you did not add to your question. However, you do state whether the catalyst is regenerated; as such, I assume you are talking about the following reaction:
$$\ce{C2H5OH ->[\ce{[H2SO4]}] C2H4 + H2O}\tag{1}$$
I.e. a dehydration of ethanol using catalytic sulfuric acid.
The first thing we have to consider is the $\mathrm{p}K_\mathrm{a}$ values of sulfuric acid and ethanol (and water; which will be present anyway unless you accidentally found dry sulfuric acid). Using a very quick ballpark, we can assume: $$\mathrm{p}K_\mathrm{a}(\ce{H2O}) \approx \mathrm{p}K_\mathrm{a} (\ce{C2H5OH}) > \mathrm{p}K_\mathrm{a}(\ce{H2SO4})\tag{2}$$
Since sulfuric acid is catalytic, we can assume that the first (complete; non-equilibrious) step will be:
$$\ce{C2H5OH + H2SO4 -> C2H5OH2+ + HSO4-}\tag{3}$$
$\mathrm{p}K_\mathrm{a,2} (\ce{H2SO4}) \approx 2$, which is only two units less than water. Therefore, — under the assumption that we added $10~\mathrm{mol}\text{-}\%\ \ce{H2SO4}$ — we might assume that some additional molecules of ethanol (and/or water) are protonated under equilibrium conditions:
$$\ce{C2H5OH + HSO4- <<=> C2H5OH2+ + SO4^2-}\tag{4}$$
Thus, part of the sulfuric acid will be present as sulfate, the majority as hydrogen sulfate. Sulfate is a stronger base (albeit still weak) and thus the final mechanistic step is probably better written using sulfate rather than hydrogen sulfate:
$$\ce{H-C2H4-OH2+ + SO4^2- -> HSO4- + H2O + C2H4 ^}\tag{5}$$
Of all the bases present at equilibrium, sulfate is the strongest as its conjugate acid’s $\mathrm{p}K_\mathrm{a} > 0$. The general order of basicity of all compounds in solution is:
$$\ce{SO4^2-} > \ce{H2O} \approx \ce{C2H5OH} > \ce{HSO4-}\tag{6}$$
Finally, note that the reaction does not have to proceed via $\mathrm{E2}$ — an $\mathrm{E1}$ mechanism is almost equally viable.

What do you mean by the proceeding of reaction via E1 mechanism also? Almost all books mention that dehydration of primary alcohol proceed via E2 mechanism.


text:
Alternative to OLI simulation software
I work with modeling a number of aqueous solutions, but mostly it boils down to single-component saline solutions. The industry standard for determining various saline solution quantities such as conductivity, activity, and density as a function of salinity and temperature is OLI.
OLI also provides capabilities that are way over the top for what I'm looking for (e.g. corrosion, scaling), and I'm wondering if an alternative exists that is less capable and hopefully less costly.
I have experience with writing and compiling software, so user-friendliness is not that important for me.

If all you are interested in are conductivity, density and activity coefficients, I would say you need lookup tables instead of a software package.  I don't know the details of the internal workings of OLI, but I would be willing to bet it is some look up table from a reference book for the kinds of values you are looking for. You may want to check the help file/documentation for OLI for such references.


text:
Among HClO4 and HClO2 which is the stronger acid if we do not know Ka values?
I have two acids: $\ce{HClO4}$ and $\ce{HClO2}$ how do I determine which is the strongest acid without knowing $K_a$? The two hydrogens are bound both to an oxygen so there isn't anything to differentiate them, how can I  proceed?

related http://chemistry.stackexchange.com/questions/40676/relative-strength-of-strong-acids

$\ce{HClO4}$  is stronger acid due to the formation of more stable conjugate base. Also there are more number electronegative oxygen atoms in case of $\ce{HClO4}$ , so acidity should be more.


text:
Why does chloromethane have a larger dipole moment than chloroform?

Why does $\ce{CH3Cl}$, methyl chloride, have a larger dipole moment than $\ce{CHCl3}$, chloroform?

Let us consider $\ce{C-Cl}$ bond moment to be $x$ and that of $\ce{C-H}$ bond to be $y$.
In $\ce{CCl4}$, the dipole moment of any three $\ce{C-Cl}$ atoms is balanced by the dipole moment of the fourth $\ce{C-Cl}$ bond dipole moment. This means the resultant of dipole moment of three $\ce{C-Cl}$ bonds in tetrahedral structure is equal to $x$, which is equal and opposite to the dipole moment $x$ of the fourth $\ce{C-Cl}$ bond.
Now, when we replace one of the chlorines with a hydrogen, the net dipole moment equals $x+y$, (as they are in the same directions). Similarly if we carry out the same procedure for $\ce{CH3Cl}$ it can be proved that its dipole moment is also $x+y$.

Where have I gone wrong?

http://chemistry.stackexchange.com/questions/48066/logical-way-to-compare-dipole-moment

@Mithoron the discussion on the first answer is on the same topic but hasn't been concluded, and I am not able to get the answer from it...could you help?

Bond lengths aren't constant.

@Mithoron At least the $\ce{C-Cl}$ bonds seem practically constant ($177~\mathrm{pm}$) – L. E. Sutton, L. O. Brockway, J. Am. Chem. Soc. 1935, 57, 473.

tetrahedral structure is a 3D structure, not a 2D one

Your conclusion is built on the assumption that the bond lengths will be constant. A bond’s dipole moment varies with bond length and thus for the vector addition to give the identical product, all $\ce{C-Cl}$ and all $\ce{C-H}$ bonds would have to be identical. They’re probably not.
Sutton and Brockway studied the bond lengths in chloromethane, dichloromethane and chloroform back in 1935 by electron diffraction.[1] Their results indicate that the $\ce{C-Cl}$ bond lengths don’t vary between the compounds in question as indicated in the table below. (Note that all chlorine atoms in each molecule are homotopic and thus equal bond lengths are theoretically expected.)
$$\begin{array}{lccc}\hline
\text{Compound} & \ce{CH3Cl} & \ce{CH2Cl2} & \ce{CHCl3}\\ \hline
d(\ce{C-Cl})/\mathrm{pm} & 177 & 177\phantom{^\circ} & 178\phantom{^\circ} \\
\angle(\ce{Cl-C-Cl}) & \text{n. a.} & 111^\circ & 111^\circ \\ \hline\end{array}$$
These results point towards the chlorine side of things not having an impact. But there is still the hydrogen side of things. Hydrogen atoms are generally very hard to locate in diffraction type experiments. X ray diffraction is observed if photons interact with electrons in a certain manner; hydrogen atoms do not have many electrons around them and are thus hardly seen. One needs to observe diffraction at very large angles to be able to locate hydrogens. I assume, although I don’t know, that it’s similar in electron diffraction. Indeed, the paper notes:

[…] and $l_{\ce{C-H}}$ was taken as $1.06~\mathrm{\overset{\circ}{A}}$.

The two carbons in chloromethane and chloroform are, however, very much not identical. One is surrounded by three electronegative chlorine atoms which will strongly remove electron density from it, polarising the $\ce{C-Cl}$ bonds towards each of the three chlorines. The other will have only one such withdrawing partner. The different electron densities around chlorine are well represented by the corresponding $\ce{^13C}$ NMR shifts: $\delta(\ce{CH3Cl}) \approx 22~\mathrm{ppm}$; $\delta(\ce{CHCl3}) \approx 77~\mathrm{ppm}$.[2] The hydrogen shifts are also different in similar magnitudes, showing that hydrogen in chloroform is much more polarised than in chloromethane.
These different electronics must have an influence on the $\ce{C-H}$ bond length of the two molecules. I can’t imagine two electronically such different systems to have equal bond lengths. My assumption is that $\ce{CH3Cl}$ should have the shorter $\ce{C-H}$ bond lengths, since carbon is more negatively charged while hydrogen remains positively charged; this should give a slightly greater attractive force and hydrogen, being such a small element, immediately feels these. Therefore, instead of the final dipole vector $\vec d$ being equal in both cases ($\vec d = \vec x + \vec y$), we have one lesser and one stronger vector.
$$\begin{gather}\vec d_{\ce{CH3Cl}} = \vec x + \vec y + \delta\vec y\tag{1}\\
\vec d_{\ce{CHCl3}} = \vec x + \vec y - \delta\vec y\tag{2}\end{gather}$$

References:
[1]: L. E. Sutton, L. O. Brockway, J. Am. Chem. Soc. 1935, 57, 473. DOI: 10.1021/ja01306a026.
[2]: Experimental spectra found for chloromethane/chloroform on SciFinder.

Since you say that C-H bond length is less in CH3Cl then why did you add a ${\delta\vec y}$ in it? shouldn't it be subtracted?

@Lalit $\delta y$ need not be positive.

Ah Ok that's the catch, thanks for replying

Citing from another answer:

[Jan] Your conclusion is built on the assumption that the bond lengths will be constant.

There is another factor: the magnitude of the partial charges. Your conclusion is built on the assumption that no matter the context, a carbon-chlorine bond will always have the same partial charge separation (or said in another way, the same electron distribution).
However, the chlorine atoms in chloroform come pretty close, 294 pm, which is closer than the sum of the van der Waals radii.
Using molcalc (not sophisticated DFT methods, but easy to access), I get the following partial charges on chlorine and hydrogen:

molecule
calc dipole (debye)
partial charge, chlorine
partial charge, hydrogen

CH4
0
./.
+0.04

CH3Cl
1.73
-0.11
+0.06

CH2CL2
1.67
-0.05
+0.09

CHCl3
1.19
+0.01
+0.13

CCl4
0
+0.06
./.

The positive partial charge on chlorine in chloroform is surprising, but a non-quantummechanical estimator (https://acc2.ncbr.muni.cz/) also gives a partial charge that is slightly positive (partial charge of chlorine in chloroform given as +0.012, as compared to in chloromethane as -0.013).

Because they are vectors and as such are not in the same direction within the molecule. from your example 4 x Cl atoms could be said to have dipole x - x = 0. You are effectively replacing 'x' with the dipole 'y' so the overall dipole for CCl3H would be 'y - x'. Similarly CH3Cl would be 'x - y' (so same magnitude but opposite direction).
What isn't taken into account with your model is steric deformation of the molecules. The three chlorine atoms missing their fourth to force a tetrahedral arrangement will adopt a much more planar arrangement; thus their contribution to the overall dipole would be reduced. The deformation and thus dipole in CH3Cl with the introduction of a single bulkier group would be less and therefore the overall dipole would be greater

Not opposite directions!

The deformation is just by 1 degree so why is there so much difference in dipole moment

If we consider deformation the dipole moment of $CH_3Cl$ is $x+1.05y$ and that of $CHCl_3$ is $1.05x+y$ which is very small and cannot account for the large difference in dipole moment.

It is a vector, parallel to the bond axis.This vector can be physically interpreted as the movement undergone by electrons when the two atoms are placed a distance d apart and allowed to interact, the electrons will move from their free state positions to be localised more around the more electronegative atom.
you can compare the dipole moment of both the molecules by these digrams. source:wikipedia.
CHCl3

CH3Cl

For polyatomic molecules there is more than one bond, and the total molecular dipole moment may be approximated as the vector sum of individual bond dipole moments. Often bond dipoles are obtained by the reverse process: a known total dipole of a molecule can be decomposed into bond dipoles. This is done to transfer bond dipole moments to molecules that have the same bonds, but for which the total dipole moment is not yet known. The vector sum of the transferred bond dipoles gives an estimate for the total (unknown) dipole of the molecule

you can reason like this:  the electron cloud is distributed in case of CHCl3.

But what is wrong in the way I have calculated it?

the way you add dipole.... i didn't understand, that's not the way it is. vectors cannot be added algebraically. like the way you did.

okay, does the resultant of the dipole by 3 C-Cl point downwards? Is it parallel to the 4th C-H bond moment? Net dipole moment is the vector addition of resultant of 3C-Cl and bond moment of C-H. Both these vectors are parallel and in the same direction. So how else do you add them?

yes you are true the resultant is downward but please see the figure it is clear that they make a large angle measured from downward. so every components get added up and creates a single dipole but still it not more than the dipole moment of one C-Cl bond.  i urge you please apply vector you will understand everything.

i think you are only confused because of the diagram you have made. search google for real diagrams.

i think i understand you. so you think like this :  in CH3Cl there is one chlorine adding to dipole moment. and In CHCl3 there is one Cl atom to add to dipole and other two must add to first and net dipole should increase, am i right? but this is only according to the diagram you've made. there's no chlorine atom opposite to H atom. again see diagram.


text:
Reduction of oximes with lithium aluminium hydride

I know that $\ce{LiAlH4}$ is good a reducing  agent and it donates hydride ions but then I am not able to understand the mechanism of this reaction.
I attempted this way

I am not understanding what to do with the extra charge on nitrogen as there is no acidic medium so that I can donate $\ce{H+}$ to it or can remove ($-\ \ce{OH}$) by any protonation.

You are leaving out one positively charged species that remains after it looses it's hydride ion.

You might be saying about Li+ and it will get donated by N-. But how will it help in getting there the group (-NH2)

How about LiAlH3+ ?

If LiAlH3+ is formed it will get donated by N- but what about further reaction? How will it help to get -NH2 there?

Comprehensive Organic Synthesis, 2nd ed. writes (emphasis mine)1

The reduction of oximes or oxime ethers and esters to primary amines is a useful synthetic transformation that is used as a substitute for direct reductive amination to convert ketones or aldehydes to primary amines. The reaction involves reduction of C=N bond and reductive cleavage of N–O bond. Thus, the transformation requires the use of strong hydride reagents or reagent
  combinations.

I would guess that after the addition of hydride to the $\ce{C=N}$ bond (incidentally, you should not draw it as free hydride) the next step is a substitution at nitrogen. $\ce{X-Y}$ $\sigma^*$ orbitals ($\ce{X},\ce{Y}$ being heteroatoms) tend to be low in energy and rather electrophilic, so this is probably no different.
If it helps you to feel better about the negative charges, you should probably draw the nitrogen as being coordinated to a lithium and the oxygen as being coordinated to an $\ce{AlH3}$ fragment, or some permutation thereof. The close association of the heteroatoms with metal fragments also encourages "hydride" to attack. Again it's not really hydride attacking but rather the $\ce{Al-H}$ $\sigma$ bond.

1 Abdel-Magid, A. F. Reduction of C=N to CH–NH by Metal Hydrides. In Comprehensive Organic Synthesis, 2nd ed.; Knochel, P., Molander, G. A., Eds.; DOI: 10.1016/B978-0-08-097742-3.00802-8.

The intermediate hydroxylamine is further deoxidatively reduced to the amine.  LAH is a blunt instrument; aluminum(III) is a potent oxophile.  Cf: ketone or aldehyde plus amine plus cyanoborohydride in mildly acidic buffer.  It only takes a tickle to reduce a Schiff base.


text:
Free radical vs radical cation
What is the difference between a free radical and a radical cation?  Are they different thing or do they mean same?

The difference is one electron ;-)

@KlausWarzecha Not always: radical dications...?

Related: The answer to http://chemistry.stackexchange.com/questions/9579 gives one example for the generation of a radical cation and how it is converted to a radical.

A radical cation is simply a free radical with a positive charge. Similarly a radical anion is a free radical with a negative charge.
In other words, radical cations and radical anions are subsets of the free radicals.
That said, the term free radical itself is not entirely clear cut, as illustrated in this question.


text:
Calorimetry, finding final temperature
I was looking for some help on understanding this practice question from my text book, I know how to figure out enthalpy, but this is my first temperature change question and I am stumped. Any guidance would be greatly appreciated.

Nitric acid is neutralized with potassium hydroxide in the following reaction:
$$\ce{NHO3(aq) + KOH3(aq) -> KNO3(aq) + H2O(l)}$$
$$\Delta_\mathrm rH= -53.4\ \mathrm{kJ/mol}\ \ce{HNO3}$$.
$55.0\ \mathrm{mL}$ of a $1.30\ \mathrm{mol/L}$ solutions of both reactants, at $21.40\ \mathrm{^\circ C}$, are mixed in a calorimeter. What is the final temperature of the mixture? Assume that the density of both solutions is $1.00\ \mathrm{g/mL}$. Also assume that the specific heat capacity of both solutions is the same as the specific heat capacity of water. No heat is lost to the calorimeter itself.

I know the answer is 29.7 degrees Celsius, I'm just not sure how to get there. I tried juggling around my equation "delta t r H= m c delta t" but since I don't know the final temperature i don't think the delta t works in this case. I've tried omitting the delta t but I get no where near 29.7 degrees Celsius.
So I took my formula Q=mc delta t and rearranged it, I now have delta t= Q / mc. Delta t= -53.4 kJ/mol / (55.0g)(4.19 J/g C) I changed the 55mL to g as the question says to assume the density is 1.00 g/mL. I also used the heat capacity as water given in my course data booklet). And I get -0.2317. I tried doubling the mass as there is two solutions but that doesn't work either

How many moles of each reactant are in the reaction mixture?  Are you aware that a negative heat of reaction means that heat is given off?

Thank you for bringing up moles!! Once I figured out the moles I was able to use my equation. I will post my answer right away. I know my editing skills need work but thanks Loong for fixing my question.

n(HNO3)= 0.055L * 1.30 mol\L = 0.0715 mol
Delta H = 0.0715 mol * -53.4 kJ/mol =3.818kJ
Delta H = mc delta t
3818J = (110g)(4.19J/g*C)(delta t)
Delta t = 8.3*C
Ti + Tf = 21.4*C + 8.3*C = 29.7*C


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What is the difference between a Chemical Formula and Formula Unit?
In my textbook, a chemical formula is described as :

"A formula that shows the kinds and numbers of atoms in the smallest representative unit of a substance"

While a formula unit is described as :

The lowest whole number ratio of ions in an ionic compound.  Thus the formula unit for sodium chloride is NaCl.

Though the definitions seem to differ, the written formulas appear identical.  A chemical formula appears to be another way to represent a formula unit. For example, NaCl is both the chemical formula and formula unit for Sodium Chloride.
Is there ever a difference between the written terms?

Molecular compounds such as water exist as discrete particles, molecules. This is due to the forming of covalent bonds where each atom has a specific partner to which it is bonded. Each molecule of water contains one atom of oxygen and two atoms of hydrogen. So, H2O is its formula. This can be more specifically called a molecular formula.
In ionic compounds such as table salt, NaCl, the atoms (as ions) do not bond to specific neighbors. Surrounding each chloride ion in a salt crystal are six sodium ions. Likewise, six chloride ions surround each sodium ion. This attraction of oppositely charged ions extends throughout the entire crystal. There is no discrete bonding of a particular sodium ion to a specific chloride ion. So the formula for sodium chloride is expressed as the smallest whole number ratio between the Na and Cl ions which is 1:1 for sodium chloride. So, NaCl is also a formula but not a molecular formula.
The reason for the term "formula unit" is that it is useful when we talk about how much of one substance is required to combine with a particular amount of another substance. For example: to make the smallest amount of hydrogen carbonate we combine one molecule of water with one molecule of carbon dioxide.
But now suppose we want to combine silver nitrate with sodium chloride. Both of these are ionic compounds, they do not form molecules. In this case you would say that one formula unit (FU) of sodium chloride combines with one FU of silver nitrate.
The one-to-one combining in both these examples is simply coincidence. It may take 3 FUs of substance A to combine exactly with 2 FUs of substance B in a different example.


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What is the difference between O & O2
What is the difference between $\ce{O}$ and $\ce{O2}$.
If C is carbon and then why $\ce{O2}$ is oxygen.

O is oxygen atom. O2 is dioxygen molecule.

Indeed. As elements on the periodic table, $\ce{H}$ and $\ce{O}$ are hydrogen and oxygen, respectively. When these are chemical species in a reaction, these ought to be called monohydrogen and monooxygen (sometimes the double 'o' is dropped in English). Systematic names for $\ce{H2}$ and $\ce{O2}$ are dihydrogen and dioxygen. The latter two are still acceptably named hydrogen and oxygen (via 2005 Red Book), and commonly used for convenience.

The common confusion here is that two very different things have the same name.  $\ce O$ is a free oxygen atom and $\ce{O2}$ is two oxygen atoms chemically bound to form an oxygen molecule.
There is no common analogy for $\ce C$, but $\ce{N2}$ is called nitrogen, $\ce{H2}$ is hydrogen and $\ce{Cl2}$ is chlorine, each having the same name as that of their constituent elements.
So, to reiterate, the difference between oxygen $\left(\ce O\right)$ and oxygen $\left(\ce{O2}\right)$ is that the former is an oxygen atom while the latter consists of two $\ce O$ atoms bound together, forming a molecule also called oxygen.

The element just has the same name as the most stable chemical modification. Nothing "unfortunate", "historical" or even ''inconsistent'' about that.

@Karl, I disagree. The term "stable" refers to the state under an arbitrary set of conditions, basically STP, etc. at the surface of our planet, and thus an arbitrary naming system.  This naming system goes back to the element's / molecule's original discovery and, for example, I doubt oxygen gas, nitrogen gas, etc would be given the same names as their constituent elements today.  There is also the chicken-and-egg ambiguity here. Are the elements named after the molecules or vise versa.  To be consistent, the answer must be the same in all cases, from hydrogen to oxygen to bromine, and so on.

All the diatomar elementar gases are thermodynamically stable from absolute zero to > 1000K and vacuum to really high pressures. Nothing "arbitrary" about that. Except for carbon, ALL elements have the same name as their most stable modification.

While I remain in at least partial disagreement with some of @Karl's points, the arguments themselves illustrate that my original answer was more complex and tangential to the question than necessary.  I have edited my answer accordingly.

Sometimes we use terminology a little loosely.
When we are just talking about the element, then just using the symbol by itself is clear. But sometimes we need to describe how the element appears in the world or in chemical reactions. Then it isn't enough to describe just the element, we need to know something about how it is found under normal conditions. Oxygen is usually found as a diatomic gas (which is why we write O2). Since this is by far the commonest way we find free oxygen in nature we often describe it this way anyway unless there is a reason not to.
Nitrogen is also mostly found in pure form as a diatomic gas (N2). Carbon, however, is usually found as a solid and never as a simple molecule (diamond and graphite are both covalently bonded solids) sort is rarely useful to describe its normal molecular form as there isn't one. We might talk about sulfur as S or, if we care about the allotrope we might specify S8, though there are others common in the lab.
It matters little which version you use to describe the element. But, if you are talking about reactions, it is usually worth describing the molecular form of the element you are talking about. Oxygen isn't always O2 but can be formed (in the upper atmosphere or in some reactions) as O3.
Chemists mix and match their terminology somewhat freely when it doesn't matter much, but try to be as specific as possible when it does.

O is a singular oxygen atom. O2 is a molecule made of two oxygen molecules. Our atmosphere is made of mostly O2, which our bodies have evolved to breathe. We breathe in O2 and cellular respiration requires it (C6H12O6+6O2→6CO2+6H2O).

O is oxygen, while O2 is dioxygen.

Examples of other oxygen molecules are: 03 which is ozone (trioxygen), there's O4 that is metastable and is actaully made up of two O2 molecules (tetraoxygen), and O8 which is a form of pressurised solid O2 (when you freeze O2 to below around 55 Kelvin and pressurise to around 10 GPa (1.45e+6 psi)it becomes known as ε oxygen or red oxygen due to the color changing from blue to red).

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Our atmosphere is mostly nitrogen, not oxygen.

Context is important. C is the atomic symbol for carbon and can represent not only the element, (but one or more) atoms of the element, as well as the element as it exists in the understood (implied) conditions. Same is true for O or N or actually any element. However, elemental O and N are both commonly found as diatomic molecules O2 and N2. I will turn your question around: since oxygen and nitrogen are commonly represented by their ground state chemical compounds (at STP) why isn't carbon represented by Cgraphite? (and why isn't sulfur represented as S8?). The reason is fairly straightforward: it is because the actual compound we have in front of us may be difficult to determine (in the case of C or S), while there's generally no question about O2 or N2 being the compound we have. So, we use the atomic symbol to represent the elemental material when it is either of indeterminate or of variable composition. Carbon, in particular, is often found or used as allotropes different than graphite. So we are as specific as we can be, without risking being wrong. (If we specify O2 then sure, there might be some trace amounts of O3 or even O. around, but we assume those complications are (also) understood.) It is much harder to do that with solid elements, so we avoid being too specific (remember, as scientists, we should be able to justify our claims, including the nature of the materials we are discussing) by using the atomic symbol. Liquid bromine, for instance might be represented as either Br(l) or Br2(l) although the second is the ground state elemental form (at STP). {Just to correct one of the answers, N2, and not O2, composes over 70% of our atmosphere by volume.}

Symbol of an element only represent one atom of that element.The chemical formula of an metal element is same as its symbol however this is not true for non metals as they exist as covalent molecules.The chemical formula for an element takes in to consideration of the different stable compositions or naturally occurring forms of that element.This is why hydrogen has the chemical symbol  ( $\ce{H}$) and chemical formula as $\ce{H2}$. Terminology is important to understand the difference between element symbol and its stable composition.It is important to note that even though atomic symbols and it's naturally occurring formula are two different things the name used to describe the two things might be same. For example oxygen symbol is  $\ce{O}$  and  $\ce{O2}$ which is its naturally occurring form is also referred to as oxygen and sometimes oxygen molecule which is the correct way to describe it.


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By what mechanism does the reaction between NCl3 (Nitrogen Trichloride) and NOCl (Nitrosyl Chloride) occur?
The reaction produces 2 Cl2 and 1 N2O. Thank you for your help!

One can rarely be sure when it comes to inorganic reactions, but I think that this would be a reasonable mechanism:

There are also other possibilities.


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Precise definition of atomic orbital
In school I was taught that an atomic orbital is the 3-dimensional region in which the electron is located with a probability of 90%.
However, aren't there infinitely many regions of space in which the total probability of finding the electron is equal? And if so, wouldn't that imply that it is not possible to assign a shape to an atomic orbit as there are infinitely many shapes with equal likelihood to contain the electron?

I think you are confusing atomic orbitals with the graphical representation of atomic orbitals. Also I wouldn't worry too much about orbitals. It could be the single most overused concept in chemistry thanks to a poor understanding by most people that argue with them. But if you are really interested, you can read up on the solution of the Schrödinger equation for hydrogen-like atoms and Hartree-Fock theory, but this is complicated stuff if you're still in school

In school I was taught that an atomic orbital is the 3-dimensional region in which the electron is located with a probability of 90%.

This is not correct.
An atomic orbital is a solution to Schrödinger's equation given the electrostatic potential. In the simplest case, we solve the one electron atom and perturb that with more electrons. An exact solution to the multielectron problem is intractable due to the chaotic nature of 3-body systems.
The concept you are referring to is a boundary surface. (See this question). Basically, it's a surface that has equal probability density for every point on the surface, and the surface encloses a region of space where the sum of probabilities is the desired value.
Because the surface is defined to have equal probability everywhere, and the wave function is smooth, it should be unique.
The boundary surface is usually how we represent the atomic orbitals that are familiar to us ($s, p, d, f$) but also molecular orbitals. Usually, we take something at 90% probability, for example, and plot the boundary, giving the orbital "shape." Otherwise, visualization can be tricky because the orbital only goes to zero probability at infinite distance from the nucleus, so there is a nonzero chance that an electron will be quite far from the nucleus.

I see, so the crucial condition I was missing is that every point on the boundary surface has the same probability density. Thank you :)

Electrons are in quantum mechanics described by wavefunctions $\Psi(\vec{r})$. These are complex physical objects which contain all the information about the state of the electrons. The square of the wave-function $|\Psi(\vec{r})|^2$ gives the probability of finding the electron at position $\vec{r}$. As you said orbitals are often considered as the smallest volume in which the electrons are expected to be with a 90% probability. To construct these one constructs isosurfaces of the mentioned probability density and tries to choose the isosurface with the smallest volume which however still includes 90% of the electrons, i.e. the integral over the probability density over the volume of the isosurface is 90%.
Of course you could construct as many possible shapes as you want, you don't have to rely on isosurfaces. However isosurfaces provide much more physical insight, since they also include information about the gradient of the probability density function which is always perpendicular to the isosurfaces. So from just looking at isosurface you get a good insight into how the function actually looks like. This information is included in atomic orbitals. While when you would choose random shapes it would give you pretty much zero information...

Thank you for this interesting and more mathematical explanation :)

According to a stone-age paper, Chem. Rev., 1986, 86 (4), pp 681–696,

Atomic orbitals are widely referred to by chemists, but are actually only loosely defined. Hydrogenic or- bitals are too different from the optimum orbitals for describing an atom to furnish a suitable definition. Slater originally introduced STO’s as approximations to atomic orbitals when the parameters n and {where chosen by certain empirical rules. Consequently, STOs are still frequently confused with atomic orbitals. In this paper we will use the term “atomic orbital” to refer to a Hartree-Fock orbital resulting from a converged (i.e., basis set limit Roothaan-Hartree-Fock or nu- merical SCF) calculation on some state of the atom with spin (singlet, doublet ,... ), symmetry (S, P ,... ) and equivalence restrictions (e.g., px = py = pz). Unfortu- nately, this definition is still somewhat state dependent as the 2p Hartree-Fock orbital for the carbon s2p2 3P ground state is not identical to the 2p Hartree-Fock orbital for the s1p3 configuration average.

That information was already well seasoned then. The work of Slater, Hartree and Fock is mostly from the 20ies to 50ies, none of them was even alive any more in 1986!


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Why is it that a little detergent cleans the clothing wash better than a lot of detergent?
People are accustomed to put a large amount of detergent in the washing
machine to do their laundry thinking that if there is not enough it
won't clean the soiled clothing.  But I have read that too much detergent
will not clean as well as putting in a smaller amount. Why is that?
If the clothing is really soiled, you would think that a lot of detergent
will attract more dirt and oil particles rather than a smaller amount of
detergent.

I suspect that the required detergent increases fairly linearly with the amount of organic soiling present.  It may be that large amounts of detergent molecules just adhere to themselves, possibly causing the detergent required for cleaning to preferentially associate with the excess detergent.  However, I suspect this is all just some urban legend and that excess detergent is just wasteful but still cleans your clothes.  Where did you read this claim?

The amount of detergent you should use varies with the hardness of the water you feed your washing machine.

If your book stated that using a small amount of detergent would remove heavy soiling better than using a larger amount for a single wash, then I say it's bunk.
If your book stated that using too much detergent on a regular basis could result in a dirtier wardrobe, then there is a good explanation for that.
Let me use a more common but analogous process to explain: carpet cleaning. Though detergents generally work best in aqueous solution, residual, dried-up detergent still tends to bind to organics (or various types of undesirable "dirt") pretty well.  So when one has a difficult to remove stain, they clean it repeatedly with highly concentrated detergent solutions that do not get thoroughly rinsed out of the carpet. Then, even if the stain is removed, that area quickly becomes soiled again due to the residual detergent sticking to more dirt than the rest of the detergent-free carpet.
Tying the carpet-cleaning analogy to laundering clothing is then pretty straight-forward.  If one always uses more detergent than can be reasonably removed during the rinse cycle, then their clothing will tend to soil again more easily than if they had used an amount that could be largely removed during the rinse cycle.  Thus, regularly laundering with too much detergent could result in a dirtier wardrobe than using a smaller amount of detergent.


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Are photolysis reactions considered exothermic or endothermic?
I know that photolysis is when a chemical reaction (usually decomposition) is forced by photons. I also note that there is a general warming in the stratosphere, due to the photolysis of ozone $$\ce{O_3 + $h\nu$ -> O + O_2}$$ and the subsequent restorative mechanism $$\ce{O + O_2 + M ->O_3 + M}$$ where $\ce{M}$ is a third body used to carry off excess energy. To me, that says that the photolysis of ozone is an endothermic reaction, and the reformation of ozone is an exothermic reaction, as the free body takes away excess energy (assumably thermal energy).
Am I correct in generalizing that photolysis is generally an endothermic reaction? If so, could the enthalpy released by the second equation be quantified by the frequency of the photon?

In general it could be either. In your example you will have to work out the heat of formation of ozone and of O atoms (presumably it is an endothermic reaction). In general what the photon is doing is providing a way to overcome an activation barrier in the ground state by opening up a pathway from an excited state to the products. This last step is generally exothermic.
The enthalpy released in the second reaction is less than the photon energy by the difference in energy of the photon and that of products, an oxygen atom in your case as the heat of formation of O$_2$ is zero.
The sketch gives the general idea, I hope, for an endothermic reaction :)

Notes: The barriers may also be called transition states. There may be no barrier in the excited state. The crossing from the excited state potential to the product may be controlled by Landau-Zener behaviour or be a conical intersection. The far right upper line (not labelled) is the potential of the excited state of the product


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Comparing basicity using inductive effect
In the following two compounds .
1)
2)
We have to compare basicity .
According to me CH$_3$ is an electron donating compound so in second compound it will increase charge density in nitrogen so second compound should be more basic.
But in the book it is written first is more basic . How

@KlausWarzecha but in that question the compounds are different

Yes, the specific compounds are different. The basis for making a decision  is not.


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How to speed up corrosion of magnesium?
For an experiment, I need a corroded magnesium, but I got fresh mill cutoffs. Is there I way to speed up the corrosion process?
Actually would like to simulate a crushed magnesium rod, that spend some time in nuclear reactor. So I would expect just some thin corrosion (oxidation) layer, basically I need to get right colour.

Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using LATEX syntax. If you receive useful answers, consider accepting one.

Could you give a little more information regarding the required physical characteristics you need?  Can it be a fine or coarse powder, flakey, clumpy, or does it need to somewhat maintain it's current shape and form?  Is it possible to be too corroded for your purposes, i.e. mostly converted all the way to magnesium oxide?  It might be simplest to just explain what you intend to do with it.

Actually would like to simulate a crushed magnesium rod, that spend some time in nuclear reactor. So I would expect just some thin corrosion (oxidation) layer, basically I need to get right colour.

You are looking for passivation? That is probably incredibly difficult to control. Speeding it up might just lead to burning it. If you just need the colour, paint it white.

Actually a "thin" layer will form rapidly on exposure to air.  You should see the fresh millings go from shiny to grey.  You could heat it in air until you think it has the coloring you are looking for.

I am going to take a bit of a jump and assume you mean you want $\ce{MgO}$, Magnesium Oxide. This should be the main product of it's corrosion in air.
If you want to speed it up merely burn it. It burns very brightly and hot, so be careful; however it is not too unsafe, just use common sense. Proper safety should always  be followed.
$$\ce{2Mg + O2 -> 2MgO}$$
Is the reaction occurring. After it has fully burnt, you will be left with magnesium oxide, what I assume you mean by corrosion.
This will result in a white powder, as corrected by Ivan.

Burning of magnesium never results in a solid chunk. Instead, you'll have a lot of fine white powder all over the place.

@IvanNeretin You are completely right. I haven't done the demonstration in a while and was thinking of sparklers.

Thanks, But not really what I need, I have updated the question...


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Type of analysis for water
What other analysis suitable for water as general? Right now, I have turbidity, pH, Total Soluble Solid and color.

Hardness
TC Total Carbon

TIC Total Inorganic Carbon
TOC Total Organic Carbon
VOC Volatile Organic Carbon

COD Chemical Oxygen Demand
BOD Biochemical Oxygen Demand

@anonymous My pleasure. If the anwer was really helpful, please consider to accept it.

Also trace metal concentrations, disinfectant byproducts, fluoride, nitrite, nitrate.  I'm kind of presuming drinking water with these though.

@airhuff Absolutely! I only listed  some parameters for which wikipedia pages with further explanation exist.


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Opening a shaken up bottle quickly = no fizz?
I recently watched this video:
https://www.youtube.com/watch?v=Lqk3lGic3D0
Summary: Opening up a shaken bottle of Carbonated drink extremely quickly results in almost zero bubbles forming in the drink and then the soda is 'flat'
How exactly does this work? The extremely fast sudden opening seems to causes the CO2 to react differently than normally expected?
Thanks.

The trick is simply waiting for the CO2 to be released ahead of time.
For a full and cold soda this trick will not work, the pressure in the head-space will be slightly lower due to the lower temperature and with less total volume of CO2 -which makes the "pop" part of the trick fail- and the CO2 initialization points within the solution will be distributed -which causes the fizzing and overflow of liquid.
With a half filled soda that's been sitting out in a comfortably warm room, the soda is warmer. This means that you are working with a (even more) super saturated solution of CO2 in water and acid. This additional instability combined with a greater head-space for pressurization allows a greater build-up of CO2 in the bottle allowing for the "pop".  If you didn't wait for the bubbles to go away you would still have fizz formation (and even flatter soda). If you did wait for the bubbles to disappear but you opened the bottle slowly, the soda would still be just as flat as with the pop-trick but with a slight hissing sound and limited additional bubble formation.


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How to reduce Al2O3 in a H2 atmosphere on industrial scale?
As known, reducing $\ce{Al2O3}$, which is the main process of the aluminium production, isn't really cheap, mainly because its energy need. Its high energy need is coming from the high binding energy between $\ce{Al}$ and $\ce{O}$.
What differs from the common $\ce{Al}$ production: the whole process has to run in $\ce{H2}$ atmosphere, where the cost of the $H_2$ gas is zero.
Thus, we need a net result of $\ce{Al2O3 + 3 H2 \rightarrow 2 Al + 3 H2O}$.
Does it make the process easier? How could it work?
Remark: actually, the goal would be to extract the $\ce{O2}$ and to produce water from it, while we have a quasi-infinite $\ce{H2}$ source. The $\ce{Al}$ production here isn't important.

The question came into my mind while thinking on terraforming.

Also if this was viable, it would be probably used.

No, in the real life the goal of the reduction is to produce Al. Here to goal would be to produce water. Furthermore, in the normal circumstances, there isn't a quasi-infinite H2 source, it had to be made from water decomposition, which is also costly. I think you can have right if the H2 production is much cheaper as the Al production (I don't have here pretty data about this).

There's even policy of not using mathjax in titles, but because of searching issues. Some elements are produced via H2 reduction, but it would be hard to turn  the equilibrium of this reaction backwards IMO.


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Good sources to understand Stat Mech with concepts
I never had a proper course on Statistical Mechanics during my undergrad. I've tried picking up the subject by reading books, but I find that many focus on the mathematical derivations and I get easily lost in the details.
I was wondering if someone could point me to some useful reference materials where I could better understand concepts that are confusing to me. For example, why do chemical systems overcome energy barriers? I know that when a molecule samples a potential energy surface, the different states are occupied with probabilities relating to their energy difference according to the Boltzmann law. But, in classical physics, a ball rolling down a hill would eventually get stuck in the energy minima? I fail to see where do chemical systems get the energy from to be constantly in motion.
Sorry if this is a vague question or ill-posed, but as you can see I have some confusion of concepts in my mind which is not allowing me to develop a more accurate question.

Without a mathematical grounding, you're going to have a hard time understanding the concepts. Words don't describe quantitative relationships like equations can.

there is an MIT course https://www.coursera.org/learn/statistical-mechanics on coursera. It might be worth it look into it.

I highly recommend Molecular Driving Forces by Dill & Bromberg.
https://books.google.com/books/about/Molecular_Driving_Forces.html?id=hdeODhjp1bUC
Contains many examples throughout to help with conceptual and mathematical understanding.


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Mathematical model for reaction rate in fixed volume
Is there a way to find the rate of reaction of a chemical reaction inside a closed container that is fixed in volume as a function of time?
Can the answer provide an example reaction (such as baking soda + vinegar $\rightarrow$ carbon dioxide + "the other stuff")?

Welcome to Chemistry StackExchange. The standard methods are stopped flow, flash photolysis, which is also called 'pump-probe', and temperature and pressure jump methods. These are usually described in physical chemistry textbooks.

You could also used spectroscopy depending upon whether or not chromophores are available.

so basically you have to find the function experimentally. I was wondering if you could derive it through mathematical models.

@user1945827 , the methods I mention normally use spectroscopy to measure changes as time proceeds.


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why molecules gain energy when heated?
molecules in solid or liquid or gas gain kinetic energy when heated. why is it like this? I request explanation  in details please

Kinetic energy of molecules is heat. Why the molecules gain heat when heated, really?

Though this question may seem simple or self explanatory to many, I try to support others in their curiosity :)

@IvanNeretin, kinetic energy of molecules together with their potential energy is internal energy, heat is one of the means of changing  this energy.

Heat is a form of kinetic energy.  Thus, the heating of a molecule is simply the transfer of energy from the heat source to the molecular bonds and electrons.

I would be careful in calling heat "a form of energy". It is one of the two means of energy transfer for the case of a closed thermodynamic system (with another one being work). Or, the amount of energy spontaneously flowing from/to a closed system due to a temperature difference between the system and it surroundings by means other than through work.

Then, transferring heat changes internal energy of a system, which, in fact, is the sum of the kinetic and potential energies of the particles that form the system.