text:
Unit of the equilibrium constant: contradiction of Bridgman's theorem?
The following equation is standard in thermodynamics:
$$
\Delta G^\circ=-RT\log(K)
$$
where $K$ is the equilibrium constant.  In dimensional analysis, Bridgman's theorem tells us that the argument of a transcendental function (like $\log$) must always be dimensionless.  But $K$ may have dimensions (depending on the particular equilibrium).  Why is this OK?

Note: working out dimensions explicitly gives:
\begin{align}
[R]&=EN^{-1}\Theta^{-1}\\
[T]&=\Theta\\
[\Delta G^\circ]&=EN^{-1}
\end{align}
where $\Theta$ is the dimensions of temperature, $N$ is the dimensions of number and $E$ = dimensions of energy = $MLT^{-2}$.  From these, we see that the quantity $\log(K)$ ought to be dimensionless, implying that $K$ should be dimensionless as well.  But it isn't always.
Furthermore, suppose $K$ has dimensions $NL^{-3}$ (for an equilibrium of the form $A+B\leftrightarrow AB$, say).  Suppose now that we scale the units for number by a factor $a$.  Then we get new values for $K,R,\Delta G^\circ$, given by:
\begin{align}
\hat K&=K/a\\
\hat R&=aR\\
\hat{\Delta G^\circ}&=a\Delta G^\circ
\end{align}
From our original equation:
\begin{align}
\Delta G^\circ&=-RT\log(K)\\
a\Delta G^\circ&=-aRT\log(K)\\
\hat{\Delta G^\circ}&=-\hat RT\log(a\hat K)\\
\hat{\Delta G^\circ}&=-\hat RT\log(\hat K)-\hat RT\log(a)
\end{align}
So the new quantities do not satisfy the old equation.  Rather, they satisfy the old equation, but with a constant factor of $-\hat RT\log(a)$ added on.  What is going on here?

The problem is that people are often sloppy with the definition of quantities. The equilibrium constant $K$ in your first equation is indeed a dimensionless quantity while the equilibrium constant $K_c$ that is usually used to describe an equilibrium in a solution is not. I will take some detour to show where they come from and how they are connected.
From thermodynamics it is known that the Gibbs free energy of reation is given by
\begin{equation}
 \Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ ,
\end{equation}
where $\xi$ is the extent of reaction and $\nu_{i}$ and $\mu_{i}$ are the stochiometric coefficient and the chemical potential of the $i^{\text{th}}$ component in the reaction, respectively.
Now, imagine the situation for a ideal system consisting of two phases, one purely consisting of component $i$ and the other being a mixed phase comprised of components $1, 2, \dots, k$, in equilibrium.
Since the system is in equilibrium and shows ideal behavior we know that the chemical potential of component $i$ in the mixed phase (having the temperature $T$ and the total pressure $p$), $\mu_{i}(p, T)$, must be equal to the chemical potential, $\mu^{*}_{i}(p_{i}, T)$, of the pure phase having the same temperature but a different pressure $p_{i}$, whereby $p_{i}$ is equal to the partial pressure of component $i$ in the mixed phase, namely
\begin{equation}
 \mu_{i}(p, T)= \mu^{*}_{i}(p_{i}, T).
\end{equation}
From Maxwell's relations it is known that
\begin{equation}
 \left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = \left( \frac{\partial}{\partial p} \biggl(\frac{\partial G^{*}}{\partial n_{i}}\biggr) \right)_{T} = \Biggl( \frac{\partial}{\partial n_{i}} \underbrace{\biggl(\frac{\partial G^{*}_{i}}{\partial p}\biggr)}_{=\, V_{i}} \Biggr)_{T} = \left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}
\end{equation}
but since $\mu^{*}_{i}$ is associated with a pure phase, $\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$ can be simplified to
\begin{equation}
 \left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i}
\end{equation}
and one gets
\begin{equation}
 \left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = v_{i} \ ,
\end{equation}
where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase.
Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get
\begin{equation}
 \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p   \ ,
\end{equation}
so that, introducing the mole fraction $x_{i}$,
\begin{equation}
 \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .
\end{equation}
Please, note that there is a dimensionless quantity inside the logarithm.
Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$
\begin{equation}
a_{i} = \frac{\varphi_{i} p_{i}}{p^0}
\end{equation}
where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition.
When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises
\begin{equation}
 \mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .
\end{equation}
Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets
\begin{equation}
 \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,
\end{equation}
where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure.
Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that
\begin{equation}
\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .
\end{equation}
So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant
\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .
\end{equation}
which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via
\begin{equation}
 K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .
\end{equation}
it is more or less proportional to $K$ and thus gives qualitatively the same information.
Edit: If the solution at hand behaves like an ideal solution then by definition it's activity/fugacity coefficient is equal to one. Furthermore the state of ideality is defined with respect to standard states: for an ideal solution this is $c^{\ominus} = 1 \, \text{mol}/\text{L}$. Using this together with the ideal gas law on the relation between $K$ and $K_{c}$
\begin{equation}
 K = \prod_i [\underbrace{\varphi_{i}}_{= \, 1}]^{\nu_{i}} \Bigl(\underbrace{\frac{RT}{p^{0}}}_{\substack{= \, 1/c^{\ominus} \, = \, 1 \, \text{L} / \text{mol} \\ \text{per definition}}}\Bigr)^{\sum_i \nu_{i}} K_{c} \qquad \Rightarrow \qquad K = \left(\frac{L}{\text{mol}} \right)^{\sum_i \nu_{i}} K_{c} \ .
\end{equation}
one sees that for an ideal solution $K_{c}$ is identical to $K$ scaled by a dimensional prefactor.
Edit: I forgot to mention there are also "versions" of the equilibrium constants that are defined in terms of partial pressures or mole fractions which provide a more suitable description for gas equlibria but all those "versions" can be traced back to the original equilibrium constant.

OK, thanks very much.  But why then do I see the equation $\Delta G^\circ=-RT\log(K_C)$ (as I do at https://en.wikipedia.org/wiki/Nucleic_acid_thermodynamics#Thermodynamics_of_the_two-state_model)?

@Donkey_2009 As I said $K_{c}$ is easier to work with and since it is more or less proportional to $K$ it gives you qualitatively the same information if the systems behavior is not too far away from the ideal behavior. But, I'd say strictly speaking the equation you quote should be incorrect but another opinion on that matter might be worthwhile. Maybe under normal conditions this equation is good enough. Unfortunately, I lack the practical expertise to say something about that.

@Donkey_2009 A little late but... because most chemist do not care about math and formalism. That equation have not mathematical sense since logarithm of anything with units is simply not defined. But normally people does not care and just neglect the units and takes logarithms. It is simply wrong, but in most practical cases works, so nobody cares.

(this is a comment to Phillip answer; upon editing it failed to get saved on due time)
Basically $K_c$ is valid as long as the solutions show colligative properties. These "concentration" constants can (and should) be defined dimensionless, as they properly emerge as a limit for the dimensionless activity in "ideal" systems, and the activity $a_j$ measures how the chemical potential $\mu_j$ of the substance $j$ changes as its concentration (or in general the chemical composition of the system) changes from some reference state: $a_j = [c]_j/c^\ominus$ where $[c]_j$ is the molar concentration and $c^\ominus = 1$ M.
Therefore to remember that inside the logarithm everything is dimensionless one should not forget to write the reference magnitude $c^\ominus$. These constants can use different units for the concentration too, for instance $m_j/m^\ominus$ where $m_j$ is the molal concentration and the reference magnitude $m^\ominus$ is $1$ mol/kg.
Regarding the limits of validity, $K_c$ fails miserably for electrolytes, where the true thermodynamic constant depends on all the charges in the system (ionic strength of the solution), that is, the activity of the electrolyte is not only proportional to its number (concentration, etc.).

Thanks.  If you change the reference magnitude, then the numerical value for $K_c$ changes by a constant multiplicative factor, which means that the expression for $\Delta G^\circ$ changes by a constant additive factor.  Does this mean that $\Delta G^\circ$ is only defined up to addition of some constant related to the choice of $c^\ominus$?

Following my recipe, $K_c$ is dimensionless. Take the chemical reaction: aA $\rightarrow$ bB. Using the molar scale, $K_c = ([B]/c^\ominus)^b / ([A]/c^\ominus)^a$. There is no free multiplicative factor. You can use $K_p$ (based on partial pressures) or $K_x$ (based on molar fractions), etc. These $K$'s are different. The relation is such that $\Delta G^\ominus$ is the same. Because $\Delta G^\ominus$ is an energy difference you cannot have any additive factor. $\Delta G$ (Phillip's answer) is a derivative with respect to the advance of reaction, $\xi$, and there is no additive factor either.

I understand that your formula for $K_c$ is dimensionless in the way the term is commonly understood.  However, it relies on the (somewhat arbitrary) choice of $c^{\ominus}$ to make it work - choose a different value for $c^{\ominus}$, and we get a different numerical value.  What do we call a dimensional quantity with an arbitrary fixed value that we measure other quantities against?  Oh yes - a unit.  Calling it $c^{\ominus}$ rather than $M$ doesn't change that.

@Donkey_2009 The choice of $c^{\ominus} = 1 , \text{mol}/\text{L}$ is not entirely arbitrary. In thermodynamics there is a model system called the ideal solution, which is (in essence) defined as a solution having a standard concentration of $c^{\ominus} = 1 , \text{mol}/\text{L}$ but behaving like an infinite dilute system, i.e. no interactions between the solute molecules. It's the analogue of the ideal gas for solutions and yields analogues equations making it very nice to work with. Though the definition shows that it's a very crude model (as is the ideal gas law) it nevertheless...

...describes the behavior of very dilute solutions quite good. But as @perplexity said this doesn't hold for electrolytes and rather concentrated solutions because there the interaction between the solute molecules is too strong to consider the system as "infinite dilute". But if the conditions for the ideal solution model are nearly fulfilled $K_{c} \approx K$ (see edit in my answer).

Thanks.  So what would be different if the concentration of the ideal solution had been defined to be, say, $c^\ominus=2\textrm{ mol}/\textrm{L}$ instead?  Would it be $\Delta G^\circ$ that changed?  Is $\Delta G^\circ$ only defined up to choice of $c^\ominus$?

@Donkey_2009 I'm a bit on slippery ground here since I don't work in the field and my knowledge is thus not particularly detailed. Nevertheless, I think the following is true: The standard state in solution is defined to be an infintely dilute system with a standard concentration of $1 , \text{mol$/$L}$. If you have a compound in its standard state, then the activity is equal to $1$ by definition. From this definition then results a certain value for $\Delta G^{0}$ as it is simply a value of $\Delta G$ for special conditions. Therefore I would say, ...

...if you change the definition of a standard state, e.g. by changing $c^{\ominus}$, then $\Delta G^{0}$ should change as well (as you said). This means, of course, that you change your $K$ values as well, because the values of the activities (for a certain temperature, concentration etc.) is changed by the choice of the standard state.

Exactly as Phillipp said. You have to look at his answer. The most important equation is (slightly edited -I skipped the often negligible pressure dependence if there are no gases): $\mu_{i} (T,\mathrm{composition}) = \mu_{i}^{*}(T,\mathrm{reference , composition})+ RT \log a_{i}(T,\mathrm{actual , composition})$. Summing over all the substances $i$, the first term gives you $\Delta G^\ominus$. You can obtain it by using thermodynamic tables, as the reference composition is always based on a solutes independent of each other...

The second term gives you the changes of the chemical potential from the reference state to the actual composition at equilibrium. This will give you $RT \ln K$ by definition of $K$ when you sum over all $i$. Of course if you change the reference state (for instance, use the molal instead of the molar scale) you change $\Delta G^\ominus$ and $K$ in such a way that the composition at equilibrium is the same (this does not depend on your scale).

The ideal gas law is not a crude model, it is actually extremely effective at low pressures...

@JohnGowers Indeed, the choice for $c^\ominus$ is arbitrary. In biochemistry, a modified standard state is often used where a different standard concentration is taken only for protons with value $c^\ominus_{H^{+}}=10^{-7}$ M (pH 7). So any chemical reaction where protons are involved with a stoichiometric number $\nu_{H^{+}}$ has a modified standard Gibbs free energy of reaction $\Delta G^\oplus$ which differs from the usual $\Delta G^\ominus$ by a term $RT \ln ((c^\ominus /c^\ominus_{H^{+}})^{\nu_{H^{+}}})$.

Yes, K is unitless, always, if defined by your equation:
$$\Delta G^\circ=-RT\log(K)$$
There is no way around this, unless you use some strange units for free energy, R or T.
Beyond the obvious mathematical answer, if you dig deeper you find that equilibrium constants are unitless because the expression relating $K$ to $ \Delta G^o$ requires the definition of reference states. This is because $\Delta G^o$ is defined as the free energy change for the transformation of reagents to products in the specified reference states. As a result, equilibrium constants contain ratios of activites (which can be expressed as functions of concentration) of reagents and products in their reference states and at equilibrium.
For instance, for a species i in the gas phase terms in the equilibrium constant involving that species consist of ratios of fugacities $(f_i/f^o)^\nu$ where $\nu$ is a stoichiometric factor, $f_i$ is the fugacity of the species, and $f^o$ is the fugacity in the reference state (fugacities become equal to pressures in the limit of zero pressure). The resulting equilibrium constant is a product of such unitless ratios, and therefore also unitless.
That equilibrium constants are sometimes reported with units is contrary to accepted conventions, but is convenient: specifying a set of units together with the equilibrium constant implicitly declares the reference state(s).


text:
Activity of hydrogen ion
I am not from chemistry background. Could you tell me in simple words what the activity of hydrogen $a_{\ce{H+}}$ is? It arises in the context that the negative logarithm of the activity of hydrogen gives the pH: $\text{pH} = -\log_{10} a_{\ce{H+}}$.

Building on @Philipp and @chipbuster's answers-
What is Activity?
When we have dilute solutions or small quantities of a gas or other substance, the substance behaves as if the particles are just points, rather than having a size. In these settings, we often use a simplified set of equations.
For example, with the hydrogen ion, we use $\mathrm{pH} = -\log_{10}[\ce{H+}]$ where $[\ce{H+}]$ is the concentration, i.e. the number of particles (in moles) divided by the volume (in litres).
Once you get a lot of these particles, they bang into each other and take up room (think of billiard balls on a pool table or a room full of tennis balls flying around).
In these circumstances, we talk about ‘activity’, which is a way to correct for this. Basically, it is what the concentration or amount appears to be if we were using the simplified laws (the effective concentration). For the hydrogen ion, it is also what the concentration appears to be to physiological systems. In other words, it uses a correction factor for the simplified laws.
Activity, Concentration and Activity Coefficient
The thermodynamic definition of activity is:

$a_{i} = \operatorname{e}^{\frac{\mu_i - \mu_i^⦵ }{RT}}$

Which in low partial pressures and dilute solutions is simplified to:

$a = \gamma\cdot\frac{c}{c^⦵}$ for a solution
$a = \gamma\cdot\frac{p}{p^⦵}$ for a gas

where:

$a$ is the activity
$\gamma$ is the activity coefficient (a constant number based on what the substance is, what the solvent is if there is one, the temperature and other factors)
$c$ is the concentration (for something dissolved); $p$ is the partial pressure (for a gas)
$c^⦵$ is the standard state concentration; $p^⦵$ is the standard state pressure

pH and Activity
The original version of pH as defined by Sørensen is:

$\mathrm{pH} = -\log_{10}[\ce{H+}]$ (hereafter referred to as $\mathrm{p}[\ce{H+}]$)

However, the modern definition that takes into account activity is:

$\mathrm{pH} = -\log_{10}a_{\ce{H+}}$

$\gamma$ is found experimentally. In the case of the hydrogen ion, it is found using electrochemical methods.
Activity of Hydrogen Ion in Dilute Solutions
Standard thermodynamic state for $\ce{H+}$ is considered to be 1 mol/L ($\mathrm{p}[\ce{H+}]$ 0); thus:

$a = \gamma\cdot\frac{c}{c^⦵}$ (as a dilute solution)
$a_{\ce{H+}} = \gamma\cdot\frac{c}{c^⦵} = \gamma\cdot\frac{[\ce{H+}]}{1\ \mathrm{mol/L}} = \gamma\cdot [\ce{H+}]$

Also, in dilute solutions, activity can be approximated/calculated using an extension of the Debye-Hückel equation, the Güntelberg equation, which applies to concentrations of < 0.1 mol/L:

$\log_{10} \gamma_i=\frac{-Az_{i}^{2}\sqrt{I}}{1+\sqrt{I}}$

where:

$A$ and $B$ are constants that depend on solvent and temperature (0.51 and $0.33\times10^8$ respectively for water at 25 °C)
$z$ is the charge of the ion (1 for $H^+$)
$I$ is concentration of the ion

For $\ce{H+}$ in a solution of water at 25 °C this would be:

$\log_{10}\gamma=\frac{-0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$

Substituting, we get:

$a_{\ce{H+}}=\gamma[\ce{H+}]$
$\log_{10}a_{\ce{H+}}=\log_{10}[\ce{H+}]+\frac{-0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$
$-\log_{10}a_{\ce{H+}}=-\log_{10}[\ce{H+}]+\frac{0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$

and so for dilute solutions only:

$\mathrm{pH} = \mathrm{p}[\ce{H+}]+\frac{0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$

Seem to be some problems with this answer.  First problem: the jump from I (ionic strength) to [H+] is not explained; seems the final equation would only be true for a solution where there is H+ and a univalent counter ion, and no other ions (like HCl solution).  Second problem, says "A and B are constant", but there is no B in the equation.

The activity is some sort of dimensionless measure for the concentration. In thermodynamics, a lot of important equations are derived for the so-called ideal gas, which is a model system where the particles are treated as if they had no own size (point particles) and don't interact with one another (apart from elastic collisions).
This model is a very crude simplification of reality but nevertheless gives very good results, when the temperature is not too high and the pressure is low. So, the equations one can derive from this model are rather simple and yet quite usable for real systems. But of course, the applicability of these equation has limits, since in real systems the particles interact a lot with each another (via electrostatic and quantum mechanical forces) and have a certain size. But, in order to retain the simple form of the equations derived for ideal gases, some physical quantities like the pressure or the concentration are scaled by empirical coefficients to reflect the deviation from the ideal behavior and get the right results for real systems. An example is the activity $a$, which is equal to the dimensionless concentration, the so-called mole fraction $x$, scaled by the activity coefficient $f$
\begin{equation}
a = f x \ .
\end{equation}
So, for example, instead of the concentration (or more correctly the mole fraction) one uses the activity in some equations derived for ideal gases to describe real systems accurately.

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How to find the number of valence electrons?
I need to understand the following:
Considering an element Sulfur - S which has 16 electrons.
How do we calculate the number of valence electrons of S?
Please correct me if I am wrong:2+2+6+2+4=16. So, the valence electron = 4+2=6
Is it that always valence electron is the addition of the last two shells? In that case for Na, sodium, if we take 11 and arrange it as 2+2+6+1, then the valence electron is 1 or 6+1?
Now, considering the fact valence electron is 1, if we go by, up down up down rule, then as it is 1, hence the spin is up?
For 6: is it up(1) down(2) up(3) down(4) up(5) down(6) as it is both up/down hence it would be +-1/2 spin?
For Phosphorous 2,8,5, what would be the valence electron? 5 or 13?
If 5 then up down up down up, will it have an up spin?
Is it that for even numbers like 4,6 the spin will be both up/down i.e. +-1/2?

On the matter of topicality, the underling QM is certainly on-topic on physics.se (and especially the difficult calculation needed to determine the structure of outer shells in high Z atoms), but the notion of "valence" is something that I usually associate with chemistry. Thoughts from the crowd?

'p' and 'd'.
There is one type of 's' orbital. It is spherical. So an s orbital can hold 2 electrons only.
There are three types of 'p' orbitals. They are the same shape (like a dumbell) but are oriented in different directions - along the x, y and z axes.  So a 'p' orbital can hold $2$x$3$ = 6 electrons.
There are five types of 'd' orbitals (10 electrons) and seven types of 'f' orbitals (14 electrons).
Fill up the orbitals in the following order (following the red arrows), which is increasing order of energy. Lowest energy fills up first.

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p

So let's take the example of sulfur.
The total number of electrons is 16. We'll distribute them in the following manner.
$1s^2, 2s^2, 2p^6, 3s^2, 3p^4$, where the superscripts of each orbital represent the number of electrons in it. Now, we have a total of 6 electrons in the last (3rd) orbital. Therefore the valence number is 6.
It is not simply the last 2 numbers while writing the electronic configuration.
For Phosphorous, we have:
$1s^2, 2s^2, 2p^6, 3s^2, 3p^3$, which gives us 5 electrons in outermost (3rd) orbit.
For Lithium (3 electrons), we have:
$1s^2, 2s^1$, which gives us a total of 3. Outermost orbit is the 2nd orbit and the total number of electrons in it is just one. Therefore valency of Lithium is one.
However, it is much simpler to do it the way I've shown above as with Sodium. Don't need to bring in all the orbitals
For the spin, just remember that you have to fill up the entire orbital with up spin electrons before you start putting the down spin electrons. Don't just alternate up and down.
Let's talk about the 'p' orbital. Each 'type' of the p-orbital must have up-spin electrons before adding the down-spin. So there will be 3 up spin electrons before you start adding the down-spin ones. Also, you can not move on to the next orbital without completely filling the previous one with an EQUAL number of up and down spins.
For Phosphorous; let's say we've filled the 1s, 2s, and 2p orbitals. Now we add one up spin to the 3s, followed by a down-spin. Which leaves us with 3 more electrons to be added to the valence orbit (3rd). It is a 'p'-orbital, which needs to have 3 up spin electrons before any downspin ones can be accommodated. But we have only 3 remaining! So all will be up-spin electrons in the valence orbit. So we'll have a total spin of +3/2.
Hope I didn't complicate too much, but I've actually tried to make it as short as possible. I've still left out quite a bit.

Out of curiosity, the "electronic configuration", like $(1s)^2 (2s)^1$ for Li or any more complicated case, is a result of Hartree-Fock calculation or from some model parametrized from experimental atomic spectroscopy? If it is the later one, how it is done in detail? I usually find book just gives result of electronic configuration with some explaination why it is like that, does not say how do we know that

I actually don't know, haha. We haven't studied the reasoning behind this stuff yet. I agree, no where does it give the reason. Have you tried Wikipedia?

wiki  "It is through the analysis of atomic spectra that the ground-state electron configurations of the elements were experimentally determined."  No details about how the electronic configuration is determined. My question is about, the determination may be ambiguous, both $2s^2$ and $2p^2$ can form $^1S_0$ state.

@user26143 Well, the wave-mechanical model (which includes all this) is just a theory. As was Neil Bohr's theory, and Thomson's theory, and Rutherford's theory. They all were proved to be wrong - or incomplete. For now, this model explains our observations to a satisfactory extent. If another theory/model comes up which can explain them better, or can improve on the drawbacks (if any) of this model, then that'll be accepted as the new one. I don't think there's any proof to it.

Well, Bohr's theory, quantum mechanics, and quantum field theory all have mathematical formulation with falsifiable predictions. I am not even sure what is the underlying mathematics about the "electronic configuration" and what is the falsifiable prediction it has... Besides, I don't know how do people write the electronic configuration like copper is as wiki and/or other sources, but in not other form. As wiki said, "... were experimentally determined" there should be a definite procedure for such determination, rather than proof (We don't need to prove an "experimental determination" ;))

It is worth mentioning the case of the rare earths, in which the nominal valence band actually lies closer to the nucleus (spacially) than one of the already filled shells resulting in a group of materials with low reactivity and very little chemical variation from element to element.

I think a more 'robust' definition for valence electrons are 'those that can form bonds'.  'Outermost electrons' can be slightly ambiguous in some cases I think.

It is not so difficult I could not understand your method of finding valence electrons, easiest way I could explain is try to find the atomic number of the element. For example chlorine it's atomic number is 17 now find it's electronic configuration which is 2,8,7. Now we know that there are 3 shells in it hence third shell would be it's last shell and valence electrons are the electrons of valence shell therefore valence electrons would be 7.

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text:
Lines on inside of hexagonal shapes of structure diagrams
just a simple question from me today...
I was looking at the structure of Norepinephrine for a school project and was curious about the diagrams on the Wikipedia page for it.

I was wondering why hexagons are used in this diagram and what the purpose of the three lines on the inside of the hexagon mean.

Snipergirl has a nice detailed answer, but the short answer is - that's what the structure is. This molecule and many other contain six carbon atoms in a ring with that bonding pattern. The parent compound is benzene.

Here is some more information on skeletal org chem diagrams: http://www.ivy-rose.co.uk/Chemistry/Organic/How-to-draw-skeletal-formulae-of-organic-molecules.php

That ivy rose link is not available from my location

Noradrenaline/norepinephrine is an aromatic organic compound containing a benzene ring (the hexagon). Essentially, the vertices in the diagram are carbon atoms and the number of lines represents the covalent bond order; 1 line (eg connecting an OH to the benzene ring) means a single covalent bond. Any 'spare' bonds are where a C-H bond is. So it can be rewritten like this:

HOWEVER, when it comes to a benzene ring (the hexagon), you do not simply have alternating single and double bonds. In fact you have what is known as a 'resonance structure', where the electrons/bond is shared around the entire ring. One way of thinking of it is that there are 1.5 bonds between each carbon atom in the ring. Another way of thinking of it is that pi-orbitals from each carbon atom merge to form a ring combined orbital structure, so instead of existing in that figure-8 type area around one carbon nucleus, those 6 electrons can be in a donut shaped area above and below the carbon nuclei:

This is why you often see a hexagon with a circle inside it to represent a benzene ring as it is less misleading than the alternate single and double bonds. In that case, you would redraw the noradrenaline diagram like this:

The advantage, though, of having the alternate double and single bonds is in drawing diagrams when working out reactions.

The reason most chemists don't draw aromatic rings with a circle is that it makes it a right pain when you do reaction mechanisms with curly arrows! The circle is a better representation conceptually, but the double bonds being drawn inside the hexagon are more useful for mechanisms.

Good point, i'll adjust the post!


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What is the colloidal osmotic pressure?
The oncotic pressure or colloidal osmotic pressure is the osmotic pressure developed due to the presence of colloids in a solution. But since the colloids are not true solution, why should the colloids be termed as solutes soluble in the solvent and consequently capable of producing any osmotic pressure? Should they not behave like insoluble substances who do not affect the osmolarity of the solution? And theses colloidal osmotic pressure plays a very important role in trans-capillary transfer dynamics.
If the colloids do affect the osmotic properties, is the expression to measure the colloidal osmotic pressure same as that for true solutions(vant-hoff equation)?

For a real world example of colloidal osmotic pressure, think of the albumin in human blood. This certainly exerts an osmotic pressure and among other things prevents most of the water in blood from entering the extracellular fluid compartment outside the circulation and causing oedema.

I think, first I should clarify what causes the osmotic pressure:
Osmosis occurs when two solutions of different concentrations are separated by a membrane which will selectively allow some species, e.g. the solvent, through it but not others, e.g. the solute.
So, there is a concentration gradient between the two solutions which would lead to a diffusion from the side with the higher to the side with the lower concentration.
But since the solute cannot pass through the membrane the diffusive flow cannot even the concentration differences out and thus constantly presses against the membrane exerting a force on it which leads to a pressure.
The pressure originating from this tendency for osmotic flow to occur is called the osmotic pressure $\Pi$ of the solution.
So, the question is: Is the theory behind diffusion applicable to colloidal solutions?
To reason about this, I will first address the matter of whether colloids can be considered "real" solutes or not. Colloidal particles are usually some polymer particle coated with a tenside layer. So, it is the molecules of those tenside layers that interact with the solvent molecules and they are chosen in such a way that the colloidal particles neither lump together (coalesce) nor sink to the ground or float to the top - that keeps the colloidal solution stable. So, a colloidal particle is surrounded by solvent molecules which solvate the tenside layer. Of course, the situation is a bit different from a "normal" solution since the colloidal particle is a lot bigger than, say, an alcohol molecule and thus the the ratio between solute particle size and the size of its solvation sphere is much smaller. But, in essence, a colloidal particle and an alcohol molecule are solvated in pretty much the same way.
Furthermore it can also be assumed that interactions (e.g. electrostatic forces) between the colloidal particles are weak.
This means that colloidal particles are thus not subject to external forces which influence their movement. They simply move through the solution by Brownian motion. This is important because then their motion can mathematically be described by a stochastic process which is the basis for the thermodynamic description of diffusion.
And, as I described above, the reason for diffusion to happen is the same as the origin of the osmotic pressure.
So, to answer your last question: Yes, I think, the description for osmotic pressure caused by colloidal particles should be the same as for "normal" solutes.

So, according to this reasoning, the process of surface film formation on the colloidal particles is absolutely required for them to exhibit osmotic pressure. That means, hydrophobic colloids and colloids forming a layer on top or bottom cannot be described by the same equations, right?

@SatwikPasani It depends on the colloidal particle whether you need the tenside coating or not. The key feature they must posess is solubility in the solvent. Because only if they are in solution there can be a concentration gradient leading to the osmotic flow. If a colloid simply sediments, then their is no appreciable of it present in solution, thus you have no gradient.


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Variation of cone angle with concentration of fuel-air mixture
A stoichiometric fuel-air mixture flowing in a Bunsen burner forms a well-defined conical flame. The mixture is then made leaner. For the same flow velocity in the tube, how does the cone angle change?

The cone angle of the Bunsen flame is determined by the flame speed of the mixture and the average flow velocity. To first approximation one can write: $$ \sin \alpha = \frac{S}{U} $$
where $\alpha$ is the half cone angle, $S$ is flame speed and $U$ is flow velocity.
When the mixture is made fuel-lean, depending on how flame speed changes the cone angle will change accordingly.
Usually flame speed for fuel-lean mixture is less than that for stoichiometric mixture. Therefore we can expect the cone angle to decrease.


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Do ions also form special electron configurations? (just as Copper and Chromium do)
Do ions also form special electron configurations? (just as Copper and Chromium do)
I know that because copper has 1 electron in its s-orbital (highest) because it wants to have a full d-orbital (d^10)
But do ions also act this way?

What do you mean by special?

that it will remove an electron from the s orbital to the d-orbital to have a stable half filled s-orbital and stable fully filled d-orbital

Check out the NIST database for electronic configurations for atoms and ions in the gas phase. If you play around a bit, you will find that yes, some ionizations will not follow the expected trends from the aufbau principle. For example, search for rhenium and compare the configurations between $\ce{Re^{11+}}$ and $\ce{Re^{17+}}$, and you'll see some of them are unexpected (and rather hard to explain). Osmium shows an anomaly going from $\ce{Os^{+}}$ to $\ce{Os^{2+}}$. There are other elements with ions containing aufbau anomalies, but I can't remember many from the top of my head.

In short, yes, the electrons will change orbitals to fill shells (even if it is an ion). This is why the transition elements can have more than one ion.


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Size of dispersed phase particles in colloid?
I am really confused about what the size of particles of dispersed phase in a colloid is: one of my textbook says its between $1 - 100nm$, another says its between $1-1000nm$, and Wikipedia says its between $2-500nm$. Is there any reliable source giving the exact information?

Its between $1-1000nm$ as stated on wikipedia in a paragraph name IUPAC definition.

Colloid: Short synonym for colloidal system.
  Note: Quotation from refs.[3][4]
  Colloidal: State of subdivision such that the molecules or polymolecular particles dispersed in a medium have at least one dimension between approximately 1 nm and 1 μm, or that in a system discontinuities are found at distances of that order.[5]
  Note: Quotation from refs. [3],[4]

Edit: Since it is mention under IUPAC definition that size lies between $1nm-1000nm$, therefore this must be the correct size.

Does the term "polymolecular particles" mean cluster of molecules?


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How should I modify my molecular dynamics simulation so that known chemical reactions can be modelled?
I have written some molecular dynamics code that I want to use to model known chemical reactions.
By known I mean that it has been observed in a lab that $\ce{A +B->C}$.
I am not interested in finding out if it is feasible that two things react (i.e. is there a favorable energy configuration?), rather I am interested in finding out what the concentration over time looks like given a volume, concentration, temperature and a pair of chemicals that I know will react.
At the moment I just check if they are very close to each other. This produces reasonable looking concentration curves.

Is this is a correct way of doing things? I am looking for publications so it has to be defend-able, not just OK.

I am also interested in modelling large bio-molecules in a coarse grained simulation.

Are there any particular potential functions that I should use for large biomolecules, or will LJ do?

Usually MD only simulations can't handle reactions, it is some hybrid of QM and MD. Alternatively you could consider a reaction probability given an intermolecular distance (i.e., if the molecules are inside a reaction radius, grab a random uniform number. If it is > .6 then reagent molecules are removed and a product is put in its place.) You'd have to be careful with artifacts in your velocity and acceleration terms in the MD simulation with a heavy handed technique like that. MD/QM is probably the best bet but computationally expensive.

What MD package are you using by the way?

I wrote my own package, as i'm trying to simulate a very specific thing. If I matched the reaction probabilities to published results do you think that would work? I am going to be simulating well understood reactions

I suppose it depends on what you're trying to get out of the MD simulation. If you're just looking for a concentration curve, MD is overkill. If you want to have a complete time history and see where the reactions are taking place, then MD is a good option. I worry about conservation of momentum if you're just popping molecules in and out and how that might introduce artifacts into the simulation.

If you want to publish this thing, you may want to be a little more familiar with existing solutions, literature etc. Your question suggest you try to re-invent the wheel which can be fun hobby, but a waste of time if it is your research.

Since you've written your own code, you will probably be able to judge this best, but I'll give you some pointers as to where I'd start.
As far as I've gathered, you simulate a NVT ensemble of a pair of chemicals.
First of all you have to decide whether you want explicit representation of the solvent or not. Hint: The latter is computationally far more expensive and I doubt that you have implemented all the optimizations that you'd find in larger MD software.
This is because this choice can drastically influence the diffusion properties of your chemicals. In a coarse-grained force field they might still be able to diffuse around pretty much unhindered, while when using explicit atomistic representation of the solvent, the properties may change to a larger degree.
The way I'd do it would be to check whether the two areas of the molecules that react with each other are close to one another, meaning that some distance criteria $(d<d_\text{rxn.})$ has to apply. When this criteria is met, you can make the reaction happen by creating new bonds and changing the atom types that are bonded on-the-fly. It wouldn't be so tricky to implement this part, I guess. Caveat: This will only work if you have a reaction whose mechanism is close to the proposed scheme. You might introduce nontrivial errors otherwise.
Then you can model the "real" concentration curves of all the substances, and simultaneously change your system to reflect the change in number of molecules. I wouldn't be surprised that you'd find that the pressure would slowly decrease as the reaction takes place (since you keep the volume constant).

Would it not be even better, if the reaction parameters are known and the transition state is known, to put a criterion on a function of distance and velocity along the reaction coordinate? This might allow to include also the temperature parameter within the MD simulation. The other problem with the distance approach is that it would not allow that there could also be a reverse reaction (equilibrium situation), whereas one could also check the distance / velocity criterion along the dissociation coordinate of C.

@PLD If the thermalization algorithm is good enough, you would already have enlarged diffusion and thus a higher rate of contact between A and B, so the temperature is already factored in here. Additionally, you could simulate the reverse reaction by assigning a probability to the formation of the bond. Or even assign a probability to the breaking of the bond, so to the reverse reaction directly.


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Why does chlorine undergo disproportionation in alkaline media?
One reaction used to form sodium hypochlorite (A type of bleach) is as follows:
$\ce{2NaOH(aq) +Cl2(g)->NaOCl(aq) + NaCl(aq) + H2O(l)}$
Provided the sodium hydroxide is dissolved in some substance.
I know that the sodium hydroxide dissociates to form the $\ce{Na+}$ and $\ce{OH-}$ ions, but I do not know why the $\ce{Cl2}$ breaks down into 2 separate atoms instead of just staying the way it is. My questions are: what are the driving forces behind this reaction and why does the $\ce{Cl2}$ break down? I see how it happens, I just don't know why.

Do this reaction requires heat/ occurs in presence of sunlight?

I don't think so. Here is a link to how it is made: link. Alternatively, there is the Wikipedia page.

At constant pressure and temperature, the spontaneity of all chemical reactions is controlled by the difference in Gibbs free energy ($\Delta G$), as described by the following equation:
$\Delta G = \Delta H - T\Delta S$
Where $\Delta H$ is the enthalpy change of reaction, $\Delta S$ is the entropy change, and $T$ is the absolute temperature. For a reaction to take place spontaneously, the free energy change must be negative (i.e., $\Delta G < 0$). In the specific reaction you're asking about, it's very probable that the change in entropy is negative, since a gas is being consumed and an equal total number of moles products are being formed as moles of reactants, so that the end result is an increase in order with respect to the reaction mixture. However, the formation of products results in a net release of heat (i.e., $\Delta H$ is negative), which increases the entropy of the surroundings sufficiently, such that, overall, entropy is increased in a global sense.
Ultimately, what drives this reactions forward is its exothermic nature (under standard conditions). In this reaction, the formation of stable chemical bonds releases energy as heat. If one were to attempt this reaction at very high temperatures, it would eventually fail, as the increase of the $T\Delta S$ term would at some point dwarf the enthalpy change.
There are other factors at work as well, including chemical equilibrium. By Le Chatelier's principle, the process can be made favorable by, e.g., maintaining a continual excess of reactants, or removing products formed from the reaction mixture as it proceeds.
Edit: Here's one source which indicates 348cal/g energy released, to put a specific number on the process. You could also calculate the enthalpy change using tables of standard thermodynamic data.

+1  plus the production of NaOCl is carried out at -40 degree Celsius as standard procedure.

@blackSmith, thanks. To your knowledge, is the reaction violently exothermic, or is that temperature merely to prevent the degradation of the hypochlorite/formation of other chlorates?

Formation of chlorate is the main issue for production, but it also favors the dissolution equilibrium in the forward direction. You've already mentioned Le Chatelier.

@blackSmith, thanks, I appreciate the response.

Thank you, but why does the chlorine break down from Cl2 in order to be used in NaCL and NaOCl?

@user2514631, I'm not sure what answer you're looking for. Thermodynamics is essentially the overriding factor in all chemical reactions at the aggregate scale. An immense number of molecules are colliding and interacting in the course of the reaction. Some interactions result in the breakage of chemical bonds, and some in their formation. Once an energetically favorable state is achieved (i.e., that minimizes free energy), a reversal of that state is unlikely, because the products are at the bottom of an energy curve. Given enough time, the preponderance of molecules will reach this state.

@user2514631, that's a simplified description of the process from a statistical standpoint and on a macroscopic scale, and there's really no deeper answer, at least not that I can provide. If you want mathematical rigor, you must study thermodynamics and physical chemistry. If you're interested in the mechanism of the reaction on the scale of individual molecules, I could plausibly imagine that it would begin with nucleophilic attack by the hydroxide on the chlorine, with chloride as the leaving group. This forms hypochlorous acid, which is deprotonated by the second hydroxide, making water.

@user2514631, I'm not certain that that mechanism is correct, but I wasn't able to find one online.

Also, why does it react with the OH− ion after being broken down?

Does it? Chlorine is not stable in water, it undergoes disproportionation :
$\ce{Cl2 + 3 H2O <=> OCl- + Cl- + 2 H3O+}$
Performing the reaction in alkaline media simply shifts the equilibrium to the right (Le Chatelier principle) and prevents a possible decomposition of hypochlorous acid ($\ce{HOCl}, pK_a = 7.49$) according to
$\ce{2 HOCl <=> Cl2O + H2O}$

According to the Frost Diagram of Chlorine in basic medium, it can be seen that Cl is at a convex side. So it will undergo disproportionation.


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Where does the extra electron go in ionic compounds?
In ionic compounds, where does the extra electron go? Or, in case of a cation, from which shell does it come?
Consider, for example, $\ce{OH^{-}}$, where did this extra electron go in the electronic configuration? And, what about $\ce{OH^{+}}$? There exists no such ion. Why?

In $\ce{OH-}$, that extra electron is on the Oxygen atom.  Oxygen has a very large electronegativity, so you can be sure that extra electron is with the $\ce{O}$ not the $\ce{H}$.  A lone oxygen has 6 electrons in its valence shell (2s2, 2p4), and wants to get to 8 (2s2, 2p6).  In the $\ce{OH-}$ ion, both atoms are relatively "happy", the $\ce{H}$ has 2 electrons in its valence shell (1s2) shared with Oxygen in a bond, and the Oxygen has 8 electrons in its valence shell (2s2, 2p6), 6 unpaired electrons and 2 shared with hydrogen.
In general, in an anion the extra electron will be going onto the most electronegative atom.  In a cation the positive charge will go to the lest electronegative atom.
Here's a helpful picture, your extra electron is in red:

picture reference: http://chemistrypartner.blogspot.com/2009/10/drawing-lewis-electron-dot-structure-or.html
As for your question about $\ce{OH+}$, it would never be stable because that oxygen is going to tear some electrons off of the first thing it sees, because it REALLY wants that full octet.  If you somehow made some $\ce{OH+}$, my guess would be that the $\ce{H+}$ would just dissociate, and your $\ce{O}$ would find another $\ce{OH+}$'s $\ce{O}$ and make $\ce{O2}$ immediately.


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Why does sodium require energy to ionize?
If you look at $\ce{Na}$, it has a very low ionization energy, but the question is, why does it have an ionization energy at all? If it can get to the electronic configuration of $\ce{Ne}$, by losing an electron, which is a very stable configuration, then it should release energy. Consider the analogy to gravitational potential energy, you can consider $\ce{Na}$ to be analogous to standing on a very high ledge. And $\ce{Na^+}$, is the position at the ground. Obviously, getting to the ground is much more stable. And if I jump down from the ledge I convert my potential energy to kinetic energy and ultimately to heat, thus releasing energy. Why does sodium require energy to ionize?
And, consider ferric chloride $\ce{FeCl_3}$.

Why does iron form this bond? I mean, it makes sense from chlorine's point of view to share an electron with iron. All the chlorines have a full outer shell. But what about iron? It had $2$ valence electrons, now it has $5$. It would make much more sense for iron to bond with chlorine to form $\ce{FeCl_6}$, since then it, too would have a full outer shell. Or, $\ce{Fe}$ could give away its $2$ electrons to $2$ chlorines and form ionic bonds. Why does this not happen?

There are two very different questions here so let's break things down a bit
Why does sodium have an ionization energy?
Since electrons are negatively charged and nuclei (owing to the presence of protons) are positively charged, electrons will typically be drawn to the nucleus.  The situation gets a bit more complicated by (a) the quantum nature of energy and (b) the presence of other electrons.  Complication (a) is the reason why electrons do not behave like planets orbiting the sun because if they did the electron would eventually be pulled in to the nucleus.  (More details probably requires a different question.)  The second complication is more relevant to your question, because additional electrons will both repel the electron we are trying to eject through ionization and also shield some of the positive charge of the nucleus.  (Think about wearing sunglasses; they don't make the sun less bright, they just make less light reach your eyes.)
Your comparison of Na to that of an object on a cliff is not correct; the sunglasses analogy I used above is better (although it's flawed as well).  The reason why sodium, or any other atom/ion for that matter, has an ionization energy is because it requires energy to separate two oppositely charged particles.  The attraction may be weak, as it is in sodium, but it is still non-zero.
Why does $\ce{FeCl3}$ bond the way it does?
First, it is a good idea to look at the wikipedia page for iron(III) chloride, which shows that the molecule actually binds in an octahedral fashion, where two additional chlorides from neighboring molecules bind in a bridging fashion.  This is a type of bonding that is typically not discussed until one's third year of chemistry education.  Valence shell electron counting works really well for identifying "stable" electron configurations of ions and covalent molecules made out of lighter elements (C, N, O, for example).  The presence of d-orbitals, plus the ionic nature of the interaction between Fe(III) and chloride ions, plus the Lewis acid/base concepts that do a better job at describing and rationalizing bonding in transition metal complexes, makes the concept of valence electrons not the best model to explain bonding in $\ce{FeCl3}$.
Without access to molecular orbital theory, it is perhaps best to explain the bonding in iron(III) chloride from a strictly ionic bond nature.  Fe(III) is a +3 cation and therefore requires three -1 anions to balance the charge.

+1, I like this answer. I would just note that another problem with the cliff analogy is that it incorrectly presupposes that Na+ is inherently more stable than Na without considering where the electron has gone. In other words, it fails because it treats the Na as if it were in isolation. The crux, it seems to me, is really the relative difference in energies between the two sides of an overall chemical process. Na+ is only more stable (at least for any appreciable length of time) if that electron is transferred to a counter-ion with high electron affinity and electronegativity.

And even then, of course, the initial ionization still requires input of energy, as you noted.

@GregE. I agree.  I thought of adding the "where has the electron gone" component but it ends up making this a long-winded answer.  (Not that it isn't already...)


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Does oil break down condoms?
It seems that every website on sexual health advises against using oil-based lubricants with condoms. It is claimed that "oil breaks down latex". One source claimed that a latex condom completely breaks down in only 60 seconds. It made me curious, so I made an experiment.
I took a piece of rubber latex condom and soaked it into regular canola oil I found in the kitchen. I checked the condom after 1 minute, nothing changed. So I let it soak for about 5 minutes more, and then 5 hours more, still nothing. It was able to hold a large amount of water without leaking or breaking.
So I am wondering, is it really true that oil degrades latex? What sort of chemical reaction is supposed to happen? What properties of the latex material and the oil influence this reaction?

As far as I know the issue only applies to latex condoms (made from natural rubber). Many oily solvents degrade rubber's integrity. But not all condoms are latex; some modern ones are polyurethane and have totally different characteristics with solvents and lubricants.

First off, may I say that I applaud your decision to test this through an experiment. It's rare to see that than I would like.
Now, on to the matter at hand. It's fairly well known from industrial chemistry that non-polar solvents degrade latex quite heavily.
I work with latex seals a lot, and the hexanes we use routinely break the seals down in under a day. Of course, if you're lubricating your condoms with hexanes, you're a) an idiot or b) absolutely insane.
A paper I managed to find suggests that there really isn't too much direct data on condoms, and it muses that the warnings might have arisen from industry, where nonpolar solvents decidedly do degrade latex.
To find out, they did a burst experiment with condoms that had been treated with various oils. Glycerol and Vaseline-treated condoms showed a very, very minor decrease in strength, while mineral oil/baby oil-treated ones burst at less than 10% of the volume of an untreated condom.
They also found that 10 month-old condoms have half the burst volume of 1-month old ones, so you could argue that using 1-month-old condoms that have been slathered in Vaseline is still much safer than using older ones.
As for the actual chemistry of the weakening, I honestly don't know. If I were to hazard a guess, I would note that the latex looks like a bunch of ethylenes glued together,
so my guess would be that the solvents get between the chains and force them apart, weakening them. For this to happen, the solvent must be nonpolar, but still small enough to slip between the chains of the polymer.
That's probably why vaseline and canola oil don't have much of an effect---they're just too big to fit between the chains. Again though, I don't know for sure, so don't quote me on this last paragraph.

Rubber isn't that much like polyethylene. It is, in effect, the polymer cis-1,4-polyisoprene (which contains unsaturated bonds in the polymer backbone). It doesn't contain aromatic rings as in the example drawn which looks more like the copolymer of styrene and butadiene, an artificial polymer unrelated to condoms or natural rubber.

Biologist here. Mineral oil is composed of alkanes. Coconut oil is largely triglycerides. Other than the fact that we call both of these products "oils," I can't see why they'd exhibit similar reactiveness with latex. In fact, I'd suggest the triglycerides are more similar to silicone oils than to mineral oil. However, I am neither a materials scientist nor a chemist.

@Dr.JohnnyMohawk Chemist here. You can basically consider any triglyeride a long, triangular (i.e. once branched) alkane chain. The main difference are six oxygen atoms somewhere in the middle that don’t contribute much to the overall hydrophobicity ;)

If you put a jelly baby in water, it does not dissolve, but swells to five time it's original size, and becomes very soft and brittle.
Same happens to rubber in oil. It is a crosslinked polymeric material, and when all the bonds in the network are completely stretched due to a solvent that has diffused in, they cannot take any additional strain, and break very easily. No chemistry is involved, the rubber just tries to dissolve, but cannot get away except by breaking.
Depending on the oil you use, and the temperature (37°C might be worse than 20), the effect can be more or less pronounced. Hexane or petrol will destroy the condome most efficiently, trigylcerides, which are slightly more polar and cannot efficiently diffuse into the rubber network because they are large and have a bulky side chain (=the middle fatty acid), somewhat less.

Not a chemist, but doesn't this depend on the polarity of the lipid solvent? Which would be why temperature would have a significant effect?

@DaMaxContent It does, as mentioned. Depending on the type and polarity of the rubber. Why would that mean temperature has an effect? The swelling bevaviour of rubbers is always also temperature-dependent.

I meant that the effect that the temperature has is correlated to how polar the lipid is. Take a comparison of 2 solvents with same-length carbon chains for example. In theory, the carbon chain with the highest polarity should show a more pronounced change in dissolution-rate as temperature changes, right? Like saturated lipids won't change dissolution rate as fast as unsaturated lipids? Is that clear? It was a hard question to phrase.

@DaMaxContent "In theory, the carbon chain with the highest polarity should show a more pronounced change in dissolution-rate as temperature changes, right?" What theory would predict that? The change with temperature even depends on the current temp. Polymers can e.g. have a maximal solubility at a certain temperature, and it drops on both sides.


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How can I calculate the energy of a solution by knowing all of it's elements?
How can I calculate the energy of a solution by knowing all of it's elements?
For example:
If I have 2 liters of salting water $\mathbf{NaCl_{(aq)}+H_2O_{(l)}}$ contains 1.5 g of salt, how can I determine the internal energy stored in it?

It would be helpful if you would put some of the ideas you've already thought about into your question.  That way, others aren't just giving you the answer, they are helping you reason it out.

You need to precisely specify what form of energy you're referring to. If you mean internal energy, then measuring it directly is impossible for any real system, one can only calculate the change in internal energy that occurs due to heat transfer and work done on/by the system in the course of moving from one state to another.

@GregE. How can this be impossible??! While I can know the energy for any food in the super market using the "Nutrition Facts"??

@MohammadFakhrey, that's food energy, which is only one specific measure of energy, namely the heat energy released during the process of metabolism. Salt water has zero food energy, as it's not consumed in the process of cellular respiration.

@GregE.OK how can I determine if the salt or any another element has food energy or not? -- Thank you

@MohammadFakhrey, first, your changing the post to refer to internal energy is not correct. As I explained in my first comment, measuring internal energy directly for any real thermodynamic system is not possible, and internal energy is emphatically not to be conflated with food energy. Second, if you want a complete answer to how to measure the caloric content of food, I suggest first researching the subject independently (perhaps start with the Wikipedia article I linked), and then asking about specifics in a separate post if necessary, or modifying this post.

@MohammadFakhrey the only other possibilities I can think of is that you're referring to free energy? Try googling that or reading about it and then edit your question.

@GregE. I don't know about your "measuring internal energy statement"  Wikipedia says it can be done with path integral Monte Carlo methods (whatever the heck that is) and we all know that if it's on Wikipedia, it must be true....

@bobthechemist, funny how that one section claims a computer simulation can be used to "determine" an intrinsically uncertain quantity of a real system. Approximate might be an apter choice of word. Otherwise somebody should go resurrect Joule, Boltzmann, Carnot, Kelvin, et al. post haste to let them know we can now measure the precise velocities and positions of individual microscopic particles.

@GregE. well at we know that through this process that entropy is heading in the right direction; all of them must be spinning in their graves.

@bobthechemist, yep, and all their constituent atoms are dancing in beautiful and deducible harmony as they do so.

Since you ask specifically for internal energy, would it be that you already have some ideas?
So what is internal energy? It is the energy due to the interaction among the species plus the energy due to the interaction between the species and any external field. That is, the energy it takes to "assemble" the system.
Then for gaseous species, it would be mainly the bonding energies of the molecules. Those are usually inferred from experiments. And you can look up tables in textbooks and IUPAC website.
For ionic crystals, you would reason this way:  first ionize the species and them find the Columb interaction among them in crystal. Ionization energies are measured. Columb interaction is probably calculated and is a nightmare to calculate.
The further problem is from NaCl(s) to (aq) where the solvation involves quite a bit of interactions between the ions and water. For the case of Na+ Cl-, water molecules do not form regular cages around them, even though we CAN study the ideal case of regular cages. I think there has been some minimal models which can possibly guide a simulation study for the solvation energy. Though I don't think there has been any simulation study or any theoretical study that is sufficient to predict solvation energy.  I can be wrong. On the other hand, direct measurement should at least be able to measure the heat change during solvation.
So in digression, I don't think you can find the complete energy too well.

You can't calculate the internal energy of any real thing. The reason is that the total internal energy is the sum of all kinetic and potential energies of every particle in the system - there is no way to know that. You could calculate it for a simulated system of particles.
What you can do for real systems is figure out the change in energy. If you can compare it to a "ground state" (a state we define as having zero energy) then you can measure the change using the law of conservation of energy. For a system closed to mass transfer:
$\Delta E = q + w$
Change in energy is equal to heat transfer plus work done on or by the system.
Heat is a thermal transfer of energy (kinetic energy of molecules) and work is a mechanical transfer of energy (potential energy of molecules) and we can measure both on the macroscopic scale. Since those are the only two ways to transfer energy, we know that if we account for both, we also have the total energy change.
Incidentally, this is the reason that you almost always see $\Delta$ in front of $G$, $H$, and $E$ in thermodynamics. There is no "universal" ground state for energy, and so we have to look at changes in those quantities to solve practical problems. Entropy, $S$ is a little different - we can define a universal zero entropy state, and so when you see $S$ used in practice, it really means "entropy relative to the zero entropy point" - which is a perfect crystal at absolute zero.


text:
What utility does the tau bond model of orbital overlap have?
In his book on molecular orbital theory, Molecular Orbitals and Organic Chemical Reactions, Ian Fleming notes that Pauling formulated an early alternative model to Hückel theory for explaining the bonding of simple conjugated polyenes. Evidently, this came to be called τ-bonding.
Fleming describes it as a modification to, or offshoot from, the hybridization model, in which orbitals similar to sp3-hybridized orbitals are combined. Fleming says that the τ-bond model makes the extent of the conjugation less obvious, but that the model

[…] might have some virtues, not present in the Hückel model, especially in trying to explain some aspects of stereochemistry.

I don't have access to the relevant primary literature, and the treatment of the subject in Fleming's book amounts to one paragraph with an accompanying diagram. A search online was relatively fruitless.

Is τ-bonding just the historical precursor to the bent/banana-bond model, or does it have some unique independent significance that makes it directly relevant to modern chemists?
What are the specific "virtues" that Fleming might have been referring to, and what, if any, advantages does the τ-bond model have over Hückel theory and/or other approaches grounded in MO theory?

τ-Bonds provide an alternative description of electron density in alkenes and alkynes valid even in the modern chemistry.
A good example is the conformational preferences in propene with the lower energy of the “eclipsed” conformation [1].

Reference

Deslongchamps, G.; Deslongchamps, P. Bent Bonds, the Antiperiplanar Hypothesis and the Theory of Resonance. A Simple Model to Understand Reactivity in Organic Chemistry. Org. Biomol. Chem. 2011, 9 (15), 5321. DOI: 10.1039/C1OB05393K.

While the question has been already answered, I think it worth to mention some controversial aspects of Pauling's theoretical work [1, p. 751]:

But first on Pauling and the ways he damaged his own credibility: he ignored MO theory to a degree that was clearly perceived by the community as blind, if not unethical. His interests (and great, great creative powers) also shifted to biological problems. And, finally, he lost touch with the explosion in structure and reactivity, to an extent that the third edition of The Nature of the Chemical Bond is … an embarrassment.

Reference

Hoffmann, R.; Shaik, S.; Hiberty, P. C. A Conversation on VB vs MO Theory: A Never-Ending Rivalry? Acc. Chem. Res. 2003, 36 (10), 750–756. DOI: 10.1021/ar030162a.

Thanks for the reference, it certainly gives an interesting historical perspective. However posterity may evaluate Pauling, and whatever the faults of VB theory, I'll always have a nostalgic fondness for his book General Chemistry.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How will the transference number of H+ ion in a dilute solution vary as temperature increases?
I know that transference number depends on Ionic mobility, which depends on temperature. But is there any equation which describes the relation of Transference number and temperature, any ions like H+.


text:
How much energy is needed to break a water molecule?
How do you calculate how much energy is needed to break a water molecule into its constituent atoms?
note
This is not homework. I just want to know what is required to get hydrogen out of water.

Please see the homework policy. Indicate what work you've done so far in trying to ascertain the answer. To get you started, you might find it useful to read up on bond energies, enthalpy of formation, and Hess's law.

@GregE: it's not homework. If you don't want to answer, that's ok, maybe someone else will. I thought the whole idea of stackexchange is that you can ask questions you want to know about, and people help you out. Note that I'm not asking the solution, I want to learn how you do it.

@JohanA, the tag is intended for a broad type of question, not just ones that are actually assigned by a teacher/professor in class. I don't mean any offense, I simply think your question fits the template of a "homework" question. I won't edit the tags again, but, in my view, it would be very good if you did some research prior to asking. This is not at all an obscure topic.

@JohanA, the other advantage of attempting to answer the question yourself first and showing the work is that somebody interested in answering your question can better gauge your prior knowledge. If you have no general chemistry knowledge, then what you're asking for is either (a) an indication of the relevant topics to read up on, which I gave you in my first comment, or (b) a concise summary of a hodgepodge of introductory gen. chem knowledge, which, AFAIK, isn't appropriate for this forum.

The answer is given by the standard enthalpy of formation, $\Delta H_\mathrm f^\circ$ for water:
$$\begin{alignat}{2}
\ce{H2 + 1/2O2 &-> H2O(l)}\quad&\Delta H_\mathrm f^\circ&=-285.8\ \mathrm{kJ/mol}\\[6pt]
\ce{H2 + 1/2O2 &-> H2O(g)}\quad&\Delta H_\mathrm f^\circ&=-241.8\ \mathrm{kJ/mol}
\end{alignat}$$
The energy required for the following reaction, your point of interest:
$\ce{H2O -> H2 + 1/2O2}$ is simply a positive value for either one of the above $\Delta H_\mathrm f^\circ$ values depending on the state of the reactant. It means that per mole water splitting requires at least $241.8\ \mathrm{kJ}$ of energy.
Any method(thermal,electrolysis etc) followed, this much of energy will be always required. When using electrolysis, it may be convenient to express it in $\mathrm{eV}$.
Regarding calculation, I have to mention that $\Delta H_\mathrm f^\circ$s are experimental values. To know how they are calculated, here is a nice example of $\ce{NaCl}$.

Just out of curiosity: One could regard the question in terms of the Gibbs free enthalpy as well. Then the situation would change as a favourable entropy term would have to be taken into account as well. What energy would we calculate then?


text:
Are phosphomonoesterases solid?
What state of matter are phosphomonoesterases?  I am not a biochemist; while I've found various lists of such enzymes, and plenty of (to me) barely comprehensible articles describing their production, it's not clear to me whether the final product in isolation is a solid, liquid, or gas.
More generally, is there some reference work in which one can look up the state of matter (at standard temperature and pressure) for a given chemical?

Enzymes are very, very large in comparison to the typical molecules you encounter in chemistry (excluding any polymers). A 10 kD protein is considered to be small, and there are proteins of several hundred kD.
If you remove all the water from a typical protein you end up with a solid. But for most enzymes this would be an irreversible step, and they would lose their activity.
Usually enzymes are kept in aqueous solution, often with some glycerol or other cryoprotectants, and frozen.
Strictly speaking, they would probably be solid, but likely inactive in that form so I'm not sure if it counts. The relevant form of an enzyme is dissolved in water.

Phosphomonoesterases are proteins, i.e. long polymer chains of amino acid monomers. When they are active in living cells they are dissolved in an aqueous solution (or found in lipid membranes, which is sorta like a 2D liquid).  You can get purified proteins in solution with a variety of techniques, and if you were to evaporate away all the water you would get dried protein solid.  Most all biopolymers are solids if you remove the water the are naturally found in.  You never have pure protein liquids or protein gasses, if you heated the solids I think you would break down the polymers into the constituent monomers long before you got a proper liquid.


text:
Electrode potentials at interfaces?
My questions relates to the fundamental concept of electrochemistry, more specifically the electrode potentials.
1) First, why is there a potential difference at the interface of two phases? Considering the simplest case of a metal rod dipped in the solution of its ions, why should there be a potential difference. Out of many ways, one of the ways of understanding this provided by this site here,is in terms of equilibrium between the metal atoms and ions. But then, why is the presence of the other phase (ions for the metal piece) required to effect this equilibrium? For example, for the equilibria of ammonia and nitrogen/hydrogen, even if we take pure ammonia, the equilibrium is set up according to the surrounding conditions. Next,then can this equilibrium be described in terms of the equation $\Delta G=-RT\ln(K)$. Lastly, if this is an equilibrium, then there must also be an equilibrium between the metal and the hydrogen or the hydroxyl ions in solution as these might also take up the electrons from the metals?
2) When we measure the difference in electrode potentials of two different half cells, we are just measuring the potential difference between the two metal electrodes. How can we get an idea of the potential difference of the metal piece and its solution as the absolute value of potential of the two solutions might also differ. That is, if the two solutions are at different electrostatic potentials ( though not measurable), the potential difference between the two electrodes will not be the difference of their electrode potentials(which are the potential difference between the electrode and the electrolyte in each case). So, in order for the potential difference of the electrodes be equal to the difference in electrode potentials of the two half-cells, the solutions should be at the same potential?

I'm sorry, I'm having trouble making heads or tails of your question.
For Question 1, I don't see where hydroxyl ions come into play with the metals. For Question 2, I'm not sure what you mean by absolute potential. Potentials are always relative to each other. The closest thing to an "absolute" potential is the vacuum, and if you want to know the potential relative to the vacuum, you just add 4.44V
http://en.wikipedia.org/wiki/Absolute_electrode_potential

@chipbuster In question 1, i am asking that in the presence of any other ionic species apart from the metal, that species should also be free to take the electrons accumulated over the electrode and hence enter an equilibrium with the metal electrode. In second part, i am just asking, unless the absolute electrostatic potentials of the two solutions are same, how can the voltage difference between the electrodes be the difference between their electrode potential.

There are a bunch of of sub parts to these questions; let me see if I can tease some of them out.
Why does a potential difference occur at an interface?
The easiest way to answer this question is through thermodynamics.  If we consider the situation of a reactive metal rod (e.g. magnesium) being exposed to water, the reaction $$\ce{Mg -> Mg^{2+} + 2 e-}$$ is favorable.  In other words, the $\Delta G$ for this reaction will be less than zero.  I've done nothing more than summarize the website you reference in your question, which does a nice job at describing what happens when a piece of metal comes in contact with water, so I don't feel the need to repeat that work.  We then get to your next question:
Why is a second phase required to generate a potential difference?
If we have a universe that consists just of magnesium, then there would be no place for the above reaction to occur.  In this trivialized case, there is no driving force to make magnesium give up its electrons.  That doesn't mean it would not happen, only that any dynamic equilibrium that is established would highly favor the pure metal.  The meat of question one, then, is (may be?):
Why doesn't a potential difference between a metal and gas occur?
Actually it does.  There's an interesting paper (which is behind a paywall if you are not at a university that subscribes to this journal) that describes the measurement of potentials across metal/gas interfaces. The challenge in studying metal/gas potentials is the ability of controlling this potential externally.  With the metal/liquid scenario, an electrochemist (or budding electrochemist) can add another electrode to the solution and apply a potential between the electrodes. It's much harder to do the same thing with a gas. (Well, technically it's *not harder to connect two electrodes to one another and hold them in air, but the redox work functions, resistance and capacitance issues that arise in gas-phase electrochemistry make this a really challenging feat.)
The paper I reference above uses chemisorbed species that have active IR or Raman vibration modes to measure the surface potential using spectroscopy and something called Stark tuning (not the best reference, but it's also not behind a paywall) which states that the vibrational frequency of an harmonic oscillator is proportional to an applied electric field.  Not many metal/gas interfaces have been explored using this method, but those that have display surface potentials fairly close to 0 vs. SCE or 5 eV vs. vacuum.
What about the rest of the stuff in solution?
Back to the metal/liquid scenario, the last part of your first question asks about other ions in solution. They most definitely play a role in defining the potential of a metal in solution. In order to accurately predict electrode potentials, one must know concentrations and possible reactions (including states of matter) for all species present.
What about question 2?
I'm afraid question 2 might be based on the false premise that we can measure the potential of an electrode in the absence of anything else (liquid, gas, ions or molecules).  When electrochemists refer to a 'half cell', they have in mind both the reduced (in this case, metal electrode) and oxidized (metal ions) forms of a redox couple contained within the same system.  (Think about a beaker with a copper wire placed in it and the beaker is filled with a solution containing 0.1 M $\ce{CuCl2}$. That half-cell or system has a given potential based upon (primarily) the standard reduction potential of $\ce{Cu^{2+}}$ to $\ce{Cu}$ and the concentration of $\ce{CuCl2}$. If I have misread your second question, please edit and I will update my answer as needed.

That explained quite a bit. I have edited my second question. I require a couple of clarifications. Is the requirement of the second phase just another way to tell that the ions would constitute a very small concentration in case of only one metal? Next, is it the standard equilibrium we are talking about,( standard as in describable by $\Delta G=-RT\ln(K)$, or a mixture of adsorption effects into the equilibrium?

I also read about the double layer but couldn't fit it in the equilibrium concept. Can you help?

@SatwikPasani The edit to your 2nd question still refers to electrostatic potentials of metals and solutions individually, which is similar to asking for free energy for the reaction $\ce{H2O ->}$.  Without knowing the products and reactants one cannot answer the question. The electrode double layer arises from aligning opposite charges at the interface.  Lastly, interfacial potentials can be described using the relationship between $\Delta G$ and $K$ as you mention, although adsorption does play a role in advanced treatments.


text:
Have we ever created artificial metals?
I really don't know anything about chemistry, but I was wondering about the following vague question: has any substance been created that does not contain any metallic elements, but is metallic?  In fact, the only nonelemental metals I know of are alloys, so I may as well ask whether we have ever created any metallic compounds, with or without metallic constituents (without preferred).  Both Wikipedia and the tag wiki here claim that metallic compounds do exist but don't give examples.
My vague understanding of what makes an element metallic is that it has something to do with the sharing of electron clouds between atoms, and this question is motivated by wondering whether this kind of quantum effect can be induced from nothing in an appropriately constructed molecule.

If I understand correctly, you have 3 questions:

Are there any compounds created from non-metal elements that behave like metals
Is it possible to alloy a non-metal with a metal
Do metals form compounds

Is that about right?

Yes; 2. No, that was just a comment about what I know about non-elemental metals; 3. No, by "metallic compounds" I meant what you wrote in #1.  I understand that metals form compounds (e.g. rust).

What properties of metals are you looking for? There are compounds that conduct like metals, shape like metals, have heat capacities/melting points of metals, etc. etc. but I don't know of any that share all the properties that we think of as "metallic."

Metals have several general properties in common (to some degree):

Luster - metals are shiny
Electrical conductivity - metals conduct electricity
Thermal conductivity - metals conduct heat
Ductility - metals can be drawn into wires
Malleability - metals can be beaten into shapes
Fusibility - metals can be melted and forged

All of these properties derive from the "metallic bond", a delocalized sharing of electrons throughout the metal lattice, which is the key necessity for displaying macroscopic metallic properties. All metals do not behave the same way for these properties. For example, gold is more malleable than iron.
Metalloids are those elements with one or more allotropes that have metallic characteristics. Few if any metalloids display all of the metallic characteristics (otherwise they would be metals!). For example, graphite is lustrous and electrically conductive, but not thermally conductive, malleable, ductile, or fusible. The metalloid with the most metallic behavior is tellurium, which is fusible and ductile.
As for compounds, many are like the metalloids - they have some properties of metals but not all. For example, silicon carbide is lustrous, but that's about it. Otherwise it is hard, brittle, and an insulator.
Polyacetylene and derivative polymers are sometimes call "synthetic metals" because of their electric conductivity and sometimes luster. However, again, they are thermally insulating. They also tend to burn instead of fuse.
Metalloid elements and compounds are not quite metallic because their electron delocalization is in only one (polyacetylene) or two (graphite) dimensions, not all three.

Update!

Based on Snipergirl's comment, it appears that we are closer than I originally thought to preparing true synthetic metals. This highlight at phys-org suggests that supramolecular lattices of redox-coupled conjugated organic ions can exhibit metallic behavior at low temperature. The full article is at Nature Communications and has been made open access. These compounds appear to be close to achieving delocalization in all three dimensions. The behavior of this organic material is not surprising: metals and metalloids become even more metallic at cryogenic temperatures. The induced superconductivity allows for such cool things as magnetic levitation and MRI.

Just to add to your excellent answer, check this out: http://phys.org/news/2013-02-metallic-behavior-material-heralds-electronic.html !

This answer explains well what to look for in an artificial metal; I gather from it that the answer to my question is probably "no", though Snipergirl's link is fascinating (and rather timely).

@Snipergirl Do you have any details to add on the subject of organic metals?  They allegedly share all properties of metals, so conceivable answer my question.

I'll need to read the article referenced in the post @Snipergirl linked and look at the data to be certain, but the research looks promising. Conductivity and other metallic properties do increase at cryogenic temperatures. However, neither I nor my institution subscribe to Nature Communications, so it'll be a while for me to get to it.

Ooh also there was something about metal oxides becoming metallic at high pressures? http://phys.org/news/2012-08-elusive-metal.html#inlRlv

Carbon Nanotubes and graphene can be metallic in the sense that they conduct electricity very well, they are produced mainly by humans (not seen much in nature), and they are created from carbon, which in it's graphitic form is not very conductive at all.
This answer doesn't answer all of your questions but it takes a small stab at the first.


text:
Why Add Salt to Vinegar to Clean Pennies?
One common cure for dirty pennies is to put them in vinegar and salt. The vinegar should react with the copper oxide on the penny in the following way:
$\ce{2CH3COOH + CuO -> (CH3COO)2Cu + H2O}$
Nurdrage, on YouTube, did a video on it, and he did say that salt speeds up the reaction. However, he did not mention the way in which it sped up the reaction.
The same exact question is here
However, the answers did conflict and there was no clear correct answer. I am re-asking the question, in hope of getting a clear(er) answer.


text:
Balancing the equation when MnO₂ is reduced to MnO by H₂(g)
I am studying for a final tomorrow and I was going through some problems on older exams and came across one which I wasn't really sure how to solve.
The problem is to write the balance equation when:
managnese(IV)oxide is reduced to manganese(II) oxide by hydrogen gas.
I took a few approaches to this, because at first I did not realize it say reduced "by" hydrogen gas. This part threw me off because I initially added $\ce{H2(g)}$ to the product side.
So far I have:
$\ce{MnO2(s) + H2(g)-> MnO(s) + H2O(l)}$ which is not balanced yet. I noticed in this case the conventional type of balancing the equation does not work. Would this be a redox rxn in which I find each half reaction and then balance?
I am completely stuck at this point, and I don't think my formula is correct. Does anyone have any insight or good examples ?

This is balanced: I count the same number of elements on each side, which is the definition of balanced. What is not explicitly stated is that your hydrogen goes from $0$ to $+1$, and is oxidized, thus reducing your manganese from $+4$ to $+2$. But this all balances!


text:
When should we not draw a Lewis dot structure?
My friend once told me that technically , we should not draw Lewis dot structure of molecules containing no central atom(for example for benzene,
acetic acid).
Is there an explanation for this?

Short answer: no, there's nothing preventing you from drawing arbitrarily complex molecules according to the conventions of Lewis dot structures. They are not special, and the rules for drawing them are settled by convention and have evolved since they were first developed in Lewis's original paper. They're just one type of diagrammatic rendering of molecular structure which (probably due to their simplicity and explicitness) has caught on in pedagogy. They don't see much use beyond that. Read on for further details.
It's worthwhile to precisely define our terms here. The IUPAC Gold Book provides the following definition for a central atom:

The atom in a coordination entity which binds other atoms or groups of atoms (ligands) to itself, thereby occupying a central position in the coordination entity.

By that definition, the term is reserved almost exclusively for coordination complexes with a distinct metal atom at the center and ligands bound to it (with some exceptions for purely molecular Lewis adducts, e.g., amine oxides), where the electrons of the bond(s) originate on the ligands, rather than each atom donating an equal number of electrons in the formation of the bond(s). For non-coordinate covalent bonds, if the bonds were to be split homolytically (i.e., such that an equal number of electrons are distributed to each atom), the result would be separate, neutral atoms. Were coordinate covalent bonds to be split this way, that would not be the result.
Obviously, that definition is highly limited, and it's not the prevailing definition in my experience. The other usage of the term is essentially to refer to any non-terminal atom (i.e., one that has two or more bonds to separate atoms). There's nothing inherent to Lewis structures that prevents them being used to represent molecules containing multiple such central atoms. They're simply one representation of bonding. In practice, they don't see much use, chiefly because they're substantially less economical than condensed structural formulas or the skeletal bond-line structures (introduced in their prototypical form by Kekule), and because pure Lewis dot structures, as they were originally conceived, are exceedingly poor at representing stereochemistry.

Lewis dot structures are rhetorical devices to understand bonding and molecular structure (using the VSEPR model). Any molecule can be drawn using a Lewis dot structure if it helps you  correctly draw the structure. Often molecules (especially large molecules) are not drawn with Lewis dots to de-clutter.

Adding to GregE's answer, one particular class of molecules where a Lewis diagram could be deceptive are resonance structures.  In such molecules, there are multiple valid Lewis diagrams because the actual bonds are delocalized, meaning that the electrons are free to roam over several bonds.  One of the most vivid examples of resonance is benzene.
Here are some representations of benzene from the wiki (licensed under Wikimedia Commons):

Now look at the diagram in the middle of the top row.  It shows that there are two Kekule structures, which could be rendered as Lewis diagrams easily.  But the reason you need both is because the double bonds shown between the carbons are notional: there is no reason why either structure is preferred.  And the reality is that a benzene molecule is more like the "average" of the two structures.  The bottom row shows various representations showing the delocalized electrons, and all of these are "better" than the static Lewis or Kekule diagrams.  But the Kekule diagram is a lot easier to write down, and any chemist will understand that resonance is involved.


text:
Balancing Oxidation-Reduction equations
Anode: $\ce{2 H2O_{(l)} -> O2 + 4 H+_{(aq)} + 4e^-}$
Cathode: $\ce{ 2 H2O_{(l)} + 2 e- -> H2_{(g)} + 2 OH-_{(aq)}}$
Can someone show me the steps for adding the two. For the overall reaction, it is $\ce{2 H2O_{(l)} -> 2 H2_{(g)} + O2_{(g)}}$

It is not entirely inconceivable that a Galvanic cell could be created with an acidic anode and basic cathode.  Having both half-reactions occurring in the same beaker, however, is unlikely.  That has implications when determining the thermodynamics (cell potential) of the reaction but not in balancing the half reactions.  Given:
$$\begin{align}
\ce{2 H2O_{(l)} & -> O2 + 4 H+_{(aq)} + 4e^-} \tag{1}\\
\ce{ 2 H2O_{(l)} + 2 e- & -> H2_{(g)} + 2 OH-_{(aq)}} \tag{2}
\end{align}
$$
Multiply equation 2 by 2 and add:
$$\begin{align}
\ce{2 H2O_{(l)} & -> O2 + 4 H+_{(aq)} + 4e^-} \tag{3}\\
\ce{4 H2O_{(l)} + 4 e- & -> 2 H2_{(g)} + 4 OH-_{(aq)}} \tag{4} \\
\ce{6 H2O_{(l)} + 4 e- & -> O2 + 2 H2 + 4 H+ + 4 OH- + 4 e^-} \tag{5}
\end{align}
$$
The key here is that the  $\ce{4H+ + 4OH-}$ showing up on the same side equation 5 will yield $\ce{4H2O}$.  Combining acid and base (to make $\ce{4 H2O}$ on the product side) and canceling the $\ce{4e-}$ and $\ce{4 H2O}$ that occur on both sides of the chemical equation gives the desired equation.

You cant really combine those because for the top one you're making an acid, and for the bottom one you're making a base.  You've either got to make an acid on one side and consume it on the other, or consume a base on one side and make it on the other.  Here are the two ways:
Anode: $\ce{2H2O(l) ⟶ O2(g) + 4H+(aq) + 4e-}$
Cathode: $\ce{2H+(aq) + 2e- ⟶ H2(g)}$
Multiply bottom by two and the $\ce{H+}$ and $\ce{e-}$ cancel out with the top leaving the overall reaction.
Or:
Anode: $\ce{4OH- (aq) ⟶ 2H2O(l) + O2(g) + 4e-}$
Cathode: $\ce{2H2O(l) + 2e- ⟶ H2(g) + 2OH- (aq)}$
Multiply bottom by two and again get the overall reaction.


text:
Difference between Reaction Rate and Rate Law?
Lets say you have the reaction
$$\ce{$a$A + $b$B -> $c$C + $d$D}$$
The rate of the reaction is given by the change in initial concentration over the change in time.
$$-\frac{\Delta \left[\ce A \right]}{a\Delta t} = -\frac{\Delta \left[\ce B \right]}{b\Delta t} = \frac{\Delta \left[ \ce C \right]}{c\Delta t} = \frac{\Delta \left[\ce D \right]}{d\Delta t}$$
However, there is also the rate expression, which is of the form
$$\text{reaction rate} = k \left[\ce  A \right]^{m}\left[\ce  B \right]^{n}$$
What is the difference between these two, and what purpose does each one serve?
Further, what does the $k$ represent?

Let's use the following gaseous chemical reaction as an example:
$$\ce{H2 + I2 -> 2HI}$$
When we study reaction kinetics, we oftentimes can't observe the changes in concentration of each reactant and product.  Therefore the rate expression helps us to relate the rates of appearance/disappearance of the species present:
$$-\frac{1}{1}\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\;
= \; -\frac{1}{1}\frac{\Delta \left[ \ce{I2} \right]}{\Delta t}\;
=\; \frac{1}{2}\frac{\Delta \left[ \ce{HI} \right]}{\Delta t}\tag{rate expression}\label{rateex}$$
Note that each rate is modified by the inverse of the stoichiometric coefficient.  I explicitly put $\frac{1}{1}$ in the expression just to emphasize this point.  We can see that multiplying the coefficients of the reaction will not change the rate expression in the following example: let's multiply the coefficients in the chemical reaction by 2 and then write the corresponding rate expression.
$$\ce{2H2 + 2I2 -> 4HI}$$
$$-\frac{1}{2}\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\;
= \; -\frac{1}{2}\frac{\Delta \left[ \ce{I2} \right]}{\Delta t}\;
=\; \frac{1}{4}\frac{\Delta \left[ \ce{HI} \right]}{\Delta t}\tag{new expression}\label{newex}$$
The $\ref{rateex}$ and $\ref{newex}$ are numerically equivalent.
Full disclosure here: the term rate expression is ambiguous and is also used as a synonym for the rate law. I've also seen the rate expression referred to as the rate relationship.
The rate law or differential rate law relates the rate of a reaction to the concentration (or pressure) of the reactants.  The rate of a reaction is proportional to the concentration (or pressure) of the reactants modified by some experimentally determined number called the reaction order.
$$\begin{align}
-\frac{\Delta\left[ \ce{H2} \right]}{\Delta t}
&\propto\left[\ce{H2}\right]^m\left[\ce{I2}\right]^n\\
-\frac{\Delta\left[ \ce{H2} \right]}{\Delta t}
&= k\left[\ce{H2}\right]^m\left[\ce{I2}\right]^n\tag{rate law}
\end{align}$$
Here, $m$ and $n$ are the reaction orders for $\ce{H2}$ and $\ce{I2}$, respectively, and the sum, $m + n$ yields the overall order of the reaction.  The proportionality $a\propto b$ can be replaced with an equality $a = k b$.  In kinetics, this constant of proportionality is referred to as the rate constant.

In summary: the rate expression tells us how the appearance/disappearance rates of the products and reactants relate to one another; the rate law tells us how the rate is related to the concentrations (or pressures) of reactants and the rate constant is a constant of proportionality that comes out of the rate law.
This answer is a summary of information found at the boundless website and from Tro's most recent general chemistry textbook.

How do you calculate the rate constant? Is Arrhenius the only equation for calculating it?

@Paul Have a look here for information on determining the rate constant.  It is easy to do if you have a first-order reaction, but becomes more complicated for higher order systems.

@bobthechemisWhich of the following formula describe the rate law (with explaining):
$$ -\frac{1}{2}\frac{\Delta [ \ce{H2} ]}{\Delta t}= k[\ce{H2}]^m[\ce{I2}]^n$$
Or
$$-\frac{\Delta[ \ce{H2} ]}{\Delta t}= k[\ce{H2}]^m[\ce{I2}]^n$$

Reaction rate $\dfrac{dA}{dt}$ is the rate at a specific concentration and a specific time.
Rate law is an equation that shows how (velocity) a rate varies as concentration changes. It describes rates at ALL concentrations and NOT just one specific rate at one specific concentration.

I am sorry, but I seem to miss how this answers the question, could you add a little more explanation?

The question is literally asking whats the difference between Reaction Rate and Rate Law