text:
(Effective) number of molecules interacting with light to produce optical activity
The question is: How can I get a sensible and concrete (not relative) value for the number of molecules light has interacted with to cause an optical activity phenomenon?
Specifically, my issue in answering the question is that although initially I thought of calculating the number of molecules contained in the volume that the light is occupying, this does not seem correct to me. The reason is that the optical activity is independent of the volume, it only depends on the path length but not the transverse area. So it seems to come down to some kind of "effective" number of interactions, which is invariant to the transverse spread of the light.
Thanks for any help you can give me!

I think the picture above is self-explaining.
Now you must know the diameter and lenght of the lightpath so you can calculate the illuminated volume. Taking this volume and the concentration c you get the mol quantity (6.022 * 10^23) of molecules which have caused optical activity.

Can you please type out whatever text there is in the image and quote the source?

Thank you for writing this.
My issue is exactly concerning such a volume calculation, as explained in the question: It does not make sense to me since the optical activity does not depend on the diameter of the light.
However, maybe the concern was a bit hidden before. I have now shortened the question to make it more apparent.


text:
What is the reason behind optically activity of compounds
I am currently being taught optical activity of compounds and I had this question:

Optical rotation (optical activity) is the turning of the plane of linearly polarized light about the direction of motion as the light travels through certain materials.

What happens inside these compounds (for example sucrose) that causes them to behave on this manner. By that I mean to say that what takes place in the compound due to which they turn the plane of light when it travels through these materials?

Look at your two hands. They are mirror images of each other. That is also why you can't put your right shoe on your left foot. Then see: https://en.wikipedia.org/wiki/Chirality_%28chemistry%29

It's a good question, but asking about physical reason how optical rotation really works could be better for Physics SE - it seems many chemists are satisfied with practical side of this phenomenon.

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text:
Physical implications of the simple harmonic oscillator quantum mechanical solutions
Take a look at the wavefunctions for the different energy levels of a simple harmonic oscillator (a crude approximation for a diatomic).

The wavefunctions seem to make sense: they tend to zero as x tends to plus or minus infinity so it is normalisable. The number of nodes increases by one with increasing energy levels etc.
However, when you take a look at $|\psi(x)|^2$ which represents the probability density of the particle to be found at that point in space (or the diatomic to be in that particular state/bond length) I struggle to see how a node can exist. This is because a diatomic will stretch and contract through the full range of extension/contraction - the quantum representation does not seem to conflict this in the sense that there is a non zero probability when the potential is equal to the total energy. However, to reach this state from, for example, $x=0$ in the $n=2$ energy level it must pass through a node. What does this mean? surely a bond can't cease to exist as it stretches from equilibrium to some extended or contracted state? What's really going on here?

However, to reach this state from, for example, x=0 in the n=2 energy
  level it must pass through a node.

To start, ask yourself: "What should pass through a node?" The answer is: it is the probability that "passes through the nodes", not atoms (or, I should better even say nuclei). Probability is just a mathematical quantity ant the zero value is allowed for it, so there is nothing to worry about. Yes, quantum harmonic oscillator has some forbidden regions of displacement from the equilibrium. But it can also penetrate into classically forbidden regions. Quantum world is weird.
The root of this (and most of the other) apparent quantum paradoxes is the classical way of thinking: you think that you have some funky quantum balls on a funky quantum spring out there, but you don't. Quantum harmonic oscillator is a quantum system that essentially has the Hamiltonian similar in its form to the Hamiltonian of a classical harmonic oscillator. And this is it. No balls, no springs, just a similar Hamiltonian.

It should be mentioned as well that atoms (again, more precisely, nuclei) are not vibrating in molecules even in a classical sense: their motion can be only approximately separated into vibrational and rotational ones. Besides, the vibrational motion can only approximately be treated as being harmonic.

I do understand that this is a rather crude and simplistic approximation but please just suppose this was the realistic situation in a diatomic - i.e the potential well is described by $-1/2kx^2$ and the diatomic (reduced to a one body problem by utilising the reduced mass) stretches and contracts from one extreme of contraction to the extreme of stretching - i.e $x$ goes from $-x_{extremity}$ to $+x_{extremity}$. At some point in this action, it must surely pass through these nodal displacements between the extremities. I'm asking how is this so?

In your simplistic approximation where you are talking about a nuclei moving from one extreme to another (in a 1D world), yes it must pass through those points of 0 probability, but the whole point of this answer is that your simplistic approximation is the wrong way to interpret the results at this level of detail.  Quantum mechanics does not treat the vibration of a diatomic molecule as a spring and mass moving one way then another, why then would we expect its predictions to retain characteristics of the spring-mass system?

The conception of moving nuclei in a harmonic motion is simply wrong. At best, it shows what geometries are possible for a given mode but don't think that molecules move like this.
The $|\Psi(x)|^2dx$ tells you what the probability is to find your geometry at displacement $x$. Forget get the idea of the moving nuclei, this idea is just a simple concept to visualize normal modes but it is not suitable for quantitative assessments.
Nodes also don't mean that your molecule can't pass through that point. Generally you can never measure a single point, meaning it is impossible to measure your geometry exactly. You always have some uncertainty in your measurement apparatus which means that you always measure a range of x-values and the probability around a node does not vanish as soon as you look at small interval for x around your node.
There is one case when the approximation of a vibrating molecule is acceptable and that is for highly excited vibrational states. These states give probability distributions which are similar to classical harmonic oscillator motion. But even then it is still an approximation and you still have nodes. But the "motion" of harmonic oscillator groundstates and lower states is nothing like the classical motion.

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text:
Is it really appropriate to measure entropy change using enthalpy?
One statement by Peter Atkins in his book Elements of Physical Chemistry confused me quite a bit:

If a chemical reaction or phase transition takes place at constant pressure , we can identify $Q$ with the change in enthalpy of the system, and obtain
For a process at constant pressure: $$\Delta S = \frac{\Delta H}{T}.$$

Is it really correct to measure the entropy change using change in enthalpy?
I'm telling so because enthalpy change only measures how much heat energy enters or gets expelled from the system. It doesn't measure how much energy gets expelled out of the system when the system does $PV$ work.
As Peter Atkins points out:

If $10~\text{kJ}$of energy is supplied as the heat to the system that is free to change its volume at constant pressure, then the enthalpy of the system increases by $10~\text{kJ}\; ,$ regardless of how much energy enters or leaves by doing work and we write $\Delta H = +10~\text{kJ}\; .$

I know it is the heat energy that is concerned in the definition of entropy change. But, suppose the system is heated which means it is in more disorder; however if it does work, then the energy the system got as heat would decrease which would therefore decrease the chaos & disorder of the system. If we use only the enthalpy change, wouldn't we exclude that work which decreases the heat energy gained by the system?
I'm rather confused on the use of enthalpy for defining the entropy change. Could anyone please help me explain this how enthalpy change actually measures the entropy change?

Please articulate your understanding of how the change in entropy of a closed system is determined, starting from thermodynamic equilibrium state A and ending at thermodynamic equilibrium state B.

@Chester Miller: It should be $$\int_{A}^{B}  dS = \int_{A}^{B} \frac{Q_\text{rev}}{T} dT $$ where $Q$ is the net heat energy received by the system reversibly.

Yes.  So to get the change in entropy, you need to identify a reversible path between thermodynamic equilibrium state A and thermodynamic equilibrium state B.  If it is a phase change, you start out with liquid  at temperature T and equilibrium vapor pressure P(T) (state A), and end up with vapor at the same temperature and pressure (state B).  This can be done reversibly by putting one mole of a liquid in a cylinder with a massless frictionless piston and holding the external pressure on the piston constant at P while adding heat very gradually (reversibly) until all the liquid changes to vapor.  During this process, the volume of material in the cylinder increases from that of one mole of liquid to that of one mole of vapor.  Since the pressure is held constant throughout, the work done is $P(V_v-V_l)$.  From the first law, the change in internal energy is equal to the heat added minus the work done:  $(U_v-U_l)=Q-P(V_v-V_l)$.  Rearranging this equation gives:$$(U_v+PV_v)-(U_l+PV_v)=H_v-H_l=\Delta H_{lv}=Q$$So the heat that you add reversibly is equal to the change in enthalpy in going from the liquid at T and equilibrium vapor pressure P(T) to the vapor at T and equilibrium vapor pressure P(T).  Since this is a reversible process at constant temperature, $\Delta S=\frac{Q}{T}=\frac{\Delta H_{lv}}{T}$.  This completes what I wanted to say about the phase change situation.  I will move on to the chemical reaction situation after you have had a chance to look this over and ask questions.

Firstly, thanks for the answer, sir.  I conceived the maths. However,  I've a query: Basically what is entropy? Entropy is the chaos or the disorderliness in the system caused due to the heat energy supplied. First the liquid got thermal energy $Q$ which increased the chaos in the liquid (soon to be vapour). But in the course of doing so, it does work which decreases the heat energy available in the system to $Q- P\Delta V.$ And it is heat which determines the chaos & now there is only $Q- P\Delta V$ left in the system which determines the chaos & not $Q.$ ...

... So, why should the definition of entropy takes $Q;?$ As the heat energy available for the chaos is $Q- P\Delta V$ & not $Q; ?$

Entropy is a measure of the number of quantum mechanical states available to the system.  This is what we mean when we talk about chaos or disorderliness.  The thermal energy of the system is vastly increased when we convert a liquid to a vapor because we have overcome the attractive forces holding the molecules together, and many more quantum mechanical states are now possible.  We have also greatly increased the volume of the gas which also allows for vastly more quantum mechanical states.


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Joules to kJ/mol conversion
In my AP Chem class we are working on testing Hess's Law and conducted three reactions. Then enthalpy changes of the 1) $\ce{NaOH + HCl}$ and 2) $\ce{NaOH + NH4Cl}$ to predict the enthalpy change of 3) $\ce{HCl + NH3}$.
I used the equation $q = C_pm\Delta T$. We determined the $q$-values using a method given to us by our teacher. She told us to assume a denisty of $1.03\ \mathrm{g/mL}$ for all solutions, we then multiplied the density by $50\ \mathrm{mL}$ to determine the grams of each reactant present. That value is the $m$ term. We used $4.18\ \mathrm{J/(g\ ^\circ C)}$ for the specific heat of water. The temperature change for the first reaction was $14.7\ \mathrm{^\circ C}$, the second was $1.2\ \mathrm{^\circ C}$, and the third was $9.8\ \mathrm{^\circ C}$.
The first reaction yielded a q value of $6328.938\ \mathrm J$, the second yielded $516.648\ \mathrm J$, and the third yielded $4219.292\ \mathrm J$.
The next step would be to convert this $q$ values into $\mathrm{kJ/mol}$. I consulted my lab partner who did the following operation:
$$\mathrm{kJ/mol} = 6328.938\ \mathrm J \times 1\ \mathrm{kJ}/1000\ \mathrm J \times x/0.2\ \mathrm{mol}$$
In his math, there was no numerator on the $0.2\ \mathrm{mol}$ term. The mol value he used came from there being 0.2 moles of the reactants present. I believe his math to be incorrect, but as my teacher is out sick and there are no other chemistry teachers present, I cannot determine the correct conversion. I believe it to be:
$$\mathrm{kJ/mol} = 6328.938\ \mathrm J \times 1\ \mathrm{kJ}/1000\ \mathrm J \times 6.022^{23}/0.2\ \mathrm{mol}$$
Both of the conversions yield values with significant % error. Any assistance is appreciated.

Accepted book enthalpies are in the units $\mathrm{kJ/mol}$.
Simply convert $\mathrm J$ to $\mathrm{kJ}$ by dividing by $1000$. Since $1\ \mathrm{kJ} = 1000~\mathrm{J}$.
The moles of dilute aquaous solutions is best determined by volume (L) times concentration (mol/L).$$n = V \times c$$ You would use the moles of the product. If there is, in fact, "$0.2\ \mathrm{mol}$ of the reactants present" then there would be $0.2\ \mathrm{mol}$ of the products present. since the chemical reactions are $\ce{1 mol + 1 mol -> 1 mol + 1 mol}$.  (You are not interested in the number of atoms per mole, so you would not use Avogadro's number.)
$$q= C_\mathrm{sol} m\Delta T$$ where $C_\mathrm{sol}= 4.06~\mathrm{J/(g \ ^\circ C)}$ (Specific heat capacity for dilute aqueous solutions)
$$q= C_\mathrm{sol} m\Delta T / n_\mathrm{product}$$  ($n = \text{moles product}$)
Remember that Hess's Law states that

if a reaction can be expressed as the sum of two or more reactions, then the heat flow for the overall reaction is equal to the sum of the heat flows.

Hence, the sum of the first 2 reactions should be equal to the 3rd reaction.

$\displaystyle \ce{NaOH + HCl -> NaCl + H2O}$
$\displaystyle \ce{NH4OH + NaCl -> NaOH + NH4Cl}$  (Reversed and Enthalpy sign changed)
$\displaystyle\ce{HCl + NH4OH -> NH4Cl + H2O}$

Note how $\ce{NaOH}$ and $\ce{NaCl}$ cross out leaving the third reaction.


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Substitution Reactions; Stability of Carbocations
I was wondering which is more stable: a secondary carbocation or a primary carbocation that has resonance stabilization?
I was reading through the organic chemistry text, it said that resonance stabilizes carbocation.
However it did not explain the which which is more stable then the other.
This is not a homework question. I am genuinely curious.

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

it is generally admitted that resonance effects are stronger that inductive effect.

The order of stability of carbocation is 3°>2°>1° of course stabilisation by resonance is more worthy then inductive effect. You can get more detail on wikipidea regarding it

If it's the case that both primary and secondary carbonation is undergoing the same resonance then we can say 2° is more stable due to added inductive effect by the alkyl groups.
Now it can also depend on the extent of resonance (the number of resonance structures) the carbocation is undertaking. There can also be hyper conjugation stability for the carbocations.
It is also important for us to consider the stability order of carbocations.
This stability order too depends on the factors mentioned above.
Hope it was helpful.

Actually a primary carbocation with resonance stabilization us more stable than secondary carbocation because resonance stabilization effect > inductive effect.


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Solubility Product and Buffers

The ionization constant of benzoic acid is $6.46 \times 10^{-5}$ and solubility product for silver benzoate is $2.5 \times 10^{-13}$. How many times silver benzoate is more soluble in a buffer of $\mathrm{pH}$ 3.19 compared to its solubility in pure water?

Equation used to calculate solubility of silver benzoate in water:
$$\ce{C6H5COOAg <=> C6H5COO- + Ag+}$$
I've calculated the solubility of silver benzoate in water , it is $5 \times 10^{-7} \pu{mol/L}$ (since that is all I calculated, this is the only equilibrium equation I know).
But I cannot understand what to do next.

Ok, the answer is in three parts. Let's use $B^-$ to represent the benzoate anion.
(1) Solubility of Silver Benzoate in pure Water
$\ce{AgB_{(s)} <-> Ag+_{(aq)} + B^−_{(aq)}} \quad\quad K_{sp} = 2.5 * 10^{-13}$
thus:
$\ce{[Ag^+] [B^-]} = 2.5 * 10^{-13}$
assume $\ce{[Ag^+] = [B^{-}]}$ then
$[Ag^+] = [B^-] = \sqrt{2.5 * 10^{-13}} \text{m/l}  = 5.0 * 10^{-7} \text{m/l}$
!! Check !!

We want to check if $\ce{[B^{-}] >> [HB]}$ so that $[B^{−}_{(aq)}] + [HB_{(aq)}] \approx 5.0 * 10^{-7}$ and that the pH will stay at 7.
We know that:
(a) $\ce{HB_{(aq)} <-> H^{+}_{(aq)} + B−_{(aq)}}$
(b) $\frac{[H+][B−]}{[HB]}=K_a=6.46∗10−5$
(c) pH for pure water is 7
So rearrange (b) and plug in our assumptions
$ [HB] = \dfrac{[H+][B−]}{6.46∗10^{−5}} =  \dfrac{(1*10^{-7})(5.0*10^{-7})}{6.46∗10^{−5}} = 7.7 * 10^{-10}$
Since $7.7 * 10^{-10} << 5.0 * 10^{-7} $ we can safely ignore protonation of $\ce{B^-}$ to $\ce{HB}$.

(2)Solubility of Silver Benzoate in a buffer of pH 3.19
(a) $\frac{[H+][B−]}{[HB]}=K_a=6.46∗10^{−5}$
(b) pH for buffer is 3.19 therefore $\ce{[H^+]}$ = 6.45 * $10^{-4}$
(c) $\ce{AgB_{(s)} <-> Ag+_{(aq)} + B^−_{(aq)}} \quad\quad K_{sp} = 2.5 * 10^{-13}$
So rearrange (a) and plug in $\ce{[H^+]}$
$\dfrac{[B−]}{[HB]}= \dfrac{6.46∗10^{−5}}{6.45 * 10^{-4}} = 0.10 $
Thus in this acid solution we must consider the protonation of $\ce{B^{-}}$ to $\ce{HB}$ when solving for the solubility of silver benzoate.
so we another equation.
(d) $\ce{AgB_{(s)} <-> Ag+_{(aq)} + B^{−}_{(aq)} + HB_{(aq)}}$
but we know that $\ce{[Ag+_{(aq)}] = [B^{−}_{(aq)}] + [HB_{(aq)}] = [B^{−}_{(aq)}] (1 + $\dfrac{\ce{[HB_{(aq)}]}}{\ce{[B^{−}_{(aq)}]}}$) = 11 [B^{−}_{(aq)}]}$
so from (c) we get:
$\ce{11[B^{−}]^2} = 2.5 * 10^{-13}$
$\ce{[B^{−}]}^2 = \dfrac{2.5 * 10^{-13}}{11} = 2.27 * 10^{-14}$
$\ce{[B^{−}]} = 1.51 * 10^{-7}$
and thus
$\ce{[HB]} = 10 \ce{[B^{−}]} = 1.51 * 10^{-6}$
$\ce{[Ag^+]} = 11 \ce{[B^{-}]} = 11 * 1.51 * 10^{-7} = 1.66 * 10^{-6}$
!! CHECK !!

$[\ce{HB}] + \ce{[B^{-}]} = 1.51 * 10^{-6} + 1.51 * 10^{-7} = 1.66 * 10^{-6}$
$[\ce{Ag^+}] [\ce{B^{-}}] = (1.66 * 10^{-6})(1.51 * 10^{-7}) = 2.51 * 10^{-13}$

(3) Ratio of Solubilities of Silver Benzoate
$\text{Ratio} = \dfrac{\text{solubility in buffer}}{\text{solubility in water}} = \dfrac{1.66 * 10^{-6}}{5.0 * 10^{-7}} = 3.3 $

Great ! But why would I want to check if [B-] >> [HB] ? And , in both the cases , why would I want to add the concentrations of [B-] and [HB] ? Aren't they two different things ?

In the first solution in pure water you want to check for the protonation of B− to HB. That would cause more silver benzoate to dissolve as it does in the buffer solution. There is "some" HB in pure water but the quantity is minuscule to the B- species.

Re adding [B-] and [HB] - There were no B species in solution before the solid sodium benzoate was added. So $\ce{[Ag^{+}]  =  [B^{-}] + [HB]}$. in other words all the B species had to come from silver benzoate that dissolved.

So we are adding them because both of them are B species. And in the first case as [HB] is negligible , the concentration of [B-] + [HB] is considered to be the concentration of [B-] alone , but in the second case , as [HB] is not negligible , we have to consider it in the total concentration [B-] + [HB]...Right ?

Yes. Both [B−] and [HB] are "b species" which were from silver benzoate.

Also , will the reaction between B- and H2O look like this : B- + H2O <--> HB + OH- ?

Yes. In the pure water system if there had been significant protonation the pH would have risen. In the buffer solution the pH is nailed to the wall.

In fact its "as easy as pie". But I know its always easy to think about things like it was an evidence. I will give my approach to solve this kind of problems. Obviously you can do the same with complexation, liquid/liquid extraction and so on.
Fisrt you have $$\ce{BzOAg}\leftrightharpoons \ce{BzO^-}+\ce{Ag^+} \text{ with the constant }\ce{K_s}$$
But you also have a second possible reaction which is $$\ce{BzO^-}+\ce{H_2O}\leftrightharpoons \ce{BzOH}+\ce{HO^-} \text{ with the constante }\ce{K_b}$$
You know that $$\ce{K_s}=[\ce{Ag^+}]_i[\ce{BzO^-}]_i$$
At any time $[\ce{Ag^+}]_i=[\ce{Ag^+}]$ and
$$\begin{align}[\ce{BzO^-}]_i&=[\ce{BzOH}]+[\ce{BzO^-}]\\&=[\ce{BzO^-}]\left(1+\frac{[\ce{H_3O^+}]}{\ce{K_a}}\right)\end{align}$$
Then you have $$\ce{s^2}=\ce{K_s}\times \left(1+\frac{[\ce{H_3O^+}]}{\ce{K_a}}\right)$$
Now you also know that $[\ce{H_3O^+}]=\ce{10^{-pH}}$ you have:
$$\ce{s}=\sqrt{\ce{K_s}\times\left(1+\frac{10^{-pH}}{\ce{K_a}}\right)}$$
I hope it will help you. Ask me if you have questions. Have a nice day.
Friendly

Thanks ! I have 2 questions , the answers of which will for sure solve my problem. 1)What is [Ag+]i and [BzO-]i ? 2)For the expression of [BzO-]i , why have you added [BzOH] and [BzO-] ?

They are the initial quantity. And for the second question it is because we need to take all the species containing BzO- :)

Just for clarification , BzO- is benozate ion , right ?

Yes of course ;)

$$\ce{s^2}=\ce{K_s}\times \left(1+\frac{[\ce{H_3O^+}]}{\ce{K_a}}\right)$$, how did you get this?

As ion benzoate is a base in the couple $\ce{C6H5COOH /C6H5COO- }$, we can write
$$s=\ce{[Ag+] =[C6H5COO- ] + [C6H5COOH ]}$$
$$s=\ce{[Ag+] =[C6H5COO- ] (1+\frac{[C6H5COOH ]}{[C6H5COO- ] })}$$
$$s=\ce{[Ag+] =[C6H5COO- ]} (1+\frac{[\ce{H3O+} ]}{K_\mathrm{a} })$$
On the other hand,$$K_\mathrm{sp}=\ce{[Ag+] \times [C6H5COO- ]} $$
$$K_\mathrm{sp}=\frac{s^2}{(1+\frac{[\ce{H3O+} ]}{K_\mathrm{a} })} $$
$$s=\sqrt{K_\mathrm{sp}} \times \sqrt{{(1+\frac{[\ce{H3O+} ]}{K_\mathrm{a} })}} $$
So, the solubility in the acidic buffer solution is increased by the factor:$\sqrt{1+\frac{[\ce{H3O+} ]}{K_\mathrm{a} }}$

Why have you added the concentrations of Benzoate Ion and Benzoic acid in your expression for 's' ?

Of course, when bonzoate ion is in aqueous solution, it reacts with water to form its conjugated acid. This reaction is not quantitative. It's  an equilibrium. So, the initial concentration of bonzoate ion, i.e. (s) becomes the sum of the remaining concentration of bonzoate ion and the concentration of the formed benzoic acid. This equation is also known under the name: conservation of matter (benzoic ion here).

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text:
How dangerous is a megacurie (MCi)?
I was researching the Kyshtym (nuclear) Disaster, and it said that it released 20 MCi (20 megacuries) of radiation into the air in a dust cloud.
I read from this Yahoo Answers answer that:

The curie is the old unit of activity and a very big unit it was.
1 curie = 3.70 x 10^10 disintegrations per sec.
In today's units 1 curie is about 30 Giga Becquerel.
So a curie of any radioactivity (alpha, beta or gamma) would fry you.
A sealed microcurie source could be safely handled.

If a single curie is this powerful, then what would twenty million curies (a 20 MCi) be like? Could you give any examples of events that exemplify this scale of radiation and damage (if you can't think of any real ones, then maybe a theoretical example)?

“So a curie of any radioactivity (alpha, beta or gamma) would fry you.” is nonsense.

There are a lot of factors at play here:

The curie and becquerel only tell you the number of disintegrations per unit time—they convey no information of the type or energy of radiation produced.
How dangerous a radioactive source is depends on the type of radiation produced and how one is exposed to it. External exposure to alpha particles is not very dangerous because alpha particles are easily stopped by skin, but inhaling alpha emitters can be very dangerous because there is direct access to sensitive alveolar cells.
How the radioisotopes are dispersed will greatly affect individual exposure to the radiation. A large amount of radioisotopes spread over a large area may increase the incidence of cancer slightly for a large number of people, but a smaller amount concentrated in one place may be very rapidly fatal to a single person.

20 MCi (740 PBq) is indeed a very large amount of radiation (~1/7 that released by the Chernobyl accident), but in a big cloud spread over thousands of square kilometres may not be acutely dangerous to people. That many radioisotopes is equivalent to 231,000 kg of cesium-137 and all that (magically) concentrated into a point source would produce a dose of .62 Gy/s for a person 1 m away—fatal in seconds.

I didn't realize the factors that could be at play here. Thanks for the explanation!

But back to the original question (and assuming that this radiation was concentrated), what kind of an event could this be compared to? You told me the effects if you stand so close- but what could have a similar effect?

@Jonathan I'm not sure what you mean by what "event could this be compared to". Except at very high levels, radiation exposure isn't immediately noticeable. If we're talking about .62 Gy/s, that's around 42 W of power dissipated as heat in a person of average mass, so there would be some heating. I could only speculate what that exposure actually feels like. Some reports from criticality accidents (e.g. demon core) described what might have been Cherenkov radiation visible in the eye.


text:
Stability of carbocations
We know that if a structure has all equivalent resonance structures, then it will be highly stabilised. So then why benzylic carbocation is more stable than other cyclic allylic carbocations, though allylic carbocation has all equivalent resonance structures?

For example you want to know why benzylic carbocation is more stable than  cyclohexen carbocation?

Do you know that the more you find mesomer forms the more the stability is? Do you want a proof?

More the number of resonance structures (alias mesomers) more is the stability of the carbocation so benzylic carbocation with 4 mesomers is more stable than 2 structures of the allylic carbocation.

-1: While this explanation is indeed correct in the case of benzylic and allylic cations, it is, in general, not correct (see: acetate, 2 resonance forms vs phenoxide, 4 resonance forms) and I cannot agree with the answer which strongly implies that it can be generalised to all systems.


text:
Reactions of Alkenes
During my organic chemistry lesson, I was told that the reason why alkenes are low in reactivity is that their sigma bonds have very low polarity.
Why is this the case?
Thanks, Dave

Is alekenes meant to say alkenes. If so then your question doesn't seem to make much sense. If not, then what is is supposed to say. Please clarify you question.

@bon sorted! Now check the question!

Alkenes are not "low in reactivity". Alkanes are.


text:
Common Ion Effect Calculations
Given a solution with a $\ce{S^{2-}}$ (sulfide ion) concentration of $2.76 \times 10^{-15}$ mol per litre, and $K_s(\ce{FeS})=4.90 \times 10^{-18}$, calculate the solubility of $\ce{FeS}$ in this solution.
I calculated this by letting the concentration of $\ce{Fe^{2+}} = x$, therefore the concentration of $\ce{S^{2-}} = 2.76 \times 10^{-15} + x$.
Therefore $4.90 \times 10^{-18} = x(2.76 \times 10^{-15} + x)$.
Upon solving this as a quadratic equation, I arrive at the result of $2.14 \times 10^{-9}$ mol per litre.
[EDIT: $2.21 \times 10^{-9}$ mol per litre]
The answers to this problem state to ignore the contribution of $\ce{S^{2-}}$ from the $\ce{FeS}$ as it is negligible, and state that the answer is actually $1.78\times10^{-3}$ mol per litre.
Which method is correct?

The value of the answer you were given is incorrect, both in its value and its explanation of what can be ignored.
As you note:
$K_{sp} = 4.90\times10^{-18}=[\ce{Fe^{2+}}][\ce{S^{2-}}]$
Just eyeballing the equation, we would expect a solution of $\ce{FeS}$ in pure water to contain equal quantities of iron(II) and sulfide ions, at a concentration of $2.21\times10^{-9} M$ (the square root of the solubility product.)  This is on the order of a million times more concentrated than the solution described in the problem, so we can ignore the sulfide concentration in the original solution.  For the purposes of this problem, the iron sulfide is dissolving in pure water.
Additionally, solving the quadratic equation you provided gives much the same result.  The solution rounds to $x = 2.21\times10^{-9} M$; it only differs in the sixth place from the square root of the solubility product.  It's easy enough to make a small numerical error with scientific notation and solving a quadratic; I have to assume that's what happened to you here.
The only way that your instructor's answer would be correct would be if the question intended us to understand that there was some chemistry going on to keep the concentration of sulfide ions at the given concentration.  Without that, it just doesn't make sense.

Yes, solving the quadratic again does result in x=2.21×10^−9.  Thank you, a bit shocking considering the question is from an example exam paper.

Given a solution with a S2- (sulfide ion) concentration of 2.76x10^-15 mol per litre, and Ks(FeS)=4.90x10^-18, calculate the solubility of FeS in this solution.

(a) $\ce{FeS_{(s)} <-> [Fe^{2+}] [S^{2-}]  \quad\quad K_{sp} =} 4.90*10^{-18}$
Given that $\ce{[S^{2-}]}$ is specified as 2.76x10^-15 mol there is no ambiguity in the $\ce{[S^{2-}]}$ concentration. So:
(b) $\ce{[Fe^{2+}]} = \dfrac{4.90*10^{-18}}{\ce{[S^{2-}]}} = \dfrac{4.90*10^{-18}}{2.76*10^{-15}} = 1.78*10^{-3}$
Chemically what is means is that something else must be reacting in the solution to remove almost all of the $\ce{[S^{2-}]}$ that the dissolution of the iron sulfide liberates. The book statement to "ignore" the sulfide that the dissolution creates is very misleading.
The gist I believe is this. Hydrogen sulfide, $\ce{H2S}$ has two protons and would seem to have two pKa values. The first depprotonation is what would be expected with $pK_{a1} \approx 7$ in aqueous solution. The second deprotonation doesn't "really happen" in water. The $pK_{a2} > 14$ which means that the $\ce{[S^{2-}]}$ ion is a very strong Arrhenius base in water.
So there is another reaction happening which removes most of the $\ce{[S^{2-}] species:
(c) $\ce{[S^{2-}] + H2O -> HS^{-} + OH^{-}}$
For calculation purposes the second deprotonation has an effective $pK_{a2} = 17.4$ in a paper by Migdisov et al.
Migdisov, A.A., Williams-Jones, A.E., Lakshtanov, L.Z., Alekhin, Y.V., 2002. Estimates of the second dissociation constant of H2S from the surface sulfidation of crystalline sulfur. Geochimica et Cosmochimica Acta 66 (10), 1713–1725.


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Major products of the reaction of 2-bromo-2-methylpropane and propanoic acid

Here's my attempt:
the electrophile is a tertiary carbocation, so that rules out sn2. I believe the nucleophile is a rather weak base since carboxylic acids are stable. However, it is polar, plus the temperature is low. Thus it seems likely that it proceeds by SN1? So would "I" be the correct answer?

After the formation of the carbocation, IV also seems to be a likely candidate to me.So I would go with I and IV (major product's').

I think more (IV) via SN1 at room temperature, with some (I) via E1 elimination. Heat to get more (I) than (IV).

SN1 does not lead to product 1, which is an elimination product. Just because it proceeds via a carbocation does not mean it is SN1.

The commenters are correct; this reaction, based on the steric hindrance of the tertiary alkyl halide and the propionic acid, I would expect a mixture of SN1 and E1, and minimal if any second-order reactions to the mechanism.
Specifically, the propionic acid would slowly approach the bromide substituent, and the acidic carboxyl proton (pKa ~ 5) would begin to associate with the bromide, which is a good leaving group. This slow formation of a tertiary carbocation (stabilized by hyperconjugation of the central empty p orbital through the C-H sigma bonds) then leads to the propionate binding to the carbocation with no rearrangements occurring.
Even though this reaction is at room temperature, E1 should not be ruled out, as there is no reason why a carboxylic acid cannot grab a proton from the 9 available ones on the alkyl halide. Due to the necessity for the proper orientation for beta-elimination though, it is plausible that E1 is less likely.
Regardless, I would project both SN1 and E1 to be the most likely reactions to occur. However, if you were to pick only one, I would pick SN1 based on the reaction temperature (and call E1 a "secondary" major product if that makes sense).

E1 does not require the antiperiplanar conformation. That is for E2.

Alright; adjusted that part.

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Why does this reaction not proceed by sn1/sn2?
this is an assignment question but I'm stumped.
we have a polar protic solvent (also a weak base) interacting with the secondary carbon -- it looks like a good example of an sn1/sn2 to me. So why wouldn't the answer be "I and III"?

do you mean a tertiary carbon? In that case, Sn2 is out of context.

Curious - What would the name of the starting compound be?  2-methyl-4-ethylether-hexane ?!?

@Max (S)-4-bromo-2,4-dimethylhexane

dah... I meant the name of the ether in answer (I)

Well, all in all, it doesn't seem that a SN1 or a SN2 reaction can be happening at all. A pure SN2 reaction would give (III). A SN1 would give a racemic mixture of (I) and (III). Some combination of SN1 and SN2 would give a  mixture of (I) and (III) with more (III).

@Max Agreed. For the compound in (I) just exchange bromo with ethoxy.

No good nucleophile and tertiary carbon means no SN2. SN1 will probably predominate. I don't like product 2 (which is formed via E1), if anything it should be the more substituted alkene. I saw your other question as well and I'm not convinced that you fully understand what SN1, SN2, E1 and E2 are... I recommend reading it through again.

The major reasons I would give for going with the SN1 products are:

The amount of steric hindrance on tertiary alkyl halides in this context highly limits the amount of second-order reactions that can occur (which require participation from two molecules).
EtOH is not a strong nucleophile in this context. In contrast, EtO- would be, but it's not present. Had EtOH been a strong nucleophile, SN2 would be a tad more likely, but a primary alkyl halide situation, for instance, is much more likely to favor SN2. Therefore, an SN2 product cannot form as a major product. However, considering both I AND III, the product is not formed through SN2 (unless you pick only III, in which case you assume I is not formed).
III is one product in the racemic mixture that would result from forming the tertiary carbocation (which is nearly planar around carbon 3, allowing for both retention and inversion of the stereochemistry). Also, the 20 degrees C does not convince me that elimination is a major product since lower temperature makes elimination more difficult. It is harder for the molecules to orient themselves quickly enough before all reactant is used up if the environment is cold. Therefore, the answer is probably I AND III, assuming that somehow the answer key is incorrect.

I also agree with orthocresol that product II is strange. I would expect by Zaitsev's rule that the new pi bond would form across carbons 3 and 4 in the main chain since his rule says that typically, the elimination occurs by acquiring a proton from the carbon with fewer protons, forming typically the more substituted alkene.


text:
Structure of OSF4
Recently I had given a test where I was asked to specify the hybridisation of $\ce{OSF4}$ and its structure.
I guessed that the structure would be trigonal bipyramidal, but I was unsure since I hadn't dealt with molecules with double bonds.
So my first question is: Should having double bonds in the structure affect my geometry considerations of the structures or will it be the same as in structures only containing single bonds (say $\ce{PCl5}$) ?
Also, if I am supposed to draw its structure then where would I place the double bonded oxygen? Axial or equatorial position? Why?
I know that when lone pairs are considered they are placed at the equatorial position (after considering repulsions). Is there anything common in this case?

The Trigonal Bipyramidal is right. There would be a linear F-S-F through the apex of the bipyramid. In the plane I'd expect the O-S-F bond to be larger than the F-S-F bond. See: https://en.wikipedia.org/wiki/Thionyl_fluoride

So my first question is: Should having double bonds in the structure affect my geometry considerations of the structures or will it be the same as in structures only containing single bonds (say ) ?
Also, if I am supposed to draw its structure then where would I place the double bonded oxygen? Axial or equatorial position? Why? I know that when lone pairs are considered they are placed at the equatorial position (after considering repulsions). Is there anything common in this case?

The answer to your question is NO. A double bond shouldn't make any difference. It would be a Trigonal Bipyramidal where the 2 F's will occupy the axial positions according to Bent's Rule and the remaining two F's along with the Oxygen atom will occupy the equatorial position.


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Understanding of acids and bases
My teacher tried to teach this concept to us in 5 minutes. I'm lost but I've made an effort to understand it all (but I need clarification.)
If we have a compound such as LiOH, what element would be needed to neutralize it? As in, I understand that this element has to be a base but, there are SO MANY bases. Can you maybe teach how to do these using the LiOH example and another example using HCl (aq). Any examples, really. I just need to understand that.
Secondly, we haven't learned solubility tables yet, I can identify an acid (H ...) and bases (... OH) but what if there's a compound like MgCl2, what do I do now? Do I use a table to look up the states or the pH? Is there another approach?
I learned that Alkali Metal + O2 = Oxides but I can't really apply that here, can I?
Literally any help will be appreciated, I'm not cramming rather just don't understand the concept.

Your question is really bad - what are we supposed to do do, teach you whole general and inorganic chemistry?

As we are discussing acid-base neutralisation reactions they will be in solution (mixed with water).
First consider LiOH when mixed with water:
$$\ce{LiOH -> Li+ + OH-}$$
As we can see $\ce{OH-}$ ions are formed, meaning this solution is a base.  This means we need an acid to neutralise it.  (Acids and bases react to neutralise each other).
Lets say we use $\ce{HCl}$ as our acid.  Consider it mixed with water:
$$\ce{HCl -> H+ + Cl-}$$
When we mix these two solutions together we get:
$$\ce{Li+ + Cl- + H+ + OH- -> LiCl + H2O}$$
Notice how the $\ce{OH-}$ and $\ce{H+}$ react to form $\ce{H2O}$.  The $\ce{LiOH}$ is now neutralised.
This can be written in one equation like this:
$$\ce{LiOH + HCl -> LiCl + H2O}$$
Other acids could also be used to neutralise the $\ce{LiOH}$ such as $\ce{H2SO4}$:
$$\ce{2LiOH + H2SO4 -> Li2SO4 + 2H2O}$$
I am uncertain of your second question.  There is no way of neutralising $\ce{MgCl2}$ as it is a neutral salt (no $\ce{OH-}$ or $\ce{H+}$).  Solubility tables (at this level) also do not affect the pH of your solution.  If you haven't learned about it yet then don't attempt to now, you will likely just confuse yourself for your test tomorrow.
I hope this has been at an appropriate level for you.


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Determining # of calories using coffee-cup calorimter
I created a coffee-cup calorimeter at home where I calculate the total amount of calories in a tortilla chip by burning the chip and measuring the temperature change of the water.
If the formula is Q = (m) (c) (ΔT)
where,
Q = s the energy in the form of heat captured by the water, expressed in calories, and
m = mass of the water expressed in grams, and
c = the specific heat capacity of water, and
ΔT = change in temperature,
and my m=80grams, and temperature change was 20 degrees celsius, -does this mean the tortilla chip is 1.6 calories or do I have to multiply this value by 4.186 (which is the specific heat of water) to get a value of 6.69 calories?  Thank you.

It takes one calorie to raise 1 gram ( 1 milliliter) of water 1 degree C. So you had 1,600 calories. The !@#$%^ idiots doing food analysis decided that the "chemical calorie" was to small so they defined a "food calorie" as 1000 "chemical calories." // As a side note if you're burning something how do you transfer heat to the water? You'd need to account for the thermal mass of whatever apparatus was used to transfer the heat to the water.  You can get this by using an ice cube. Need to dry and weigh cube.


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Orbital conservation in Molecular orbital theory
MOT is really confusing me.
I was taught that the no. of orbitals remains conserved.That means, the number of atomic orbitals used is equal in number to the molecular orbitals formed.
But how is it possible in the following case?
For example in Case 1, I take a positive lobe of a p orbital and combine it head on with the positive lobe of another p orbital I get a bonding MO.
In a different scenario i.e Case 2, I take a positive lobe of a p orbital and combine it head on with the negative lobe of another p orbital to get an antibonding MO.
So overall I have used 4 atomic orbitals (2 in each case) to get 2 MO( one bonding and one antibonding).
So what is happening here?
Am I missing something?

You don't take lobes separately. There is no such thing. There is an orbital as a whole: take it or leave it.

I am not too sure of this, but isn't only one lobe actively involved in bonding.In pictures,the other one is usually behind.

Possibly related: http://chemistry.stackexchange.com/q/39223/186.

@KaranSingh, no, you don't combine lobes, you combine the whole orbitals. But you combine them in two different ways: one way gives rives to bonding MO, another to antibonding MO.

@Wildcat I totally understand that.But since the antibonding and bonding happen at the same time , I can't imagine how both combinations are possible at the same time.

@KaranSingh, that is because there is no any time out there. AOs are combined into MOs only on a sheet of paper, it is not a real physical process taking place in space-time. Rather, it is just a mathematical formalism.

In the linear combination of atomic orbitals (LCAO) there would be a bonding orbital and an antibonding orbital when two atomic orbitals combine to form a bond. The bonding orbital would be lower in energy that either atom contributing an orbital, and the antibonding higher than either.
One twist comes from the number of electrons. If there are just two electrons, then they both go into a bonding orbital. If there are three electrons then two go into a bonding orbital and one into the antibonding orbital.
There are three p orbitals so two atoms could be bonding with 1-3 p orbitals. If two p orbital bonds were being used then both p molecule bonding orbitals should be lower in energy than either antibonding orbital.
The whole concept of having a bonding orbital and an antibonding orbital is necessary to explain for instance electronic transitions. Imagine an oxygen atom in free space. If an electron is added, the oxygen molecule will accept the electron to form an ion $\ce{O^{-}_2}$. The electron doesn't go to one oxygen or the other, it is shared between both in a molecular orbital. Look at the diagram below for $\ce{O2}$. We'd predict that the electron will fall into one of the antibonding pi orbitals ($\pi^*$). From this we'd predict that a free oxygen atom could actually take three extra electrons and be stable. However the fourth electron would fill all the orbitals and $\ce{O^{-4}_2}$ would have no more electron binding energy than two free $\ce{O^{-2}}$ atoms.


text:
What is the correct order of boiling points for group 15 hydrogen compounds?
What is the order of boiling points of the following
$\ce{NH3}$,  $\ce{PH3}$, $\ce{AsH3}$, $\ce{SbH3}$ and $\ce{BiH3}$?
What I thought was, that as $\ce{H2O}$ has a higher boiling point compared to its group members due to hydrogen bonding, so should nitrogen. And then the order would follow according to the molecular weight. When I wrote this as an answer, it was marked wrong. What is the correct ordering and what is the reason for it?


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Can dissolving be a chemical change?
My textbook says that $\ce{SO2}$ dissolving in water produces sulfurous acid, but that seems to be a chemical change. But $\ce{NaCl}$ dissolving in water is not a chemical change. Please explain.

Possible duplicate or closely related to https://chemistry.stackexchange.com/questions/33278/how-is-dissolving-a-physical-change-and-not-a-chemical-change.

Also this answer and the question associated therewith.

Your book is glossing over some details.
There are three things occurring when you dissolve non-metal oxides in water (e.g., $\ce{SO2}$, $\ce{CO2}$, $\ce{NO2}$, etc.):

The gas simply dissolves in the water. You are correct that this step is a physical change.  How much dissolves depends on the partial pressure above the water and the Henry's law constant.
Once dissolved in water, some of the non-metal oxide will react with water to form an acid (sulfuric/sulfurous acid, carbonic acid, nitric/nitrous acid, etc.) and this is the chemical change that your book is referring to. There will be an equilibrium constant that will dictate the relative concentration of dissolved gas to the corresponding acid that is formed.
The acid (usually weak) will dissociate a little bit.  Again, the exact amount of dissociation will depend on the equilibrium constant, $K_\mathrm{a}$.

Using your example, we would have:
\begin{align}
\ce{SO2(g) &<=> SO2(aq)}\\
\ce{SO2(aq) + H2O(l) &<=> H2SO3(aq)}\\
\ce{H2SO3(aq) &<=> HSO3^- (aq) + H+(aq)}
\end{align}
Similar things happen when metal salts dissolve in water (although not $\ce{Na^+}$ or the other group I and II metals). The metal cations will react with water and form a weak acid.

According to Wikipedia H2SO3 article, existence of H2SO3 in solutions was not confirmed by Raman spectroscopy, only dissolved SO2 and HSO3-. It was detected just in gaseous phase .

$\ce{SO2}$ at normal pressure and temperature is a gas, and once dissolved follows an acid-base path. This is a chemical change.
NaCl is a salt and when immersed in water its crystal structure breaks and gives hydrated ions ($\ce{Na+}$ and $\ce{Cl-}$). This also is a chemical change.
The process of dissolution is always a chemical change because there is an interaction between solute and solvent (high or low interaction). Even when oil is in contact with water an equilibrium is established with partitioning coefficients and constants to describe the amounts of molecules inter-exchanged and rates.
What the textbook means to say is that in NaCl dissolution, no new molecules are generated. A specific chemical reaction (acid-base in this case) is taken place in dissolving the sulfur dioxide and generates a new molecule, sulphuric acid, but in both cases, there is a change in the chemical structure of solvent and solutes.


text:
What word, or phrase, do chemists use to describe the color saturation of a solution?
Outside of chemistry, I've used the word saturation to talk about how blue (or how green, red, etc.) something is. In the lab that I'm writing, I have three undesirable options:

use saturation and risk ambiguity with the other sense of the word (I could include a footnote that specifies the intended sense of the word; however, my experience has taught me that TA's rarely read footnotes.),
use an awkward construction (e.g. the degree/amount of blue exhibited by the solution), or
use an unusual term or a neologism (i.e. blueness).

None of those options are especially appealing.
Is there a word that chemists use to talk about how blue (or how green, etc.) a solution is?

And what's the problem with darker/lighter?

I'll point out that chemists don't use clear and colorless to mean the same thing. Clear means devoid of particulates or any colloidal scattering, so the solution would transparent but it could be any color. Colorless means devoid of any color and transparent. // I'd go with light/medium/dark for blues.

What about opaqueness. It is unambiguous for single-dye mixtures. If saturation is undesirable you could also use a simple derived neologism like "hue-strength".

Using a spectrometer, you could accurately measure the degree to which light is absorbed by a solution ("absorbance" or "optical density") at a certain wavelength, and in a cell of certain dimensions. So, if you had a blue dye dissolved in water (which might appear blue because it primarily absorbs orange light), the absorbance (measured using orange wavelengths of light) would be proportional to the concentration of dye in water.

that's exactly the lab I'm doing. In fact, it's for a sentence stating that there is a proportional relation between absorbance of non-yellow light and the yellowness of the solution.

I was going to suggest intensity but now you've got me thinking! It's a good question. Colors can be vibrant but I don't know if that applies either.


text:
major products of this reaction?

My attempt:
polar protic solvent, low temp, tertiary carbon, and a weak (?) nucleophile and a weak base.
It seems like this is a pretty standard SN1 reaction so "I and II" would be the correct answer?

I think your answer is correct as long as you can justify it. (IV) is clearly absurd but why do you eliminate (III) ? However in my opinion at $\ce{20 ^{\circ} C}$ there would be no reaction from a practical point of view, i.e. one with reasonable yield and reaction time.

20 C suggests a low temp which favors sn1 over e1 (which would give you III) since e1 has a higher deltaG

The alcohol is indeed a weak nucleophile. A good general rule of thumb is that an alkoxide (a deprotonated alkyl alcohol such as ethanol or isobutanol) is a much better nucleophile than its alcohol. The $20^{o} C$ is also a good indication for a slow E1 product, as john has said (higher $\Delta G$).
Since the question asks for specifically the major products, I would not include III in the answer. However, if the question had asked for all products, III is a possible answer to include. Its mechanism would involve first-order beta elimination due to the sterically-hindered alkyl halide.


text:
Is it possible to react Rb + Al2O3?
I am wondering if the Rubidium ($\ce{Rb}$) in gas form can react naturally with Sapphire crystal ($\ce{Al2O3}$)?
The temperature is $500\ \mathrm{K}$ and the pressure is around $1\ \mathrm{Pa}$ to $10\ \mathrm{Pa}$.
One of the reaction that comes into my mind is
$$\ce{4 Rb + Al2O3 -> 2 Rb2O + Al2O}$$
I was trying to calculate the Gibbs free energy of the reaction, but I could just find the Gibbs free energy for Rubidium and the Sapphire, which is $\Delta G_{\ce{Rb}}=53.1\ \mathrm{kJ/mol}$ and $\Delta G_{\ce{Al2O3}}=-1582.3\ \mathrm{kJ/mol}$. I can hardly find the Gibbs free energy for the products.
Is there any other reaction that can occur naturally?

https://en.wikipedia.org/wiki/Aluminium%28I%29_oxide  Al2O is unstable in such low temp. and rubidium is liquid.

Why don't you just try reducing Al all the way to the metal? I mean, Al(I) is pretty unstable.

@Mithoron, we are working on very low pressure, it is around $10^{-5}$ times lower than the atmospheric pressure (i.e. $P=0.7\ \mathrm{Pa}$). The vapor pressure of Rb is 1 Pa when the temperature is 434 K, so it is in the gas form.


text:
Observe from IR/Raman spectra whether molecule is in a local minimum on PES
I have done a geometry optimization of cis and trans-difluoroethene using RHF/STO-3G in Gaussian03. Although this is a simple method and a small basis set, this should ideally put place the molecule in a minimum on its potential energy curve. Within this minimum are vibrational states. I am wondering if it is possible to see, from the calculated IR and Raman spectra whether the optimisation was successful, i.e. whether the molecule was indeed placed in a minimum.
I am thinking like this: If the molecule is in a minimum, then all vibrational IR and Raman active vibrational states should be visible (given a high enough activity, of course) in the theoretical spectra. However, if the molecule is not in the minimum, then only those vibrational states of higher energy should be visible. Hence, fewer IR/Raman active vibrational modes would be observed if the geometry optimization was not complete.
According to this question, all frequencies at a local minimum would be positive, and all negative on a maximum. I assume then that somewhere in-between, you would have both negative and positive frequencies. However, given not the table of IR/Raman active modes and their frequencies, but the actual spectra instead, would I then be able to decide if the molecule is in a minimum or not?

I'd expect some relative peak intensity changes over large temperature changes. But I don't think that you would be able to get a chemical hot enough to eliminate any vibrational modes entirely. The IR vibrational nodes detected are "resonance vibrations" and a molecule can't be resonating at all frequencies at once.

My guess is that you're going to get all real (positive) frequencies as opposed  to imaginary (negative) frequencies. I say this because, as you point out, you're basically at an energy minimum so you're not gonna get imaginary frequencies which I believe you only get at inflection points and plateaus.


text:
What's the difference between pharmacotherapy and chemotherapy
Is it fair to say that pharmacotherapy (the use of pharmaceuticals to treat disease) is a subset of chemotherapy (the use of chemicals to treat disease)?
If not, why? Aside from the fact the former involves a subset of available chemicals and the latter all chemicals, what is the difference between the two?

I would like to believe that the major difference between these terms is based on their backgrounds and time however nowadays there are used interchangibly, without paying much attention to actual definitions.
Background

The term chemotherapy was coined by Ehrlich at the beginning of the
20th century to describe the use of synthetic chemicals to destroy
infective agents.  In recent years, the definition of the term has
been broadened to include antibiotics-substances produced by some
microorganisms (or by pharmaceutical chemists) that kill or inhibit
the growth of other microorganisms. It can also be  broadened to
include agents that kill or inhibit the growth of cancer cells.

Chemotherapy is the term originally used to describe the use of drugs that are 'selectively toxic' to invading microorganisms while having minimal effects on the host.
The term also embraces the use of drugs that target tumours and, in fact, has now come to be associated specifically with that branch of pharmacology (Oncology). Although, in the public mind at least, chemotherapy is usually associated with cytotoxic anticancer drugs that cause unwanted effects such as loss of hair, nausea and vomiting.
Pharmacotherapy is a more recent term which generally means the treatment of diseases through the administation of drugs. However there is more to it, the term also describes a more patient-focussed approach in which pharmacists treat patients through drug management with one or more specific outcomes.
The specific goals of pharmacotherapy (primary therapeutic outcomes include):
• Cure of disease (e.g., bacterial infection).
• Reduction or elimination of symptoms (e.g., pain from cancer).
• Arresting or slowing of the progression of disease (e.g., rheumatoid
arthritis, HIV infection).
• Preventing a disease or symptom (e.g., coronary heart disease).
Other important outcomes of pharmacotherapy include:
• Not complicating or aggravating other existing disease states.
• Avoiding or minimizing adverse effects of treatment.
• Providing cost-effective therapy.
• Maintaining the patient’s quality of life.
A systematic approach is consistently applied to each case. The steps involved in this approach include:

Identifying real or potential drug therapy problems

Determining the desired therapeutic outcome(s)

Evaluating therapeutic alternatives

Hope this helps
References

RANG AND DALE’S Pharmacology 8th edition: Basic principles of antimicrobial chemotherapy

Pharmacotherapy Casebook

In common medical usage, chemotherapy refers specifically to treating cancer with drugs. Pharmacotherapy is a more general term that includes treating any disease with a chemical substance.
This is the customary way the terms are used which has arisen because that is how people use the terms. It might seem that chemotherapy should refer to any chemical used in treatment but that isn't how the terminology evolved. Language isn't always a rational thing.


text:
What [if anything] is special about the boiling process of purifying water (excluding temperature)?
I know that water boils at $100~^{\circ}\rm C$. I also know that the standard way to purify water is to boil it for roughly a minute. What I'm curious about is, what properties of the boiling process purify the water?
Is it purely a temperature thing? Are we saying "roughly a minute" so that the temperature actually increases beyond $100~^{\circ}\rm C$, thus reaching some actual temperature needed to kill bacterial/viral stuff? Or, does the "rolling" of the boil actually add a physical benefit to the purification process?

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Unless you're adding a lot of salt to your water, you're not going to get a temperature above 100 ºC.  In fact, at high altitudes (e.g. backpacking) your boiling water will be at lower temperature than 100 ºC.

@S.Burt - very intriguing! Care to explain why this is, or is that fit for another question?

I suspect there are several questions related to boiling point change already here on chem.SE.  In short, the boiling point of any pure liquid depends only on the external pressure (hence, higher altitudes have lower boiling liquids).  You cannot exceed the boiling point by adding more heat or waiting for more time.  This boiling point can be change, however, by dissolving other things into the water (try searching for colligative properties for an explanation).

There nothing special about boiling that is doing anything.  For example, you can put a beaker of water under vacuum and get the water to boil at room temperature, but that won't kill any bacteria or render any viruses impotent.
It's purely a heat thing.  You want to destroy any pathogens that may cause illness.  Sufficient heat will denature proteins in a virus and kill bacteria, etc.  Boiling simply gives you a visual cue that you've reached a very high temperature (so you don't need a thermometer) and the ~1 minute is just to ensure that you sustain a high temperature long enough to destroy any bacteria or viruses.

An additional point of boiling via heating is that it creates convection currents which stir the water well. // I attached a link to a nice chart which has time/temperature needed to kill some certain organisms. http://cid.oxfordjournals.org/content/34/3/355/F2.large.jpg

Fantastic reference.  Some of those temperatures were lower than I would have expected, but I was surprised that hepatitis can survive up to 98 ºC.

I should have pointed out that boiling is an absolute guarantee. Some bacteria have spore forms which are not killed by boiling.


text:
Which bond angle will be larger in a compound like CH2X2?
Just thought of a certain situation.Say for example in a compound like difluoromethane $\ce{CH2F2}$ or $\ce{CH2Cl2}$ obviously the tetrahedral structure will be distorted.
But my question is which bond angle will be larger: the angle between halogen-halogen, halogen-hydrogen or hydrogen-hydrogen? Why?

Hint: Think in terms of electronegativity of the atoms.


text:
Influence of sodium cyanoborohydride on C=N bonds in DNA bases
I would like to crosslink an amino-modified DNA strand with an APTES monolayer (which also has a $\ce{NH2}$ end group) using a glutaraldehyde linker. After the formation of Schiff base linkages between the aldehyde and the amino groups, I would like to reduce the bonds using sodium cyanoborohydride to form secondary amine bonds.
Does the cyanoborohydride also affect the $\ce{C=N}$ bonds in the DNA bases or not?

Sodium cyanoborohydride?  Pretty toxic stuff...

Sodium cyanoborohydride is a very selective reducing agent. Another (safer) alternative would be sodium triacetoxyborohydride. But this doesn't really matter as far as $\ce{C=N}$ bonds of DNA bases are concerned since they are very stable.

authors: هبة سعود سليمان حمود الجابري


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Are adamantane crystals covalently bonded?
I'm writing a report and need one thing clarified after reading the Wikipedia article on adamantane:
'All the above methods yield adamantane in the form of polycrystalline powder. Using this powder, single crystals can be grown from the melt, solution or vapor phase (e.g. with the Bridgman–Stockbarger technique). Melt growth result in the worst crystalline quality with a mosaic spread in the X-ray reflection of about 1°. Best crystals are obtained from the liquid phase, but the growth is inpracticably slow – several months for a 5–10 mm crystal. Growth from the vapor phase is a reasonable compromise in terms of speed and quality.[2] Adamantane is sublimated in a quartz tube placed in a furnace, which is equipped with several heaters maintaining a certain temperature gradient (about 10 °C/cm for adamantane) along the tube. Crystallization starts at one end of the tube which is kept near the freezing point of adamantane. Slow cooling of the tube, while maintaining the temperature gradient, gradually shifts the melting zone (rate ~2 mm/hour) producing a single-crystal boule.[16]
Would I be write in thinking that these crystals are not randomly shaped diamond structures you would get if the adamantanes covalently fused with eachother? In other words, are the crystals formed from polycrystalline adamantanes (with different numbers of basic ten carbon adamantanes in one derivative crystal) bonding together without covalent bonds? Or perhaps without many covalent bonds? For example, one C-C bond between two crystals? What other type of bonding could be at work here? Is the macro-crystal structurally weak as a result?

Adamantane is a molecular compound, much like naphthalene, paraffin, and many, many others. Molecular compounds consist of molecules, which stay pretty much the same in the gas and solid states. A molecule has certain molecular formula. Each carbon in a molecule is linked to a certain number of hydrogens and other carbons. There are no dangling covalent bonds to be formed between molecules, let alone between crystals. Consequently, you can't just take some adamantanes and covalently fuse them; some people would discuss diamond crystal structure in such terms, but only in a metaphorical sense.
Adamantane molecules in a crystal (being non-polar molecules without any specific interactions) are held together only by the weak van der Waals forces. Because of that, yes, the crystal is not mechanically strong at all (again, think of paraffin wax).

I did mean it in a metaphorical sense! But thanks, you've confirmed my suspicion.


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Why does Soda Pop after Shaking
Why does shaking a soda can and opening it make the soda jump out of the can?

Soda contains carbonated water, and by shaking it, the carbon dioxide is released which generated the pressure and the soda jumps out. Even when you open the can without shaking, the gas will come out but slowly. $$\ce{H2CO3 (aq)<->H2O (l) + CO2 (\uparrow)}$$ [Not enough information so posted as comment]

Add the word "oversaturated", and that would be the complete explanation.

And the nominal oversaturation is possible due to the can being pressurised under CO2 to begin with.

Soda drinks are carbonated meaning that carbon dioxide is dissolved into it. When carbon dioxide dissolves, the following reaction occurs: $$\ce{CO2(g) + H2O(l)~ \leftrightharpoons ~H2CO3(aq)}$$ As you can see, carbonic acid (a weak acid) is formed. This is responsible for the 'fizziness' of the drink.
Note that this reaction is in equilibrium, meaning that not all of the carbon dioxide dissolves to form carbonic acid, but only a certain percentage of it. Usually, the percentage of carbon dioxide that is dissolved is constant, however this can be changed by doing certain things to the system.
For example, by decreasing the pressure, the equilibrium will shift to the left, meaning that the percentage of carbon dioxide that is dissolved will decrease. This will result in an increase of gaseous carbon dioxide in the bottle. This reason for this is explained by the Le Chatelier's principle. You can also imagine it as a piston pushing the gas into the liquid as shown in the below image. Image (a) is at low pressure while image (b) is at high pressure.

Normally in soda bottles, the pressure inside the bottle is relatively high, to ensure that more carbon dioxide dissolves into the drink. So when you open a soda can you decrease the pressure. This results in more carbon dioxide gas being produced. It is the release of this gas that causes that pfffftt sound.
Now, when you shake the bottle, due to turbulence, small bubbles containing carbon dioxide form. The formation of these small bubbles allows even more carbon dioxide gas to form. So if you shake a soda can and then open it soon after, the soda will jump up quite violently, as there is now a lot of carbon dioxide being released.
However, if you shake a soda can, and then wait a while before opening it, it will most likely just create the normal pfffftt sound. This is because you give the system some time to go back to its original equilibrium position, meaning that some of the carbon dioxide gas will redissolve back into the solution .

No worries, I am glad that I could help.

Well this isn't quite right. In truth $$\ce{CO2_{(g)} <-> CO2_{(aq)} <-> H2CO3_{(aq)}}$$ Most of carbon species in water is dissolved gas not carbonic acid.

I think the key here is fluid mechanical: shaking moves the fluid around, reducing the effective pressure, then the carbon dioxide tries to make up for it, per Le Chatelier.

Shaking does not change the pressure It simply distributes small bubbles of CO2 throughout the liquid that all expand more or less together entraining liquid. when the bubble is on the top only gas escapes. try opening the bottle upside down and see what happens.

I was watching a Vertassium episode where Dr. Muller was holding a soda bottle with a pressure gauge atop that registered a rather high pressure. He guaranteed that the bottle was still a long time and equilibrated and asked what would happen to the pressure when the bottle was vigorously shaken. I thought about it and said Nothing. He shook it and the pressure did not change.  The reason a shaken soda erupts seems to be the distribution of microgas bubbles thruout the liquid that expand when the pressure is released, not conversion of carbonic acid to CO2 or even CO2 being released from the solution. Champagne seems to require additional research. Maybe by New Years Day that research can be finished.

Soda or other cool drinks like Coke, Pepsi etc. contain water along with dissolved carbon dioxide (aerated). Since the solubility of carbon dioxide increases with increase in pressure (Henry's law), the Soda is bottled under pressures higher than the atmospheric pressure.
When we open the soda bottles or cans, the pressure drops again to atmospheric pressure and the dissolved carbon dioxide will escape out into the surroundings. It occurs all of sudden so that soda will jump out with fizz. The scenario will be even more worse if we shake the bottle before opening due to thrust created.
Remember that soda can be referenced as carbonic acid, H2CO3. For more information see carbonate and bicarbonate chemistry.

Explain thrust created!


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What do the prefixes meta, ortho, pyro mean in inorganic chemistry?
In inorganic chemistry, when is the prefix meta used? (as in metaborate and metasulphite)
What about the terms pyro and ortho (as in orthophosphorous acid)?
For example, I recently came to know about the prefixes per and hypo which are used to refer to the highest and lowest oxidation states of an element respectively (like perchlorate and hypochlorite).
Are there any terms and conventions I should be aware of?

What does meta means in the compound "meta aluminate"?

In the older literature, the prefixes “hypo-”, “per-”, “ortho-”, “meta-”, and “pyro-” were used in some cases for the distinction between different acids with the same characteristic element.
The prefix “hypo-” was used to denote a lower oxidation state, and the  prefix “per-” was used to designate a higher oxidation state. (The prefix “per-” is not be confused with the prefix “peroxo-”.)

$\ce{HClO}$ hypochlorous acid
$\ce{HClO2}$ chlorous acid
$\ce{HClO3}$ chloric acid
$\ce{HClO4}$ perchloric acid

The prefixes “ortho-” and “meta-” were used to distinguish acids differing in the formal content of water.

$\ce{H3BO3}$ orthoboric acid
$\ce{(HBO2)$_n$}$ metaboric acid
$\ce{H4SiO4}$ orthosilicic acid
$\ce{(H2SiO3)$_n$}$ metasilicic acid

The prefix “pyro-” was used to designate an acid that is formally formed by removing one molecule of water from two molecules of an ortho-acid.

$\ce{H4P2O7}$ pyrophosphoric acid $(\ce{2 H3PO4 -> H4P2O7 + H2O})$.

Various traditional names are retained for use in IUPAC nomenclature, though the number of retained names has been reduced with each succeeding edition of the IUPAC recommendations. The Second Edition (1970 Rules) of the IUPAC Nomenclature of Inorganic Chemistry (published in 1971) still included all of the above-mentioned prefixes; however, the approved use was limited to only some selected retained names. The further development and the current situation of the nomenclature are well explained in @Jan’s thorough answer.

Hey sorry but I still Did not get the ortho and meta part! What do you mean by formal content of water?

@KaranSingh For example, metaboric acid can be formed by condensation (dehydration) of orthoboric acid. Heating of orthoboric acid releases one equivalent of water to give metaboric acid: $\ce{$n,$H3BO3 -> (HBO2)$_n$ + $n,$H2O}$. Therefore, the formal difference is just the content of one equivalent of water.

@Loong          But meta phosphoric acid is not formed by loosing one water molecule from orthophosphoric acid. So what is general equation?

@Loong What's the difference between -ous and -ic?

The ending -ous is used for a low oxidation number. -And -ic is used for higher oxidation numbesr.

What does meta means in the compound "meta aluminate"?

Are there any terms and conventions I should be aware of?

Yes. IUPAC publishes the Red Book also called Nomenclature of Inorganic Chemistry which is the authoritative source of nomenclature terms. Its current version (2005; the only one I will be referencing above the horizontal rule) does not contain the word pyro meaning that that is unacceptable for inorganic nomenclature purposes.
Ortho and meta are sometimes allowed as traditionalised common names. Ortho only in the cases of ortho-periodic acid and ortho-telluric acid and their derivatives ($\ce{IO(OH)5}$ and $\ce{Te(OH)6}$, respectively) to distinguish them from periodic acid ($\ce{HIO4}$) and telluric acid ($\ce{H2TeO4}$). Meta is only allowed for metasilicate and metaborate, both describing long-chain anions whose systematic name would be catena-poly[(oxidoborate-μ-oxido)($1-$)] (or with a dioxidosilicate instead of oxidoborate) to distinguish them from silicate ($\ce{SiO4^4-}$)  and borate ($\ce{BO3^3-}$), respectively.
The usage ‘*ortho-phosphoric acid’ and the like has been deemed unnecessary and obsolete because there is no ‘other’ form of phosphoric acid that one could want to differ between. The former meta-forms are better described as what they actually are: cyclo-triphosphoric acid, for example.
Note that the prefixes hypo and per are also non-systematic as per IUPAC every instance where they occur can also be named systematically. However, they remain as acceptable alternatives in many instances due to their simplicity and commonness, e.g. peroxide, periodic acid, hypochlorite. Note that not always does per refer to the highest oxidation state: persulphate is still a sulphur(VI). However, there is a peroxido ligand attached to it.

Historically, the terms were also used in an often non-systematic way. Ortho is Greek and means true, implying that an ortho-something be the true form of something: Hence ortho-phosphoric acid for ‘the true’ monophosphoric $\ce{H3PO4}$.
Meta means after in Greek, so Aristotle’s Metaphysics was a book that was conventionally placed after Aristotle’s Physics in the library of Alexandria. Thus, meta compounds are the ones which arrived after ortho compounds by some mechanism. meta-Phosphoric acid was seen as a monohydrate of $\ce{P2O5}$ while ortho-phosphoric acid was the ‘true’ trihydrate of $\ce{P2O5}$. ($\ce{P2O5} + \ce{H2O} = 2 \ce{HPO3}$; while the formula for phosphorous pentaoxide is actually $\ce{P4O10}$ and that of cyclo-triphosphoric acid would be $\ce{H3P3O9}$.) The logic is that the ‘true’ hydrate needs to be hydrated, to afterwards give the meta-hydrate.
Para is not used in inorganic chemistry to the best of my knowledge.
Pyro, meaning fire in Greek, is a prefix for something that has to do with being burnt in some way. Hence, pyrophosphates were initially prepared by heating hydrogen phosphates to red glow. A few decades later, researchers realised that the ‘pyro’-phosphate was, in fact, a diphosphate. Likewise, pyruvic acid was originally prepared by heating of tartaric acid. According to a quick search across Wikipedia, pyro is only applicable to phosphate and sulphate and their respective acids meaning the dehydrated dimer. Personally, I have never encountered pyro-anything outside biochemical lectures (and my school’s biology classes) when connected to phosphate. You can consider all pyros entirely obsolete and replaced by di- (disulphuric acid, disulphate, diphosphate, etc.).

Also, a question:) Does it all also mean that orthovanadate and metavanadate are obsolete and shouldn't be used?

@andselisk To the best of my knowledge yes.


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Closed shell molecules can have degenerate states?
I read a paper which said that $\ce{CO2}$ despite having orbital degeneracy is stable and not subject to the Jahn-Teller effect because it is a linear molecule. But $\ce{CO2}$ is a closed shell molecule, so how can it have orbital degeneracy?

I would say the paper is badly written. ‘Closed shell’, i.e. all orbitals of one energy either completely populated or completely unpopulated does not allow for any distortion stabilisation and hence $\ce{CO2}$ does not show any.

Degeneracy is usually based on symmetry. You can map the binding orbitals for each bond onto the other bond. So you can't distinguish between the molecular orbitals and its energies.
Jahn-Teller doesn't apply for linear molecules because there is no distortion which can reduce the symmetry and reduce the system energy as well.


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Is there any software to simulate a laboratory?
I wanted to do some experiments with some chemicals but unfortunately I do not have access to them. Is there any software I can use to get the products?
I don't want to do very complicated things, I just want to gain some experience in basic chemistry, with simple chemicals like $\ce{H2O2}$, some salts, sulfuric acid, potassium permanganate, etc.
I tried to search on the net but most of the software I found is too complicated for me.
Do you know about any such software? Or do you have any alternatives?

The Powder Toy game is probably more fun than learning, but is IMO worth mentioning.

There is a product called Yenka Chemistry which simulates basic chemicals/apparatus, and allows basic manipulations (for instance to show the pH change as more acid is added to base, or the amount of gas evolved).
The program evolved from another piece of software called Crocodile Clips, which was aimed at GCSE science in the United Kingdom. For this reason it is incredibly basic, mainly focusing on inorganic reactions (acid/base, titrations, reactivity of group 1 and 2 metals), with no real capacity for demonstrating organic reactions.

I just tried it right now, Its an awesome software!

I would recommend standardizing on Mathematica. Mathematica is the best general software to do physics and chemistry. You can use Mathematica absolutely free from the web with Wolfram Alpha. Instead of learning a specific chemistry program, if you learn Mathematica you can do any kind of science with it. Its the single most valuable tool you can learn to do math and science.
Here is an example, let's say you want to know what happens when you combine hydrochloric acid and calcium hydroxide. Just type it in to Wolfram Alpha:

It tells us the result: calcium chloride and water. Mathematica can do anything you would need to do in computational chemistry, stoichiometry, rate equations, molecular modeling, you name it.
Mathematica is free or inexpensive for students.

I had seen a trial version of mathematica but I did not know it could do reactions also. Can we also get the colors of compounds and weather the compound will dissolve or precipitate?

No, but both the color and solubility of common compounds can be found on the Wikipedia.

Don't get your hopes up too high. Wolfram Alpha's capabilities to list the products of chemical reactions are currently very limited. Fun idea, though.

I don't think Mathematica would be able to manipulate organic compounds at all...

Three years later, Mathematica is still about as good in chemistry as Photoshop in hydrodynamics.


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When you increase the temperature such that change in Gibb's free energy becomes zero, does Keq = 1 at that temperature?
$$\Delta G = \Delta H - T \Delta S$$
say for example, both $\Delta H$ and $\Delta S$ are positive. So assuming $\Delta H$ and $\Delta S$ do not change or change minimally with temperature, if you increase the temperature until $\Delta G = 0$, is the equilibrium that you reach an equilibrium in which reactants and products are in standard conditions i.e. $K_\mathrm{eq} = 1$?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I have improved the formatting of your post by adding MathJax for mathematical equations. With the \ce{...} syntax, this can also be used for chemical formulae. You can find out more about that in the [help] or by visiting this or this meta-post.

Yes and no.
$K = 1$ is true, and it follows directly from the equation $\Delta G^\circ = -RT \ln K$.
However, there is no reason for it to be "an equilibrium in which reactants and products are in standard conditions". $K = 1$ does not necessarily mean that the concentrations (or activities if you prefer) of all reactants and products are also equal to $1~\mathrm{mol~dm^{-3}}$ (or $1$, in the case of activities). Remember that $K$ is a fraction. For the reaction:
$$\nu_{r1}\ce{R1} + \nu_{r2}\ce{R2}\cdots \longrightarrow \nu_{p1}\ce{P1} + \nu_{p2}\ce{P2} \cdots$$
we have:
$$K = \frac{[\ce{P1}]^{\nu_{p1}}[\ce{P2}]^{\nu_{p2}}\cdots}{[\ce{R_1}]^{\nu_{r1}}[\ce{R_2}]^{\nu_{r2}}\cdots}$$
Just like how the fractions $\displaystyle \frac{2}{2}$, $\displaystyle \frac{4}{4}$, or $\displaystyle \frac{4000\hbar\pi e}{4000\hbar\pi e}$ are all equal to $1$, the individual concentrations $[\ce{P1}], [\ce{P2}], [\ce{R1}], [\ce{R2}]\cdots$ might not be equal to $1~\mathrm{mol~dm^{-3}}$, but the result will still be $1$.

No, $\Delta G = 0$ implies that the reaction quotient (Q) = K.
K can be any value, not necessarily 1.
Only if you additionally have all the reactants and products in the standard state, and therefore know $\Delta G^\circ = 0$, can you conclude K = 1.


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Quantum Numbers
Why are the Lanthaide (atomic #'s 57-71) elements in period 6, but in terms of electron configuration, are assigned the principal quantum number of 4? Why are certain subshells assigned quantum numbers that are different than their corresponding period? (example is d-block elements as well)

http://chemistry.stackexchange.com/questions/33297/why-do-3d-orbitals-have-lesser-energy-than-4s-orbitals-in-transition-metals/33310#33310 and http://chemistry.stackexchange.com/questions/8357/why-does-the-3rd-electron-shell-start-filling-up-with-scandium


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Propagation of Uncertainties
I have five values for the volume of sodium hydroxide needed to neutralise a fixed quantity of hydrochloric acid, each trial with an uncertainty of 0.05 mL. If I take the average of these five values, in what way would the uncertainty propagate? I can think of three of the following ways, but don't understand which would be correct:

Because we can see the calculation of an average as the sum of the values for the five trials (with their uncertainties), and then a division by 5, we sum up the uncertainty to get 0.25 mL and then divide it by 5 to get the uncertainty of the average to be 0.05 mL.
Because we can see the calculation of an average as the sum of the values for the five trials (with their uncertainties), and then a division by 5 (which has no uncertainty), we sum up the uncertainty to get 0.25 mL and then divide it by 5 to get the uncertainty of the average to be 0.25 mL.
Range divided by two is often another way to calculate the uncertainty of an average value, which for my case results in a value different to the above two for the uncertainty of the average.

Which one, and why, would be the correct one to state next to the average value (with a plus-minus sign), as an uncertainty of the average value?

Related: Uncertainties after Averaging Trials

None of your methods are correct. If you have a small data set - in this case, 5 values from a titration - then the absolute uncertainty for each individual value is half the range of data. For example, say you get the following results (in ml, from a burette with a precision of 0.05ml): $4.70$, $5.35$, $4.85$, $4.60$ and $5.20$. The range of these results is $5.35-4.60=0.75$. This gives an absolute uncertainty of $\frac{0.75}{2}=0.375$ for each individual piece of data e.g. $4.85\pm0.38$ml.
However, if the absolute uncertainty calculated from half the range is less than the precision of the instrument (in this case $<0.05$ ), then the precision of the instrument should be given as the uncertainty e.g. $5.00\pm0.05$. I think you may have incorrectly calculated the absolute uncertainties.
For small data sets ($<10$ pieces of data), the uncertainty in the mean is then given by $$\Delta {x}_{avg}=\frac{\Delta x}{\sqrt{N}}=\frac{R}{2\sqrt{N}}$$ where $\Delta {x}_{avg}$ is the uncertainty in the mean, ${N}$ is the number of times the measurement was repeated, $R$ is the range and $\Delta{x}$ is the uncertainty in one measurement.
For example, using the example data above, $\Delta {x}_{avg}=\frac{0.375}{\sqrt{5}}=0.168$. The average can then be stated as $4.94\pm0.17$ml

I don't think you understand what you're asking. Ambler's answer gives you the correct procedure to handle your experimental measurement error.
But there is also an inherent error against which you could do an F-Test to see if your experimental variation was within the limits of the error of the burette.
When you say that "each trial with an uncertainty of 0.05 mL" what you're really saying is that the error is +/- 0.05 for each trial which means that the error could be -0.05, -0.04, -0.03, ..., +0.03. +0.04, +0.05 with equal probability. so over 5 reading the expected error could be as low as -0.25 and as high as +0.25.
So the expected variance over the samples combined is:
$V(n) = \dfrac{0.1*n}{12}$
the variance of the mean is $1/\sqrt{n}$ smaller so:
$\sigma_{mean} = \dfrac{0.1*\sqrt{n}}{12}$
where $n$ is the number of repetitions of the measurement.


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Ratios and mass; reaction calculation problem
I have come across the following problem:

One method used to inflate air bags in cars is to use nitrogen produced chemically from the decomposition of sodium azide. The sodium formed reacts with potassium nitrate to give more nitrogen.
$$\ce{2NaN3(s) -> 2Na(s) + 3N2(g)}$$
$$\ce{10Na(s) + 2KNO3(s) -> K2O(s) + 5Na2O(s) + N2(g)}$$
a) In what ratio by mass must the sodium azide and potassium nitrate be mixed in order that no metallic sodium remains after the reaction?
b) Calculate the total mass of the solid mixture needed to inflate a $60.0\ \mathrm{dm^3}$ air bag at room temperature and atmospheric pressure.

I have a solution to part a), which is 1:0.311, but this differs from the given answer (1:3.11). Here is my solution:
For every 10 moles of $\ce{NaN3}$, 2 moles of $\ce{KNO3}$ is needed i.e. ratio in moles is 5:1. Mass of $5\ \mathrm{mol}$ of $\ce{NaN3}$ is $5 \times 65.0 = 325$ and mass of 1 mol of $\ce{KNO3}$ is 101.1. So, ratio = 325:101.1 = 1:0.311. Is this incorrect?
I also have a solution to part b), but again this differs from the given answer $(40.9\ \mathrm g)$. Here is my solution:
Using ideal gas law to find how many moles of $\ce{N2}$ needed to inflate air bag: $n=\frac{PV}{RT}=\frac{101000\times 0.06}{8.31\times293}=2.49\ \mathrm{mol}$ If the first equation is rewritten as $\ce{10NaN3(s) -> 10Na(s) + 15N2(g)}$, then there are $\frac{15}{16}\times2.49=2.33$ moles of $\ce{N2}$ from the decomposition of sodium azide, and $\frac{1}{16}\times2.19=0.156$ moles of $\ce{N2}$ from the second reaction. Using these values gives $101\ \mathrm g$ of $\ce{NaN3}$ and $31.5\ \mathrm g$ of $\ce{KNO3}$, which clearly does not give the answer. What's incorrect here?

I suspect you're using the wrong gas equation. There is a much simpler one relating the number of moles of a gas directly to the volume.

@Beerhunter, I assume your referring to the fact that 1 mole of gas always has a volume of 24 decimetres cubed at rtp, but this gives the same value as mine for the number of moles of nitrogen .


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Should titration stop at the intermediate color or a drop past the intermediate
Using Bromcresol green as an indicator, should one stop when the blue color changes to green or to yellow for more accurate results.
Shouldn't one aim to get a green because its far closer to the end point than a yellow could be. For example, if the solution turned yellow and an additional drop was added, it would be indiscernible.

Neither of your options is correct.
Ideally an acid base indicator would be colorless at one pH and colored after the transition. Such an indicator is phenolphthalein. So when using phenolphthalein it is indeed impossible to detect when all the phenolphthalein is colored. So you titrate to a point where there is the faintest pink color.
So for bromcresol green you should either stop when the solution starts to turn from blue to blue green (ph about 4.6), or as the solution turns from green-yellow to yellow (ph about 5). It would very difficult to judge the deepest green (at about 4.8) accurately.
So which transition you would choose would depend on balancing between which side of pH 4.8 that you wanted to end on, and which transition was easier to detect visually.

You can look at a list of some acid/base indicators at Wikipedia. There are few that go between colorless to colored. Most are color-1 to color-2.

I also point out that you question contains a fallacy.

Should titration stop at the intermediate color or a drop past the intermediate

The burette column is marked "to contain" an amount of fluid along its length. So you can let a partial drop form, then touch the tip of the burette to the flask and get a partial drop of titrant. You don't have to titrate in "quantized" drops.

‘Ideally an […] indicator would be colorless at one pH and colored after the transition.’ [citation needed]

@Jan - Sorry I don't have any analytical chemistry books handy. I'll point out though that a titration is highly technique oriented. You need to watch how the titrant is dispersing in the solution at first as you're swirling the flask. The closer to the end point you watch how a single drop diffuses into the solution without swirling, then swirl to equilibrate. So end-point detection isn't just comparing two colors.

I'm just questioning why colourless → colour be better than e.g. yellow to dark blue ;)

@Jan - It is just easier to visually detect the slightest appearance of a color than to detect a slight color change in an already colored solution. As I tried to explain end-point detection is about watching the titration as a dynamic system. It isn't lining up 10 flasks and judging them from most yellow to most yellow green.

@MaxW Burettes are calibrated ‘to deliver’ (TD, Ex), not ‘to contain’ (TC, In).

@Long - I disagree. The "amount" of solution that a burette delvers is the difference between two readings of the column which has nothing to do with the tip. So there shouldn't be any part of a drop of titrant  hanging on the burette tip when you start the titration nor when you finish. // With a pipette you clear hanging fluid from the tip when full, and at the end, but whatever "hangs" inside the pipette is supposed to stay there.  // I reworded to make the point more clearly.


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Density of plutonium oxide
I am trying to find the mass density $\rho$ of plutonium dioxide, $\ce{PuO2}$. I could calculate it, but the mass density of oxygen found in the literature is the density of gas oxygen.
So I was wondering where can one find mass densities for compounds that come from experimental measurements?

you simply say density, no need for "mass density" 2) you can't calculate density of compound from data for elements in any simple way 3) simply google it

Well, actually, I've seen the term "molar density" given many times before in my Physical Chemistry textbook, with units of $\frac {mol}{L}$ rather than $\frac {kg}{L}$, so I find it reasonable to specify that sometimes. :)

There are many densities, but "mass density" is the density.

A quick Google search shows the density to be $11.5~\mathrm{g~cm^{-3}}$.
You might have had trouble finding it since it is actually called plutonium (IV) oxide, as plutonium is an f-block metal (specifically an actinide) displaying multiple  oxidation states (commonly III, IV, V, and VI, while VII is rare).

This can be done if you know the unit cell dimensions of plutonium dioxide. You also need to know the locations of the atoms in the unit cell. The unit cell for plutonium dioxide is the same as uranium dioxide and calcium(II) flouride in terms of the fractional coordinates.
First calculate the volume of the unit cell in cubic meters
Next work out the number of plutonium atoms inside the cell, you will need to use the following rules.

If the atom is totally inside the unit cell then it is not shared between more than one one unit cell.
If an atom is touching the face of the unit cell then it is shared between two unit cells, so count it as half an atom.
If an atom is touching an edge (the line where two faces meet) then it is shared between four cells, count it as a quarter atom.
If the atom is at a vertex of the cell then it is shared between eight cells, count it as an eighth.

Repeat this process for the oxygen atoms.
Next work out the mass in amu of the atoms inside the unit cell.
Convert this into kilos
Divide the mass by the volume of the cell, this will give the density.
I have used this as an undergraduate (first year) teaching excerise, it is something which takes a few minutes on a blackboard.


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What reaction occurs when etching copper to create circuit boards?
The components used for etching copper are acetic acid, hydrogen peroxide and sodium chloride. I think the equation may be similar to this:
$$\ce{2H3C2O2H + H2O2 + 2NaCl + Cu -> CuCl2 + 2H3C2O2Na + 2H2O}$$
Is this equation correct? There are a lot of bubbles created. What is the gas that is given off?

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Well your equation balances but that isn't the right reaction.
First lets break this down into half cell reactions.
Oxidation reaction

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

Reduction reaction

$\ce{H2O2 + 2 H^+ + 2e^{-} ->  2 H2O_{(l)}}$

Decomposition reaction
Hydrogen peroxide spontaneously decomposes in acid solution.

$\ce{2H2O2 ->  2H2O_{(l)} + O2_{(g)}}$

Stoichiometric reaction between copper and hydrogen peroxide
So the stoichiometric reaction between copper and hydrogen peroxide is:

$\ce{H2O2 + 2 H^+ + Cu_{(s)} -> Cu^{2+}_{(aq)} + 2 H2O_{(l)}}$

the $\ce{2 H^+}$ ions come from the vinegar. (Let's use $\ce{HOAc}$ for vinegar).

$\ce{H2O2 + 2 HOAc + Cu_{(s)} -> Cu^{2+}_{(aq)} + 2OAc^{-} + 2H2O_{(l)}}$

Sodium chloride's Role
Finally the role of the $\ce{NaCl}$ plays in the etch. The $\ce{NaCl}$ must be fairly high to drive the $\ce{Cu^{2+}}$ to a copper chloride complex. This thus reduces the "free" $\ce{Cu^{2+}}$ in solution which keeps the electrochemical potential of copper high so that the solution keeps dissolving copper. (Think of the half-cell copper reaction as if it is in a battery. The voltage of the half-cell would drop as the copper builds up.)

Before I edit it away amoung another minor typo I found, is there a reason why you write $\ce{Cu+^2}$ rather than $\ce{Cu^2+}$?

Changed it. Didn't realize that there was such a strong convention, or that anyone would care...

Will, I'm not aware of any conventions, but I only ever saw $\ce{Cu^2+}$ (or $\ce{Cu++}$ from a particularly old biology teacher) so I thought I'ld better ask before doing things wrong ;)

@Jan - LOL Pauling "The Nature of the Chemical Bond" used the $\ce{Cu^{++}}$ notation so your odd biology teacher was in good company....

I guess "++" was a good notation until the number of pluses required got larger ;)


text:
What would carbon-oxygen polymer be called?
So I have a chain molecule made of carbon and oxygen, and I'm not sure what it would be called. Would any of you lovely fellows be able to help me?

The best name I've been able to come up with so far is "polyquadcarboxate" but I'm pretty sure this isn't right.
I'm looking for an IUPAC name if possible, but I suppose other names will do.
Thank you very much!

Carbon dioxide. (Because that is what it would degrade to, give it a femtosecond.)

I'm looking for the name of the molecule itself, not what it would decompose to. But thanks anyway.

Giving names to non-existing entities is not IUPAC's primary interest.

related https://en.wikipedia.org/wiki/Amorphous_carbonia - CO2 polymerises at very high pressures, but creates 3D not linear structure

@querambique your compound could be called linear polymeric carbon dioxide

The generic name is oxocarbon. In particular, "Carbon suboxide spontaneously polymerizes at room temperature into a carbon-oxygen polymer, with 3:2 carbon:oxygen atomic ratio." This is not quite the formula you provide, but the polymer has been synthesized.

authors: Γλυκερία Ριγανά