text:
Hydration of lithium salts
Why are lithium salts heavily hydrated? I am not able to understand this because I feel salts of larger cations like Caesium should be more hydrated because of the larger size, they can able to hold more water molecules.
What is the flaw in my reasoning?

The larger $\ce{Cs+}$ ion may interact with more water molecules at once than $\ce{Li+}$, but each interaction is weaker. The prevailing interaction between water molecules and ions are ion-permanent dipole interactions (or just ion-dipole), which are proportional in strength to the amount of charge on the ion and the partial charge on the dipole, and inversely proportional to some power of the distance between the centres of the charges (via the shell theorem). Hence, the closer the water molecules get to the ion, the stronger the interaction. Since $\ce{Cs+}$ is very large, the charge centres don't come so close, and the interaction is relatively weak. $\ce{Li+}$ is much smaller, so there is strong solvation. For sufficiently small, highly charged ions, the first solvation shell comes so close to the ion that it allows covalent bonding to take place.
You may find this succinctly described as $\ce{Cs+}$ having a lower charge density than $\ce{Li+}$, which is about the same thing.


text:
Relation between speed of sound c and compressibility
We know that
$c^2=\frac{\partial p}{\partial ρ}$
The adiabatic compressibility is defined as: $\beta_S=-\frac{1}{V}\frac{\partial V}{\partial p}$ such that the subscript "S" stands for "adiabatic"
How can I show that $c^2=\frac{1}{\rho \beta_S}$ ?
I tried replacing $V$ by $\frac{m}{\rho}$ but I get for $\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}$

http://en.wikipedia.org/wiki/Compressibility has some information about this

With your result
\begin{equation}
\beta_S=-\rho \frac{\partial \frac{1}{\rho}}{\partial p}
\end{equation}
you're nearly there. All that is missing is a little mathematical trick. Simply make the substitution $f = \frac{1}{\rho}$ and use the chain rule to differentiate $f$ by $\rho$ and you see that:
\begin{equation}
  \frac{\partial f}{\partial \rho} = - \frac{1}{\rho^2} \quad \Rightarrow \quad \mathrm{d} f = - \frac{1}{\rho^2} \mathrm{d} \rho \quad \Rightarrow \quad \mathrm{d} \frac{1}{\rho} = - \frac{1}{\rho^2} \mathrm{d} \rho \ .
\end{equation}
Using this and the definition of $c^2$ you get to the desired result:
\begin{equation}
\beta_S = -\rho \underbrace{\frac{\partial \frac{1}{\rho}}{\partial p}}_{= - \frac{\frac{1}{\rho^2} \partial \rho}{\partial p}} =
\underbrace{(-\rho) (- \frac{1}{\rho^2})}_{= \frac{1}{\rho} } \underbrace{\frac{\partial \rho}{\partial p}}_{= \frac{1}{c^2}} \quad \Rightarrow \quad c^2 = \frac{1}{\rho \beta_S} \ .
\end{equation}

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text:
Stability of superoxides of alkali metals
Why are superoxide ions stable only in presence of large cations such as $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$?

I think it has a lot to with the heat of formation/..; See this link for some research on lithium superoxide: http://www.google.be/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CC0QFjAA&url=http%3A%2F%2Fcontrails.iit.edu%2FDigitalCollection%2F1965%2FAMRLTR65-126.pdf&ei=tbb3UvTsLKPy7AaTt4HoCA&usg=AFQjCNEJ5yK3cTwESsuGx4Ds9b0zFbvu9A&sig2=Y-W2XUhH1vBWnTPSJaQgSQ&bvm=bv.60983673,d.ZGU&cad=rja

According to the Wikipedia page Alkali metal oxides, the superoxide of $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$ form when these metals are burnt in air, and according to the Bodner Research Group page Active Metal Reactions, they

react so rapidly with oxygen they form superoxides, in which the alkali metal reacts with $\ce{O2}$ in a 1:1 mole ratio.

Given, according to the Chem-Guide blog article Alkali metals that

The fact that a small cation can stabilize a small anion and a large cation can stabilize a large anion explains the formation and stability of these oxides.

Which, according to the blog, results in the  $\ce{Li+}$ ion (which is small, possessing a larger positive field around it) only being able to naturally combine with a small anion $\ce{O^{2-}}$ forming $\ce{Li2O}$, whereas the substantially larger $\ce{K+}$, $\ce{Rb+}$ and $\ce{Cs+}$ have weaker positive fields and can stabilise the larger superoxide anion $\ce{O2^{-}}$ forming superoxides.

This is hardly an explanation; I don't why how this is the accepted answer. Stating a general rule from some blog is not the same as an intuitive reason, which is what the OP was looking for.

The word "stability" itself isn't very clear, so you have to figure out exactly what it is stable with respect to. If you look up the data, the reaction
$$\ce{NaO2 (s) -> Na(s) + O2(g)}$$
is not enthalpically favourable. So, $\ce{NaO2}$ is actually stable with respect to decomposition to the constituent elements.
Instead, it is actually likely to be unstable with respect to decomposition to the oxide/peroxide. Since sodium is known to burn in oxygen to form the peroxide, this hints that we should be looking at the energetics of the reaction
$$\ce{2NaO2 (s) -> Na2O2 (s) + O2 (g)} \qquad \Delta H_1$$
$$\ce{2CsO2 (s) -> Cs2O2 (s) + O2 (g)} \qquad \Delta H_2$$
If cesium superoxide is "stable" in this respect, then we would expect $\Delta H_2 > 0$. And if sodium superoxide is "unstable" in this respect, then we would expect $\Delta H_1 < 0$.
The controlling factor here is likely to be the lattice energies (which are directly related to the size of the cation). Using the Kapustinskii equation we have
$$\Delta H_\mathrm{L}(\ce{M2O2}) \propto \frac{3\cdot 1\cdot 2}{r_\ce{M} + r_\ce{O2}} = \frac{6}{r}$$
and
$$\Delta H_\mathrm{L}(\ce{MO2}) \propto \frac{2 \cdot 1 \cdot 1}{r_\ce{M} + r_\ce{O2}} = \frac{2}{r}$$
where $r = r_\ce{M} + r_\ce{O2}$ (I have assumed the radii of the superoxide and peroxide ions to be the same). In the enthalpies of the decomposition reactions above $\Delta H_i$ ($i = 1,2$), the lattice energy contribution is
$$\begin{align}
\Delta H_i &= 2\Delta H_\mathrm{L}(\ce{MO2}) - \Delta H_\mathrm{L}(\ce{M2O2}) + \cdots \\
&= 2\left(\frac{2}{r}\right) - \frac{6}{r} + \cdots \\
&= -\frac{2}{r} + \cdots
\end{align}$$
Sodium is smaller than cesium, and therefore we expect the value of $r$ to be smaller. This clearly translates into a more negative value of $\Delta H$, as expected.
Why should $\Delta H_2$ be positive, then? It is to do with the other $\cdots$ terms that were omitted. These terms basically involve the production of $\ce{O2^2- + O2}$ from $\ce{2O2-}$, which is enthalpically unfavourable and hence have a positive contribution to the enthalpy of reaction.
In the case of sodium, the $-2/r$ term is large enough to outweigh this positive term. In the case of cesium, it isn't.

Looking to other aspects of inorganic chemistry one might recall that carbonates of heavier alkaline earths are more stable than carbonates of light alkaline earths (e.g. $\ce{MgCO3}$ vs $\ce{BaCO3}$). The reaction $\ce{MCO3 -> MO + CO2}$ can be treated very similarly and the same conclusion will be reached.
The thermodynamics of the alkali metal oxides/peroxides/superoxides is treated more thoroughly in the following question: Why do the alkali metals form different products upon combustion in air?

Well according to my book 'General and inorganic chemistry volume 2, by Dr. Ivan Filipovic & Dr. Stjepan Lipanovic ', the oxidation state of superoxides is -1/2. $\ce{K}$, $\ce{Rb}$ and $\ce{Cs}$ react with oxygen and form type $\ce{MO2}$ compounds called superoxides. They are ionic crystalised compounds containing cation $\ce{M+}$ and anion $\ce{O^2-}$, and since the oxygen anion has 13 valent electrons, superoxides are colorful, paramagnetic, and unstable compounds, and they also exist only in solid state.

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text:
Baking Soda + Oil at high temps
My mom while making Potato chips put some Baking Soda in Oil with it. Just curious to know, it was first time Baking Soda in Oil at that high temperature. Is it safe to eat?
Answer ASAP before my chips get cold.

I don't see baking soda reacting with oil - so in that case it is safe. However deep frying can give temperatures of around 175–190 °C and sodium bicarbonate (which is baking powder) starts decomposing in sodium carbonate (what I don't advise eating) and at higher temperatures the baking powder will decompose very quickly into Na2CO3. Of course if there is a very small amount of baking powder added I don't see a problem (but I'm not certain that it is safe to eat...)

Baking soda is essentially sodium hydrogen carbonate, which should not be expected to react with oil because of its poor solubility. $\ce{NaHCO3}$ melts at 50 °C and starts to decompose slowly to sodium carbonate, carbon dioxide and water.
$$\ce{2NaHCO3 ->~ Na2CO3 + H2O + CO2\uparrow}$$
By the release of carbon dioxide, $\ce{NaHCO3}$ acts as a raising agent when used in baking. Under the conditions of deep frying (150-200 °C), the decomposition is fast as the reaction is entropically favored (generation of gases). $\ce{CO2}$ and $\ce{H2O}$ escape from the baking mixture, leaving sodium carbonate behind.
The ingestion of small amounts of $\ce{Na2CO3}$, which are generated from typically used amounts of baking soda, should not be a health problem. Acidic ingredients in the baking mixture will convert sodium carbonate into the sodium salts of the respective acids, carbon dioxide and water. Higher amounts of sodium carbonate, stemming from the use of too much baking soda, would cause an unpleasant metallic taste of the baked food. Sodium carbonate is actually used as a food additive (E 500) and in toothpastes as a foaming agent and abrasive.

The flavor of fine fried potatoes is from browning, the Maillard reaction.  Carbonyls (sugars) condense with amino groups (alanine residees in protein) to Schiff bases, then the fun starts.  The optimum pH for condensation is bicarbonate buffer, slightly basic.  One might then do the experiment of soaking fries in dilute bicarbonate solution, drain, and fry.  Does that taste better than untreated controls?  Push it a little harder by dissolving a jot of honey (or corn syrup) in the bicarb buffer to add reducing carbonyls from glucose and fructose.
A fine bagel is formed from dough, boiled in honey water (reducing carbonyls), brushed with egg wash (protein), then baked.  It's all about the crust!
http://en.wikipedia.org/wiki/Maillard_reaction
Either way, don't overcook toward burned - trace acrylamide formation.

While your answer contains many interesting tidbits of information, it does not address the question at hand. Generally, this isn't a helpful answer.


text:
Finding the final temperature
If $125$ Calories of heat is applied to $60.0\text{g}$ piece of copper at $21.0^\circ\text{C}$, what will the final temperature be? The specific head of copper is $0.0920 \text{ cal/gC}$.
All I have to answer this question is a formula:
$SH = \text{heat/(mass}\times \Delta T)$

Please see http://meta.chemistry.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange for how to ask homework questions.  Please outline what you have attempted towards solving the problem.

Any hint would be great. I'm really not sure how to go about the answer. I've been working on it for about 3 hours already, as silly as it sounds.

I understand that asking homework questions here will erode the quality of this site. I wrote anyways as a last resort. Any hints on how to go about the answer would be great. Anyone else looking for the answer can see your hints and benefit as well.

Rather than just us being draconian, it is helpful for those answering to see what your thought process is so they can steer you in the right direction.  What else do you know about the problem?  Do you have a formula in mind, if so, do the dimensions work out?

Split $\Delta T$ up into $T_f - T_i$, and you have all of the rest of the parameters.

Ohhh.. ok great that makes thinges easier!

And now, post your answer below so the next student finds your method instead of having to ask another question ;)

Oh, :/ I'm a bit embarrassed because I don't remember it. I just submitted my assingment. I can see why homework isn't allowed. I appreciate the help though!
I remember plugging all the numbers into the equation and then solving for Tf, or the final time. So I multiplied the denom by both sides, and I think I had to do some unit conversions. It's simply algebra from there.

The energy amount used for heating up a material can be calculated according to the following equation:
$$ E = c_\text{s}\, m\, \Delta T$$
where $E$ is the energy, $c_\text{s}$ the specific heat capacity, $m$ the mass of the object and $\Delta T$ the temperature difference that is to be achieved.
Solving the above equation for $\Delta T$ we get:
$$ \Delta T = \frac{E}{c_\text{s}\, m}$$
Plugging in the values provided by you, I get $\Delta T = 22.6\;\mathrm{K}$ (or °C, since we're talking about the difference and the scale is the same, it doesn't matter).
Adding this increase in temperature to the initial temperature gives the final answer:
$$ \vartheta_\text{f} = \vartheta_\text{i} + \Delta \vartheta = 21.0\; \mathrm{°C} + 22.6\; \mathrm{°C} = 43.6\; \mathrm{°C}$$
Note the following:

Number of significant digits: Since the data given consistently had three significant digits, I've rounded to the third digit.
Symbol for temperature: I used $\vartheta$ for temperatures in °C and $T$ for temperatures in K. I strongly suggest you follow this example as to never confuse the two.


text:
What's the voltage of a battery?
The definition of voltage I'm familiar with is the amount of electric potential energy per unit charge at a location. My prof defined the voltage of a battery to be the amount of potential energy difference an unit charge would have at the positive terminal vs. the negative terminal.
But from what I remember from high school, the "terminals" aren't points, their rods of metal:

Since the voltage is defined at a point, not cylinders or rods what does the voltage of a battery mean? Also, how does the definition connect to the "pushing" power of a battery?
I took a look at Electrode Potential but it was a bit over my head.
I'm looking for an intuitive explanation, not one using rigorous chemistry concepts. Thanks!

what does the voltage of a battery mean? The difference in electrical potential between the points on the picture you posted (one on one end of the battery's positive terminal and one on the negative). Unit is V = A*Ohm = W/A = J/C = kg m² / A s³ .

Voltage is indeed something that is measured at a particular point, but that doesn't mean its a property of only one point and in fact every point will have some voltage. In particular you can have a series of points that map out an "equipotential" where the potential is equal all along these points. A terminal of a battery is an equipotential so no matter where an electron "is" it will have the same potential energy U = e V.
Intuitively this is just like potential energy due to gravity. The battery is like a mountain with a flat top and level plane at the base. Everywhere along the top the altitude is equal and a ball will have the same potential energy. Similarly everywhere along the level base of the mountain the ball will have the same potential energy.

An electrochemical cell which causes external electric current flow can be created using any two different metals since metals differ in their tendency to lose electrons. Zinc more readily loses electrons than copper, so placing zinc and copper metal in solutions of their salts can cause electrons to flow through an external wire which leads from the zinc to the copper.As a zinc atom provides the electrons, it becomes a positive ion and goes into aqueous solution, decreasing the mass of the zinc electrode. On the copper side, the two electrons received allow it to convert a copper ion from solution into an uncharged copper atom which deposits on the copper electrode, increasing its mass.
Here the voltage of the battery means the potential difference between the two half cell, this is how the enode values are different . The bigger the difference in the enode value( its ability to lose or gain electrons) the bigger the voltage of the dry cell will be


text:
For a given teabag and dunking duration - will I get more caffeine in a bigger cup or the same amount?
When choosing a cup of tea, I can choose a big cup or a small cup. I'm wondering if I get more caffeine with the bigger cup or the same amount of caffeine. (in moles)
My question is: For a given teabag and dunking duration - will I get more caffeine in a bigger cup or the same amount?
Assumptions

Teabag is placed in cup first - and hot water poured directly onto the bag
Same amount of stirring across both cups (assume three spoon swirls)
The main varying parameter is the volume of solvent changing

Question behind the question - trying to express it another way

Is there a kind of 'fixed rate of dissolution' that limits the amount of caffeine that comes out over a period of time - ie for two cups where one is 250ml and one is 300ml - and each have a tea dissolving time of 30s - wouldn't they have the same number of moles of caffeine at the end because of an upper bound on the rate of caffeine dissolving?

This post may interest you.  I have linked a couple of journal articles in the comments to one of the answers that was given which you may find useful. http://cooking.stackexchange.com/questions/41413/all-else-held-equal-does-a-pot-of-tea-have-more-total-caffeine-than-a-mug/41420#41420

Dunking the bag up and down within the tea probably helps get the caffeine out as quickly as possible, too. Fick's law... Thanks for the caffeine solubility info! I've wondered about your question for a long time and appreciate your insights.

From what I remember from Fick's laws, I would say that the factor that limits the rate of dissolution is the concentration of the already-dissolved compound. At first, you don't have any caffeine dissolved so the rate of dissolution is fastest. As time goes by, more and more caffeine becomes solvated and that hinders the dissolution of remaining caffeine from the tea leaves. At any point of time until saturation, the 300 ml cup will have faster dissolution rate because it has more "free space" to solvate the caffeine.
Edit: However, after thinking this for a while, I have to add that in practice the difference might be too small to be detected. The solubility (~ saturation point) of caffeine in water is 20 g/l at 20°C while the concentration of caffeine in tea is 0.1 - 0.4 g/l – huge difference. It might be that from caffeine's point of view, the "free space" is more or less the same in both cups.

The same source mentions an estimate of LD(50) for humans to be in the order of 150--200 mg caffeine/kg body weight, so it were not desirable to attend the formation of a caffeine-saturated solution, either.

How do you prepare you your cup of tea?
Dunking the leaves of tea (solid phase) exposes them the liquid phase (water).  The concentration in the liquid phase of caffeine (and other chemicals/ingredients carried by your leaves of tea) will increase. Because hot water is more efficient than cold one for this extraction, temperature plays a role, too.  Stirring the liquid phase for a homogenous dissipation of extracted matter across the whole solvent volume at your disposition may be to be considered as a parameter, too.  Yet, the more volume of solvent of extraction at your disposition, the more total solvent capacity to get more caffeine out of your leaves of tea.
Yet another thought may arise, too; if you do not spend all the volume at once. As textbooks for physical chemistry, regarding multiple extractions, teach, repeated extraction with a small fraction of solvent of extraction would be then more efficient than deploying the whole solvent of extraction in a single extraction.  (However, by time the composition of the ingredients being extracted by hot water, varies, too.)

Really helpful answer - I've added some assumptions. I think the question behind my question is, is there an upper bound on the rate of dissolving - which has a lot to do with your comments about volume of solvent capacity.


text:
Ester as a solvent
I am curious as to why ester is used often as a solvent in organic chemistry? How is it different that just using water, which is also polar? I'm sorry if this is a basic question, I'm just a student.

Ethyl acetate, perhaps one of the best examples that would fit in your category "often used solvents in organic chemistry" is prepared at low cost from ethanol and acetic acid (and these two are cheap, too).
It may dissolve a wide range of organic material. Its low solubility in water presents as an advantage when you want to extract something.
It has a boiling point of 77 Celsius that is elevated enough to work at moderate temperatures; yet if you want to remove it as solvent you do not spend terrible much energy to distill it off.
Set side-by-side with other solvents, it comes with a relative low toxicity (at the level we know today).

Of course there is not a "one [solvent] fits for all [purposes/applications]", there are chemical conditions where other solvents are more appropriate to be deployed than esters. The presence of mineralic acids, or the of  organometallic reagents for example would call for a substitution.


text:
How do I calculate the enthalpy change when a gas is being used to heat water?
The question I'm trying to answer is:

Calculate the mass of butane gas that would be needed to heat $724\ \mathrm{cm^3}$ of water from an initial temperature of $7.44\ \mathrm{^\circ C}$ to $50.7\ \mathrm{^\circ C}$. The thermochemical equation for the combustion of butane is:
$$\ce{2C4H10(g) + 13O2 -> 8CO2(g) + 10H2O(l)}\quad\Delta H=-5748\ \mathrm{kJ/mol}$$

I know that the equation for enthalpy is
$$Q=m\times c\times\Delta T$$
and I calculated this for water:
$$Q=724\times4.18\times43.26$$
$$Q=130918.6032$$
I don't know where to go from here in order to figure out the mass of butane needed.  Please tell me if I have done something wrong or if you can help me to solve the rest of the problem.

If you want to be sure that your result isn't off by some orders of magnitude please check the units. Unit analysis looks boring but is helpful.

You're on the right track.

\begin{eqnarray}
Q & = & m\cdot c\cdot \Delta T\\
\mathbf{unit\ analysis} \qquad \textrm{kJ} & = & \textrm{kg}\cdot \frac{\textrm{kJ}}{\textrm{kg} \cdot \textrm{K}} \cdot \textrm{K}
\end{eqnarray}

For water, the specific heat capacity $c = 4.18 \large{\frac{\textrm{kJ}}{\textrm{kg}\cdot \textrm{K}}}$. You have that right.
The temperature difference $\Delta T = (50.7 - 7.44) = 43.26\ \textrm{K}$. We're talking about differences, so we can use kelvin as the unit. You have that right too.
Now, what is the mass of $724\ \textrm{cm}^3$ water?
Choose any unit you want, but adjust the outcome of your calculation according to the unit analysis above!
Now you know the kJ you need.
From the combustion equation, you know the amount of heat delivered by $n$ moles of butane. Note the $n$!
Now you can calculate how many moles of butane you will need.
From there, the calculation of the mass is not difficult either.


text:
Which of the these SN1 solvolysis of two alkenyl bromides reactions is faster?

The SN1 solvolysis of one of the following alkenyl bromides is in MeOH approximately 100000 times faster than the other one. Which is the more reactive one? Draw the mechanism including the transition state, determine the rate determining step and draw the reaction product of each compound.

Now what is apparently important here is the rotation of the substituent in each relative to the aromatic ring. In the second compound (with two methyl groups), the rotational profile shows destabilized molecule except when 90 degrees. That means the rotation in the second molecule helps with the leaving of Br and this (rate determining) step is much faster than when there are no methyl groups (first compound).
Is that a complete explanation or I am missing something important? Also is there anything special about the mechanism of the two reactions and the products in each case?

[...] SN1 solvolysis [...]

Let's recall what we know about these. We have a two-step reaction, consisting of

heterolysis of the starting material
association of a nucleophile $\ce{Nu}$

We remember that the heterolysis is not assisted by $\ce{Nu}$, there is no backside attack of $\ce{Nu}$, pushing out the leaving group. We also remember that the association step (addition of $\ce{Nu}$ to the carbocation intermediate) is usually pretty fast. Consequently, heterolysis, i.e. the formation of a carbocation is the rate-determining step.
Thank's to good old Arrhenius, we know that the rate is determined by the activation energy of this step.
The activation energy is lower when the resulting cation is more stable.
The cation is more stable when the delocalization of the charge is better.
Distribution of charge on the vinyl cation is better when it's actually a
styryl cation.

What is the geometrical requirement for this interaction and what might prevent it?

The stabilisation is understood. But the geometry seems to be problematic as the cation is planar and the methyl groups are preventing the interaction. So what is the role of the rotation of this component in question that I referred to earlier?

So, we agree that the stabilization is best when the double bond and the arene are coplanar? Now, what if this orientation is prevented by the methyl groups? Is the cation more or less stable then? What follows for the activation energy? Is full rotation really relevant? Or is it more important whether a full pi-overlap is possible?

If I got it right, a full pi-overlap is hindered by the methyl groups present in the second molecule. Then, stabilization most efficiently occurs for the first molecule and ${SN_{1}}$ is then faster for the latter. Is that correct ?


text:
What are the products of [Ti(H2O)6]Cl3 + H2SO4 + K2Cr2O7 --->
I am slightly lost on how to proceed with this equation. I know potassium dichromate is a strong oxidizing agent but I am unsure of how to deal with it along with the titanium complex. I was wondering if I could get tips on how to handle these types of equations.

I know potassium dichromate is a strong oxidizing agent

You have already identified one partner and its role in a redox reaction. Good start!
So which component will be oxidized to what?
Can we assume that the reaction is carried out in water? If so, let's have a look at the ions in solution.
We'll have $\ce{H3O+}$, $\ce{K+}$, $\ce{Cr2O7^{2-}}$, and "that complex".
Let's have a look at $\ce{[Ti(H2O)6]Cl3}$.
Apparently, it's a complex of titanium with 6 neutral molecules of water as the ligands and 3 $\ce{Cl-}$ as the counter ions. So, what is the oxidation state of titanium in the complex when it is neutral with 3 $\ce{Cl-}$ as the counter ions?
Let's look again: $\ce{[Ti(H2O)6]^{3+}}$. OK, it is obviously $\ce{Ti(III)}$ and it seems that we have found the victim of our stong oxidant in the system.
We remember that most titanium compounds in nature are $\ce{Ti(IV)}$ oxides and conclude that we will have to balance our redox reaction to account for a $\ce{Ti(III) -> Ti(IV)}$ oxidation.
We could now continue to conclude that the reduction of $\ce{Cr(VI)}$ to $\ce{Cr(III)}$ is accompanied by $\ce{[Ti(H2O)6]^{3+} -> [Ti(H2O)6]^{4+}}$ but then we remember that $\ce{Ti(IV)}$ - tiny and higly charged - is a hard Lewis acid and will not exist as a naked cation, barely covered with just six neutral ligands in a hexaquo complex!
Instead, the likely $\ce{Ti(IV)}$ species is the titanyl cation, $\ce{TiO}^{2+}.$
Time to collect the puzzle pieces and write the redox reaction.


text:
How do I find the hybridisation of the orbitals of carbon in a molecule?
What are hybridisation states of each carbon atom in the following compounds?

$\ce{CH2=C=O}$
$\ce{CH3CH=CH2}$
$\ce{(CH3)2CO}$
$\ce{CH2=CHCN}$
$\ce{C6H6}$

How do I find the hybridisation of the orbitals of the carbon atom in such compounds?

My method for finding the hybridization of the orbitals of an atom in a molecule:

Look at the molecular structure (geometry), apply the following:

Tetrahedral: sp3
Trigonal planar: sp2
Linear: sp

Just apply the bond method which is the simplest and most effective. Count the number of bond pairs of central atoms excluding pi-bonds and the number Of lone pairs and match the corresponding hybridisation.

Welcome to the site. Could you just be more elaborate, just? This answer would benefit from better descriptions, just, and some examples, just.

For hydrocarbons, if the carbon atom forms only σ bonds, the orbitals of it will be sp3 hybridised. If the carbon atom forms a π bond, its orbitals are sp2 hybridised. If the carbon atom forms two π bonds, its orbitals are sp hybridised.

Seeing as three of the five examples asked about are not hydrocarbons, this is not particularly helpful. You also don't explain why this is the case, which would be more useful in order to provide a general answer.

I wanted to say organic compounds. As others have suggested usual ways to identify the hybridisation, I gave a simple method to find it. ☺

Please note that atoms cannot be hybridised, only orbitals. It's common to refer to atoms to be hybridised, but technically this is not correct.


text:
What is hyperconjugation?
What is hyperconjugation and how do you know if there is hyperconjugation in a compound? Which orbitals are involved in hyperconjugation?

In the ethyl cation, one can write the following resonance structures:

The structure on the right can be actually written 3 times since there are 3 hydrogens available. Note also that the "ethylenic-like" structure on the right involves a $\mathrm{sp^3-p}$ bond (the $\ce{C-H}$ bond is $\mathrm{sp^3}$ hybridized and the cationic carbon has a $\mathrm{p}$ orbital), so it is not a true pi bond ($\mathrm{p-p}$) like that found in ethylene.  The resonance structure on the right is meant to show the "hyperconjugation" effect of the $\ce{C-H}$ bond ($\mathrm{sp^3-s}$) with the $\mathrm{p}$ orbital located on the adjacent cationic carbon.  The drawing is a shorthand way of saying that when the $\ce{C-H}$ bond is rotated into the same plane as the $\mathrm{p}$ orbital on the cationic carbon, there can be a stabilizing overlap - and it is called "hyperconjugation." In the 2-propyl cation, there are 6 of these hyperconjugative resonance structures (there are 6 $\ce{C-H}$ bonds on the 2 methyl groups adjacent to the cationic carbon and in the t-butyl cation there are 9 such resonance structures. This can be used to explain why tertiary carbonium ions are more stable than secondary carbonium ions, which in turn are more stable than primary carbonium ions.

Your answer is really very insightful. I would just like to ask if hyperconjugation only exists only if there is a positive formal charge on one of the atoms in the molecule.

No, hyperconjugation can be used to explain properties in neutral compounds too.  For example, it can be used to explain why substituted double bonds are more stable than unsubstituted double bonds (see here).

@ron can more than one $\ce{C-H}$ bond overlap with one vacant $\mathrm{p}$ orbital? for example, in the t-butyl cation, can $\ce{C-H}$ bonds from the three different methyl groups overlap with the vacant $\mathrm{p}$?

Hyperconjugation is the interaction between adjacent orbital (empty $\mathrm{p}$ for cation, $\pi$ for alkene) with $\sigma$ bond, while common conjugation is between adjacent orbital and $\pi$ bond.
As one example, one reason why more alkyl substituted alkenes are more stable than less substituted ones is because more alkyl groups provide more hyperconjugation stabilization of the alkene. Why more alkyl substituted cations are more stable is also because the hyperconjugation stabilization from more adjacent alkyl group.

Hyper-conjugation is in general a stabilizing interaction which involves the delocalization of σ-electrons belonging to the $\ce{C-H}$ bond of an alkyl group attached to an atom with an unshared p-orbital. The σ-electrons of a $\ce{C-H}$ bond of an alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared p-orbital.

@ParthChauhan Chemical markup can be applied easier and more correct using the mhchem extension. If you want to know more, please have a look here and here.


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Thermal decomposition of calcium carbonate
Thermal decomposition of calcium carbonate.
In general, we heat $\ce{CaCO3}$ to temperature of approximately 825°C it decomposes into calcium oxide and liberates carbon dioxide gas:*
$$\ce{CaCO3} \overset{825\ ^\circ \text{C}}{\ce{ ->}}\ce{CaO + CO2}$$
Is it possible to heat the calcium carbonate at temperatures below 825 degrees Celsius.

Do you mean "Is it possible to achieve the same effect heating the calcium carbonate at temperatures below 825 degrees Celsius?"

Yes, complete calcite at temperatures below 825 degrees Celsius.

Pump a hard vacuum and calcium carbonate will calcine at a lower temperature.  825°C is about one atmosphere of carbon dioxide at equilibrium.  Thermodynamics proposes, kinetics disposes - finely ground goes faster than bulk.

In the US we activate molecular sieves in a vacuum oven around 150-200 C.  In Canada, they use porcelain bowl and a Meeker burner underneath.  If nanostructure is important, go cooler.


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What is the nature of the oxide of boron?
I think it should be acid, as I am aware that a compound by the name of Boric acid exists. But how can the oxide be acidic when it is electropositive? Also, other members of its group are well known metals. Why isn't the oxide amphoteric, when its group member Aluminium has an amphoteric oxide?
My teacher told us that the oxide of boron is basic in nature? So what exactly is its nature?

The most common oxide of boron is boron trioxide, $\ce{B2O3}$, which is the anhydride of boric acid:
$$\ce{H3BO3 -> HBO2 + H2O}$$
$$\ce{2HBO2 -> B2O3 + H2O}$$
$$\text{net} \ \ce{2H3BO3 -> B2O3 + 3H2O}$$
As the anhydride of an acid, boron trioxide is acidic. It reacts with water to regenerate the acid:
$$\ce{B2O3 + 3H2O -> 2H3BO3}$$
Boric acid's acidity does not derive from the dissociation of a proton (or abstraction of a proton by water), as does many acids, for example sulfuric acid:
$$\ce{H2SO4 <=> H+(aq) +HSO4- (aq)}$$
$$\ce{H2SO4 + H2P <=> H+(aq) + HSO4- (aq)}$$
Boric acid (written now as $\ce{B(OH)3}$, which is suggestive of its structure) instead reacts with water by forming a dative complex which is the active acid. A water molecule associates with the empty $p$ orbital on boron. This difference in behavior is enabled also by the electropositive character of boron. The acidic oxides of some metals behave similarly.
$$\ce{H2O + B(OH)3 <=>H2OB(OH)3 <=>H+(aq) + B(OH)4^- (aq)}$$

This isn't completely true. Boric acid does also dissociate normally, to similar extent.


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Does anyone know where I can find the reaction rate coefficient and design equation for the electrolysis of brine and sodium hydroxide solution?
equation 1) $\ce{2 Cl^{–} -> Cl_2 + 2e^{-}}$
equation 2) $\ce{2 H2O + 2 e^{–} -> H2 + 2OH^{–}}$
These equations are the key reactions in the chlor-alkali process. I want to model a reactor for this process based on a CSTR (continuous stirred tank reactor). However I really need reaction rate coefficients (k values) and design equations. Any help would be much appreciated!

Do You have the meanst knowledge on electrolysis? Reaction rate is determied by the current you apply, up to some saturation value, in this case the saturation is reached when the electrodes are covered by the gases evolved. CSTRs and electrolysis is pure nonsense.

Actually, the book I am referring too, 'Electrochemical Engineering by H.Wendt' explains in section 6.5.2 that a CSTR is a good approximation for the cell design as the cell is intrinsically well stirred by gas evolution. The reaction rate constant can be determined via the applied current.

You should look in some kind of reaction rate database, e.g. NIST. An example from the database:


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Symmetry element in the coversion of cyclopent-2-enyl anion to pentadienyl anion?
I'm trying to draw the state correlation diagram for the reaction of cyclopent-2-enyl anion to pentadienyl anion but I don't know which symmetry element is preserved in this reaction. What could it be?

Let's look at the back reaction, the closure of the pentadienyl anion to the cyclopentenyl anion.  Once you understand this you'll see that the exact same reasoning applies to the reverse reaction that you asked about.  So here's a picture of the reaction we're talking about.

And here is a picture of the 5 molecular orbitals for the pentadienyl system

Since the pentadienyl anion has 6 pi electrons, the first 3 molecular orbitals (from the bottom up) are filled, each with two electrons.  That makes the third molecular orbital (the one in the middle of the diagram), let me call it $\ce{\psi}_3$, the highest occupied molecular orbital (HOMO).  The HOMO is so important because it is the orbital that will control our process.  If we look at the symmetry of the $\ce{P}$ orbitals on $\ce{C_1}$ and $\ce{C_5}$ in $\ce{\psi}_3$, we see that in order to get constructive overlap (overlapping the parts of the $\ce{P}$ orbitals with the same sign) as we rotate the $\ce{P}$ orbitals on these two carbons to form the bond that forms the cyclopentenyl ring, we must rotate the tops (or bottoms) of these two $\ce{P}$ orbitals towards each other.  If we were to rotate the top of one and the bottom of the other towards each other we would overlap a plus side of one $\ce{P}$ orbital with the minus side of the other $\ce{P}$ orbital and this would not result in constructive or bonding overlap.  The type of rotation that leads to bonding overlap in this example is termed "disrotatory".  It preserves a plane of symmetry bisecting the pentadienyl system (e.g. it contains $\ce{C_3}$) and is perpendicular to the plane of the screen.  This is the symmetry element that is preserved during the ring closure.  The exact same arguments also explain the ring opening of the cyclopentenyl anion to the pentadienyl anion.

I would suggest that you phrase your answer in terms of overall symmetry of the state wave function (capital \Psi) rather than the individual molecular orbitals (lowercase \psi).  Also, upon change in geometry, the molecular orbitals themselves also change, and so the frozen orbital approximation may not be appropriate.

Thank you for the suggestion, but orbital symmetry, Frontier Molecular Orbital theory, the Woodward-Hoffman rules - whatever name you choose - focuses upon the transformation of the HOMO during the process under consideration, so focusing only upon $\ce{\Psi_3}$ seems appropriate. You are correct that the orbitals transform during the process.  In this case the 3 MOs of the allyl system and the 2 MOs of the newly formed sigma bond are created.  For brevity I left out the MO's of the final product.  Although it is nice to view them, they are not necessary to explain the transformation when

you start from the pentadienyl side.  Just google "Woodward-Hoffmann rules" for more detail.

Maybe you should have done the same.  A quick look through "Woodward-Hoffman Rules" on Wikipedia reveals: "However, as reactions do not take place between disjointed molecular orbitals, but electronic states, the final analysis involves state correlation diagrams."   That is state (capital Psi) and not molecular orbitals (lowercase psi).  Your answer does not correct symbology; and given that there are MO considerations and state symmetry considerations, this could be confusing.

Please post an answer so that I can better understand your comment.  As to correlation diagrams, in MO theory they show how each of the pentadienyl MOs transform (correlate) into their respective cyclopentenyl MOs.  Again it would better help me understand your comments if you posted more detail.

PS The thing orbital symmetry has going for it is that it is a relatively simple and straightforward method that allows for the accurate prediction of reality, e.g. the disrotatory closure of the pentadienyl anion and the conrotatory closure of the pentadienyl cation, etc.

The allyl anion is delocalized and planar.  You have a mirror plane through its middle carbon normal to the plane of the ring, or a ${C_2}$ axis through that carbon depending on how the other two carbons are canted.


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Enthalpy of Zinc and Sulfur Reaction
I was working on my homework which was to write a description of what would occur if zinc and sulfur reacted and what steps would a scientist have to make in order to make them react. When I researched I bit, I saw that it would be a exothermic reaction, and I wanted to show this by using enthalpy. I haven't been taught this however, and as a result, I'm confused. I know that enthalpy equals the heat transfer and there is some sort relationship with bond energy, but I don't know how to apply this to the reaction.
Zinc Powder + Sulfur Powder = Zinc Sulfide + Energy
$\ce{Zn_{(s)} + S_{(s)} -> ZnS_{(s)}}$ + $205.98 \text{ kJ}$
Because $205>0$ this is an exothermic reaction. But where did this $205.98\text{ kJ}$ come from?

The free energies of formation of the elements in their standard states is zero, by definition.  You can look up the (variously reported) standard enthalpy of formation of the product, −204.6 kJ/mol (exothermic!).  It is roughly the binding energy of the crystal lattice less the ionization energies of the inputs.  So, by the numbers, do you obtain spalerite or wurtzite?  ZnS has crystal polymorphs and polytopes with differently populated unit cells
$\ce{Zn_2S_2}$ Wurtzite-2H
$\ce{Zn_4S_4}$ Wurtzite-4H and Sphalerite
$\ce{Zn_6S_6}$ Wurtzite-6H
$\ce{Zn_15S_15}$ Wurtzite-15R

If the reaction was Zn(s) + S_8(s) --> 8 ZnS(s), how would the enthalpy be affected? Also, you said that the free energies of the elements in their standard states is zero. So is the standard state of sulfur S_8 or S?

Balance the equation.  Enthalpy of formation is per mole product.  8x input giving 8x output is the same number.  Sulfur has some 30 allotropes, plus crystal polymorphs.  "Elemental sulfur" is an S_8 ring formula unit in a = 10.45, b = 12.845, c = 24.46 A; Z = 16 (128 atoms); V = 3283.27 A^3 D_{calc} = 2.08 g/cm^3; orthorhombic space group Fddd.  http://www.gps.caltech.edu/~vijay/Papers/Chemistry/Meyer-76.pdf


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Electrophilic aromatic substitution reaction of dihydrofuran-2,5-dione with anisole

I am having trouble understanding the reaction of anisole with dihydrofuran-2,5-dione (succinic anhydride) in presence of $\ce{AlCl3}$ and acid works or how an electrophile is formed. I feel like the product should be something like $\ce{C11H12O4}$.

This looks like a Friedel-Crafts acylation to me. Maybe it all becomes clear after visiting the Wikipedia page.

Nice! You've brought everything you need.
If you can't identify your electrophile immediately, do it the other way around: Assign function to all components. The one that is left must be your electrophile - or its precursor.
Do you remember the role of $\ce{AlCl3}$ in electrophilic aromatic substitutions? Where could it possible attack here?

Here is another hint: What is the trivial name for dihydrofuran-2,5-dione?


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Standard solution and extracted solution : same concentration?
Summary :
I am currently testing how a certain type of foam adsorbs HAP, PCB and OCP, and to this end I am depositing some on the foam, then extracting it. To know how much was extracted, I need to compare with a standard solution of HAP, PCB and OCP. The problem is that when I look at the chromatograms, the extracted solution's peaks' areas are about 10 times the standard solution's peaks' areas. I don't know where in the experiment I am doing something wrong.
Details :
I have a solution of HAP, PCB and OCP at 10 mg/L. I put 100 µL of this solution on the foam, so that the deposited mass is 1 µg.
The foam is ASE-extracted with acetonitrile, then the acetonitrile is rotary-evaporated. What is left in the round-bottom flask is then diluted in 1 mL acetonitrile. So at maximum I have 1 µg pollutants in 1 mL acetonitrile.
To this 1 mL I then add distilled water until I have a 20 mL solution. The concentration of this solution should at maximum be 50 µg/L, right ? (If all pollutants are extracted from the foam and none lost to evaporation.)
The standard solution is made by also taking 100 µL from the 10 mg/L solution (so 1 µg pollutants), and adding distilled water until I have a 20 mL solution. So this solution's concentration should also be 50 µg/L, right ?
But when I put the two solutions through GC-MSMS, the peaks' areas of the extracted solution are about ten times greater than those of the standard solution. (Not always ten, but it is usually the case.)
Have I made a mistake somewhere ? Are my calculations/manipulations wrong somewhere ?
The internal standard used is naphtalene d8, and in each 20 mL solution I added 10 µL of a 10 mg/L naphtalene d8 solution. (So its concentration is 5 µg/L : is it a problem that the standard is not at the same concentration as the pollutants ?)
The experiment was repeated several times, with different foam extracts and different standard solutions, and the results were the same. It is not a problem of contamination, as everything has been extensively washed with distilled water, detergent and acetone.
In case it is relevant : when a solution is done, I sample 4 mL that I extract with a SPME fiber, and then I inject into the GC-MSMS.
Thank you in advance, I am still testing and calculating, so any help is appreciated.

To tschoppi (since I don't have enough reputation to comment) : I haven't tried with another kind of foam, but it's a good idea (to see at which step the problem lies). I'll see what we have in the lab. I am also pretty sure the foam I used isn't contaminated, because it was given to us straight from the package, and I washed it three times with different solvents and analysed the wash-solvent (which was clean)(but I may check again to be sure).

Have you tested a foam where the adsorption (and desorption) properties are known, and do you still get the same result? Is the foam free from analyte contamination?

What is the ASE procedure you are using? The concentration of the solutes are 10 mg/L  HAP, 10/L mg  PCB 10 mg/L o OCP?

I found the answer to my question, and I am going to share it just in case.
Apparently, when extracting the pollutants from the foam, I am also extracting some foam-specific materials that influence the pollutants' behaviour during the SPME part of the extraction.
I tested that by extracting the blank foam, then adding a standard amount of pollutants to the extraction solvent. This gave the same result as extracting a spiked foam, meaning higher peak areas than a standard solution.
In short : matrix effect from the foam. I don't yet know why or what, but it's a matrix effect.


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How is Langmuir adsorption isotherm different from Freundlich's?
I see no difference between the two. Both say that the degree of adsorption reaches a limiting value at high pressures and at very low pressures degree of adsorption is directly proportional to the pressure.
Where does the difference lie?

There is a fundamental difference between both models.
Langmuir's model was a theoretical construct, while the Freundlich isotherm is empirical.
In the Langmuir model, it is assumed that at maximum coverage, there is only a monomolecular layer on the surface. This means that there is no stacking of adsorbed molecules. The Freundlich isotherm does not have this restriction.

I do know it.But my question was 'if both the isotherms gave same results,how was Langmuir's betteroff than Freunlich's

Freundlich gives you the amount absorbed per unit mass of absorbate, and has this absorbtion fraction be the nth root of the pressure, with n being a material-specific constnat. Langmiur gives the surface coverage, so it only allows the surface of a material to be used for sorption. Also, the functional relationship is not a root, but an inverse of the pressure multiplied by the pressure.
Qualitatively, in the limits, both seem similar, but they differ in their details and will give different answers.

The difference is that cation exchange reactions involving clay minerals are analogous to the Langmuir isotherm because the concentration of cations adsorbed can never be greater than the cation exchange capacity of the solid.


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Reaction between zinc and sulfur
Would the reaction between zinc and sulfur be
$$\ce{Zn_{(s)} + S_{(s)} -> ZnS_{(s)}}$$
or
$$\ce{8 Zn_{(s)} + S_8\ _{(s)} -> 8ZnS_{(s)}}$$
I know that $\ce{S}$ and $\ce{S8}$ are allotropes of sulfur, but which one should I use for a reaction with sulfur at room temperature?

Either reaction is valid for representing the reaction between zinc and sulfur. The important thing is that both equations stipulate the same stoichiometry (one atom of zinc for one atom of sulfur). The mathematics of chemical equations work just the same whether you add structural information about the reactants or not; the reaction equation is just shorthand for a system of mathematical equations involving the amounts and types of atoms.
In truth, neither equation is a correct description of the reaction at the microscopic scale (though I again restate that's not really the point of a reaction equation). Free zinc atoms do not interact with free sulfur atoms or $\ce{S8}$ rings. Rather, after a little bit of initial heating to get the reaction going *, liquid sulfur (a very complex mixture of sulfur chains ranging from $\ce{S6}$ to $\ce{S_{n}}$, where n is as large as 100,000) is in contact with a surface of zinc (or possibly the molten metal), and electrons are exchanged by some mechanism which probably hasn't even been elucidated, as electron transfer reactions are very quick and hard to monitor.
* You mention considering the reaction at room temperature. Even though thermodynamically, the reaction between zinc and sulfur is spontaneous at room temperature, it is kinetically very slow because both reactants are solid. Even if you mix them very very well, the reaction is diffusion-limited in the solid mixture; the atoms can't "reach" each other very well. Heating a little bit creates a small amount of liquid sulfur, which can then mix better with the zinc.

So the reaction would occur like this: 8 Zn(s) + S_8(s) + Heat (activation energy?) --> 8 ZnS(s) + 205 kJ

@bandicoot12 Usually any added heat is represented with a capital delta above the reaction arrow, like so: $\ce{Zn_{(s)} + S_{(s)} \overset{\Delta}{\rightarrow} ZnS_{(s)}}$

Crystalline sulfur, at room temperature, exists only as $\ce{S8}$.
So basically, that's the only option you have.
Breaking down $\ce{S8}$ chains is doable but over 160°C, followed by polymerization.
At 200°C, polymer chains start to break.
Back to your question, $\ce{S8}$ is the most stable form for sulfur at room temperature.
From this paper.


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Getting to use the lab!
Today I finally mustered the courage to go and talk to the director of the labs to politely ask him if I could use the labs out of class hours. Long story short, our lab sessions "are bad" because we are given a protocol, and like babies we just need to follow it. No brainer, and I am against that. So I got permission to use the lab for experiments that don't pose a mayor risk, as well as a slight restriction of the reagents.
I'm not sure of which reagents we have in the lab, but there aren't that many there. So, now to the point. I was hoping that you could tell me which experiments I should try. I came up with the idea when we worked with ferric ammonia ferrocyanide $[Fe(CN)_6][Fe(NH_4)_2]$ and I had to balance the chemical equation of the reaction that we used. However, I had no idea of what the byproducts where, because I had no way to test for their presence! My idea is to bring an hypothesis and then, starting pretty much from scratch (except reference tables or such) prove or disprove it.
Could you share a nice example of something I could test in the lab? If not, where can I find stuff that I can work with, without getting the answer right out of where I got the info. Did I get to express my idea correctly? More or less?
Thanks for the help!
P.s.: I kindly asked a professor to supervise my work, so I would get help regarding methodology.

like babies we just need to follow it. No brainer, and I am against that - Sounds like you should get yourself involved with a research group.  Many groups may have spots open for temp. work or summer positions.  If you are an undergraduate, look into summer REU programs.

Please keep in mind, that the labs should make you comfortable with common experimental procedures. For this there are protocols, which are known to work that well, that even untrained student is able to achieve their goal. It is difficult to think of what to do, what method to use, which glassware to use, how to heat, how to put another chemical in, if it is moisture sensitive, etc... You should acquire such skills step by step, so please follow the no brainer for a while, to let your hands and eyes learn the tools.

http://books.google.com/books/about/Crystal_Growth_in_Gels.html?id=pQ-JuNXEL6wC
  The classic text.  Set them up, come back in a few weeks.  (Potentially any gel - polyacrylamide, agarose, agar, gelatin...)  Meanwhile,
The Prussian Blue prep works across the transition metals, with different colors.  Alas, it is not crystalline (but maybe in gel...).  Vanadium oxidation states are a spectrum of colors.  Vanadyl, V=O, is especially fine (but not as a food supplement).
Cobalt can be preparatively more fun than iron.  Co(II) is substitution labile - put anything on it you want, in real time.  Co(III) is substitution inert.  Anything you put on Co(II) then oxidize to Co(III) stays forever (absent a trace of reducing agent to reopen the door).   There is vast and colorful Co coordination chemistry.
http://onlinelibrary.wiley.com/bookseries/10.1002/SERIES2146
  Everything in there works to spec.  Fine some interesting things to do (and scale down to grams).
Cobalt sepulcrate and especially sarcophogate (Alan Sargeson)!   Tris(ethylenediamine)cobalt(III) is a propeller, left- or right handed. It can be resolved afterward (Inorg. Syn. 6 186 (1960)) or made entirely as one hand.  Net charge is 3+.  Tris(oxalato)cobaltate(III) is a propeller, left- or right handed. It can be resolved afterward (Inorg. Syn. 8 204 (1966)) or made entirely as one hand.  Net charge is 3-.  Tris(ethylenediamine)cobalt(III) tris(oxalato)cobaltate(III) as a salt with same or opposite hands could be interesting.  The crystal will certainly be tightly bound.

Walking through the oxidation states of vanadium is a great idea!

You can try this manual. The manual is used at ICT Prague as a guide for students during inorganic laboratory exercise. The description of experiments is described there with The description of lab. equipment needed for the experiments.

Can you recommend any experiments directly related to the OP's area of interest?


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Effect of periodic acid on trans diols of 3, 4 and 5 membered rings
Does periodic acid cleave the following diols?

trans-cyclopropane-1,2-diol
trans-cyclobutane-1,2-diol
trans-cyclopentane-1,2-diol

I seriously doubt that any of the trans diols is cleaved.
The reaction is assumed to proceed through a cyclic intermediate, which is unlikely to be formed in any of the cases.

Are you 100 % sure that trans diols aren't cleaved. According to the book by Kurti and Czako (http://www.amazon.com/Strategic-Applications-Reactions-Organic-Synthesis/dp/0124297854/ref=sr_1_2?ie=UTF8&qid=1392294686&sr=8-2&keywords=czako%2C+kurti) a similar reaction, the Criegee Oxidation, in which vicinal diols are cleaved by $\ce{PbOAc4}$, works with both cis and trans diols (although the reaction with trans diols is slower).

@Philipp Very interesting! But was it shown for 5-membered rings too? The successful cases for trans diols I know are pyranoses, whereas I've not seen it for furanoses. But I'm not too old to learn :)

I haven't dived in the referenced papers and couldn't find anything for the ring diols. But in the book by Smith and March (http://www.amazon.com/Marchs-Advanced-Organic-Chemistry-Mechanisms/dp/0471720917/ref=sr_1_3?ie=UTF8&qid=1392295682&sr=8-3&keywords=smith+march+organic) it is also claimed that cis and trans react, while the reaction with cis is faster (see pp. 1734-1735). There you can even find that (4ar,8ar)-decahydronaphthalene-4a,8a-diol (sorry, I don't know the trivial name) is cleaved by $\ce{PbOAc4}$ (though other glycols that cannot form cyclic esters are not cleaved).

Apperently, there is the possibility of a mechanism involving a cyclic transition state (without the cyclic intermediate), that allows for trans compounds to be cleaved (see p. 1735 or Criegee, R.; Höger, E.; Huber, G.; Kruck, P.; Marktscheffel, F.; Schellenberger, H.
Liebigs Ann. Chem. 1956, 599, 81.).

No problem, the IUPAC name is fine, even on a train ;-)

"IUPAC names on a train" sounds a bit like a horror movie to me :)


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Why does acetylene, having an acidic proton, not react with strong bases like NaOH and KOH?
In acetylene (ethyne, HC≡CH), the carbons are sp-hybridised, which make the attached protons more acidic than those in alkenes or alkanes. (See also: Relative acidities of alkanes, alkenes, and alkynes)
Why doesn't acetylene react with strong bases like NaOH or KOH?

The $\mathrm{p}K_\mathrm{a}$ of acetylene is $25$. It is very weakly acidic, and hence, $\ce{KOH}$ is not sufficiently alkaline to react with it. Even methanol, with a $\mathrm{p}K_\mathrm{a}$ of $15.5$ does not react much with hydroxide ion. On other hand, $\ce{NaH}$ or $\ce{NaNH2}$ have more basic anions than hydroxide, and weak acids such as acetylene or alcohols can react with them.
Predicting the relative basicities can be done qualitatively by invoking the Brønsted conjugate acid-base theory. The rule to remember is, a base gives its corresponding conjugate acid in solution, and stronger the conjugate acid, weaker its conjugate base (and vice versa). This can be understood from the following
$$\ce{HA<=>H^+ + A-}$$
Here, $\ce{A-}$ is the conjugate base of the acid $\ce{HA}$. Now, if $\ce{A-}$ is highly unstable, it will have a stronger tendency to react with a proton and get converted to its conjugate acid which is relatively more stable. Since the equilibrium lies more to the left in that case, we say that the conjugate base is strong (and consequently the conjugate acid is weak). Now consider the following anions for the sake of example.
$$\ce{HC#C-} > \ce{NH2-} > \ce{OH-} > \ce{F-}$$
These are listed in decreasing order of stability, and hence also their strength as bases. This is because as the electronegativity increases from carbon to fluorine, the negative charge becomes increasingly stable on the atom. More electronegative elements are increasingly stable with the extra electron due to higher effective nuclear force of attraction. Correspondingly, by the dictum "stronger base is weaker conjugate acid", the order of the acidity is:
$$\ce{HC#CH} < \ce{NH3} < \ce{H2O} < \ce{HF}$$


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Figuring out the concentration of weak polyprotic acids in presence of strong acid
We have a $\pu{0.1 M}$ solution of $\ce{H2S}$ with a $\pu{0.1 M}$ solution of $\ce{HCl}$. We need to figure out the concentration of $\ce{S^2-}$ ions at equilibrium.
$$\ce{H2S <=> HS- + H+}$$
$$\ce{HS- <=> S^2- + H+}$$
Given $K_{\mathrm a_1} = 9.1 \times 10^{-8}, K_{\mathrm a_2} = 1.2\times 10^{-13}$.
I have no idea how to do this problem. I can't relate the $\ce{S^2-}$ concentration to the first equation. I first tried determining the $\rm{HS^-}$ concentration from the first equation, then putting that into the second one to obtain the required concentration, but then I realized that as the second reaction approaches equilibrium, the concentrations of the $\ce{HS-}$ ions will change, which will then throw the first reaction out of whack.
I tried taking the required concentration as $\pu{y M}$ and then tried to establish a relationship, but the only thing I could get was that the concentration of the hydrogen ions will go up by $y$, which is of no use at all.

You should probably be able to assume that the second equilibrium doesn't significantly shift the first. Any $\ce{HS^{-}}$ consumed to make $\ce{S^{2-}}$ will be almost completely regenerated to the previous amount, since there is a very large excess of $\ce{H2S}$.

The first step in solving this kind of problems is to write the reactions that take place in your solution, and the equations of the acidity constants :
$$\ce{H2S + H2O <=> HS- + H3O+}$$
$$K_{\mathrm a_1}=\frac{[\ce{HS-}][\ce{H3O+}]}{[\ce{H2S}]}$$
$$\ce{HS- + H2O <=> S^2- + H3O+}$$
$$K_{\mathrm a_2}=\frac{[\ce{S^2-}][\ce{H3O+}]}{[\ce{HS-}]}$$
Since $\ce{HCl}$ is a strong acid ($\mathrm pK_\mathrm a < 0$), the dissociation equilibrium for it will be totally shifted to right:
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
And the dissociation of water, occuring in any solution:
$$\ce{H2O + H2O <=> H3O+ + HO-}$$
$$K_\mathrm w = [\ce{H3O+}][\ce{HO-}]$$
*The concentration of water is not taken into account because it remains approximately constant as it is much bigger than the concentration of the $\ce{H2S}$ dissolved. It is also right to write the equations as you did above, considering the dissociation of the acid, but it's better to use the hydronium ion, $\ce{H3O+}$, as we know $\ce{H+}$ is quite unstable on its own. (In some books $[\ce{H+}]$ is used to ease writing)
As you saw, you cannot calculate the concentrations at equilibrium (written in square brackets, i.e. $[\ce{HS-}]$) considering only the first reaction (unless you consider the shift of the second reaction doesn't affect significantly the first, as Nicolau said, and so we use the analytical concentration of a compound (in this case $\ce{H2S}$ and $\ce{HCl}$) which means the specific concentration of the solute when the solution is prepared ($\pu{0.1 M}$ for both acids, if these are the concentrations in the mixture of acids solution and not the initial concentration of each acid solution). It is usually denoted $c_{\ce{H2S}}$ and $c_{\ce{HCl}}$ respectively.
Looking at the above reactions, we can write the formula for the analytical concentrations :
$$c_{\ce{H2S}} = [\ce{H2S}] + [\ce{HS-}] + [\ce{S^2-}]$$
and
$$c_{\ce{HCl}} = [\ce{Cl-}] $$
As you might notice no $\ce{H3O+}$ concentration appeared in any of these formulas, because it depends on both $\ce{HCl}$ and $\ce{H2S}$ concentrations.
However, this is not enough to solve our problem, and so the electroneutrality principle is used, stating that the concentration of the negative ions must be equal to the concentration of the positive ones, and in our case may be written as:
$$[\ce{HS-}] + 2[\ce{S^2-}] + [\ce{Cl-}] + [\ce{HO-}] = [\ce{H3O+}]$$
*The concentration of $\ce{S^2-}$ has the coefficient $2$ because of the two negative charges, unlike the other ions, which have only one negative or positive charge and don't need any.
Using the equations of the analytical concentrations in the electroneutrality formula, we get this :
$$c_{\ce{H2S}} - [\ce{H2S}] + [\ce{S^2-}] + c_{\ce{HCl}} + [\ce{HO-}] = [\ce{H3O+}]$$
And, using also the acidity constants and the water constant :
$$c_{\ce{H2S}}-\frac{[\ce{H3O+}][\ce{HS-}]}{K_{\mathrm a_1}}+\frac{K_{\mathrm a_2}[\ce{HS-}]}{[\ce{H3O+}]} + c_{\ce{HCl}} + \frac{K_\mathrm w}{[\ce{H3O+}]} = [\ce{H3O+}]$$
The equation has 2 unknown concentrations ($[\ce{H3O+}]$ and $[\ce{HS-}]$) and you may approximate the concentration of $\ce{H3O+}$ to $c_{\ce{HCl}}$ since the acidity constants for $\ce{H2S}$ are very small:
$$[\ce{H3O+}]=c_{\ce{HCl}}$$
So, the concentration of $\ce{HS-}$ can be calculated:
$$[\ce{HS-}]=\frac{c_{\ce{H2S}} + \frac{K_\mathrm w}{c_{\ce{HCl}}}}{\frac{c_{\ce{HCl}}}{K_{\mathrm a_1}} + \frac{K_{\mathrm a_2}}{c_{\ce{HCl}}}}$$
However, since $\frac{K_\mathrm w}{c_{\ce{HCl}}} << c_{\ce{H2S}}$ and $\frac{K_{\mathrm a_2}}{c_{\ce{HCl}}} << \frac{c_{\ce{HCl}}}{K_{\mathrm a_1}}$, the above equation can be approximated to:
$$[\ce{HS-}] = \frac{c_{\ce{H2S}}K_{\mathrm a_1}}{c_{\ce{HCl}}}$$
The approximations lead us to a simple formula, which could be denoted directly from the first equilibrium, but without understanding exactly why. From this the concentration of $\ce{S^{2-}}$ can be calculated easily, just like you did first. Even though some would just ignore the second equilibrium for $\ce{H2S}$, that is NOT the way to solve analytical chemistry problems. First you consider all of the equilibriums and processes occuring in a solution and then (if possible) approximate, and not the reverse.


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Can we goldify metals?
I've read in some book that one of the old chemists was able to transform cheap metals into gold.
I know that gold is an element (Au) so we can't transform a metal into Au. But as I read further it says that the chemist doesn't actually transform metals into gold but into something very similar to gold (similar physical and chemical characteristics )
such that you can't differentiate between it and between gold.
Is this chemically possible?

A particle accelerator can be used to turn lead into gold but it is not cost effective.

No it is about transforming cheap metals like iron or Thinning

This is a book for silly readers. You should have heard from "goldmakers" is synonymous for snake oil vendors.

In pre-colombine America they would make things with an alloy with a little bit of gold. On treating it with an acid paste only the gold was left on the surface so it looked like solid gold.

Possible duplicate of Turn lead into gold via radioactive decay

@Mithoron This can't be a duplicate, it was on the site 6 months before the other question.

@A.K. Age is practically non-issue for two old posts when considered which one should be the "duplicate". Better Q&A should stay opened.

@Mithoron , perhaps yes, but this is also a different question. The other one is explicitly nuclear while this one is implicitly non-nuclear. Is this chemically possible?

@A.K. Ah, perhaps I didn't read it carefully, it's not particularly clear. Perhaps too-broad or unclear would be better close reason.

In the days before chemistry, alchemists searched for the Philosopher's stone which would enable them to transform base metals into gold. They spent a lot of effort searching for it but we know now that their search was futile as no chemical process can cause elemental transformation. Only nuclear processes can do that but they cost far, far more than mining the element ever will so this doesn't really count as a solution.
But plenty of people faked a result or were conned by their lack of understanding. Fool's gold, for example looks a lot like gold, but doesn't share many other properties but, presumably, the name arises because some people are gullible. There are also many metal alloys that look like gold but are not pure (including many alloys that contain some, but not much, gold). It is also possible to use a variety of chemical or electrochemical processes to coat the surface of many metals with a thin film of gold.
But careful analysis can reveal most of these methods of fakery (including the alloying of gold with other metals which was very tempting when coins were made of gold). One of the most famous stories of Greek science is based on one of the most reliable methods. Archimedes was faced with the task of distinguishing debased coinage or other gold objects and, in the legend, realised the answer while having a bath, leapt out the bath and ran down the street naked yelling "eureka". What he realised was that an object immersed in water displaces a volume of water equal to its volume and this can be measured by weighing the displaced fluid. Combined with the weight of the object this can be used to estimate the density of the object. Since the only other metal with a density close to gold is tungsten (gold is 19.28 g/cm3 and tungsten is 19.25 g/cm3) knowing the density is a good way to test the purity of the metal. Common metals used in alloys, like silver (density 10.5 g/cm3) will substantially lower the density.
So, in short, you can make a lot of things that look like gold, but it is very hard to make something that shares enough properties to convince a skeptical observer who will test more than the appearance.

No alchemist can perform transmutation of lead into gold with alembic! But now day we can perform nuclear transmutation  and gold can be now synthesized  by irradiation of mercury or platinum in a nuclear reactor or with an accelerator. However this is not cost effective!
In fact there are many alloys that looks like gold and an alchemist could make easily most of them. Brass is one metal that can be similar to gold  varying the copper zinc ratio you can make Aich's alloy and Prince's metal two types of brass with an aspect very similar to gold. In ancient rome aurichalcum was a metal (brass type alloy probably) with an aspect similar to gold. The problem is that usually you don't see if something is made of gold you feel it! If  you try to cheat someone who deal with gold with these metals he suddenly tell you that they are to light to be gold! Gold is in fact heavier than lead! This is why density esteem works great and simply coating the surface with gold doesn't work. Ethiopian  fake gold fail for that!
So to make a good gold sophistication you have to find an isobar of gold, an element with the same weight and density of gold. Tungsten is a perfect candidate, and sometimes you can read about founding of gold plated tungsten bars or other theory about FortKnox. Here you can read a good article about gold sophistication. In this case a layer of gold over the tungsten is needed to make it looks like gold and to pass the exams with superficial elemental analysis.
For a good looking gold surface I prefer iron(III) oxide mica pigments very nice color without cheating nobody!


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why CO2 is not organic
this might be very childish question but i am comfused because they say that organic molecules are those that contain H and C if that is so how can you call carbontetrachloride organic, even it hasn't any H atoms

Don't listen to them. I would say $CO_2$, $CH_4$, $CCl_4$ are all organic. This type of classification has no real use anyway.

Related: Is carbon dioxide organic or inorganic?

Carbon dioxide
"is" neither organic nor inorganic, it's just carbon dioxide.
This distinction is nothing but a convention to separate things
between the textbooks and professors.
Carbon dioxide forms carbonic acid in water and the usual salts.
This was/is reason to think of it as inorganic.
On the other hand, urea, the diamide of carbonic acid was thought
to be organic when Wöhler synthesized urea for the first time
in lab, today it is categorized as inorganic.

An organic compound
1) must be a (stoichiometric) compound.  Graphite, diamond, buckyballs, polycarbyne, etc. are not compounds.
2) must contain reduced carbon, at least one C-C or C-H bond.  Carbon tet is inorganic, chloroform is organic.  Urea, guanidine, cyanuric acid, isocyanic acid, melamine are inorganic.
Carbon monoxide, carbon dioxide, phosgene, carbonic acid are inorganic.  Ethylene carbonate, vinylene carbonate, oxalic acid, formic acid, dichloroacetylene are organic.  Aluminum carbide, $\ce{Al4C3}$, is inorganic (hydrolyzes to methane).  Calcium carbide, $\ce{CaC2}$ is organic (hydrolyzes to acetylene).

Buckyballs aren't compounds? Are you sure about that?

The C-H bond has a lower bond energy than
the carbon-oxygen bond in carbon dioxide,
making carbon dioxide more stable/less
reactive than the typical organic compound.
So, when you're determining whether a carbon
compound is organic or not, look to see
whether it contains hydrogen in addition to
carbon and whether the carbon is bonded to
the hydrogen.

According to this definition, hexachlorobenzene and tetrafluorethen are not organic.


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How to make something solid that dissolves faster than bar soap in water
What is a soap-like material that holds a solid shape, but dissolves completely in water over 10-15 minutes. Think disappearing Rubber Ducky.

Potassium soap is soft.  Vary the sodium-potassium ratio to get what you want.  Leave some glycerin in.  Make the soap from polyunsaturated fats.


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What are the main differences of CTAB and DTAB surfactants? And what are practical applications of them?
Just doing a little bit of research of both surfactants but I can't seem to find too much online about them. Hopefully you guys can help out!

CTAB, aka Hexadecyltrimethylammonium bromide
DTAB, aka Dodecyltrimethylammonium bromide
Sort of a vague question, don't know where to start. To begin, CTAB has a hydrocarbon chain length of 16 (as its name suggests) whereas DTAB has a -CH2- chain length of 12.
Otherwise they have the same hydrophilic/polar "head"; all this is obvious, you can tell just by the two compounds' names. They also have the aforementioned hydrophobic "tails", which refers to the hydrocarbon chains because they are non-polar.
CTAB and DTAB can be used to create micelles. It's like how a biological cell is separated from its environment with a phospholipid bilayer where the hydrophobic and hydrophilic components of the phospholipids align such that you get an enclosed structure that separates a non-polar "organic" region in a polar environment (e.g. water). The wikipedia link has good diagrams.
The reverse is called a reverse micelle, where the heads create a polar region separated from a non-polar environment (e.g. when the solvent is hexane).
For one example of practical usage, I did a lab just today using hexane as a solvent where I used CTAB to create an environment to limit the formation of CdS quantum dots. This worked by limiting the volumes where conditions aren't non-polar. In this case, reverse micelles were formed; tails pointing out, head surrounding to form a more polar center.
Another: In DNA extraction CTAB is used to pick up insoluble particles (e.g. polysaccharides) and exit the working solution.  You can relate surfactants to soap, which picks up insoluble (i.e. oily) substances and can easily be removed by water due to its hydrophilic outer layer, while containing the oily substances in its hydrophobic center.

I made the links in your post prettier. You can do this for yourself using the link tool (looks like a chain-link) or by using the following notation [text to become a link](url). For example [a link to Wikipedia](http://en.wikipedia.org) becomes a link to Wikipedia.

As rch has already described, CTAB and DTAB are cationic surfactants that form microheterogenous environments in solution. In aqueous solution, these micelles consist of an unpolar core and a polar boundary that interacts with the bulk (water).
The main difference between CTAB and DTAB is the surfactant concentration, at which these aggregates start to form. The CMC (critical micellar concentration) , determined by ESR spectroscopy and by fluorescence spectroscopy are in the range of 1 mM for CTAB and 14 mM for DTAB, respectively.
Note that these are just the lower limits for the formation of mostly spherical aggregates. At higher surfactant concentrations, larger rod-shaped micelles are typically observed.


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Limiting Reactants without a given mass
This is a question I have for Chemistry homework, but I don't understand how to solve it without a given mass. I narrowed the choices down (see below _) I know a limiting reactant well limits the reaction (so if there are 2 reactants) and the react left is the excess reactant.
I thought Propane would be the limiting reactant because the oxygen is in excess (though that was incorrect).
Then, I though Oxygen would be the limiting reactant (naturally) since its a combustion reaction (though that is incorrect).
Determine whether each reaction depends on a limiting reactant. Explain why or why not and identify the limiting reactant. (Select all that apply.)
(c) Propane (C3H8) burns in excess oxygen to produce carbon dioxide and water.
Yes, one reactant is naturally limiting.
Yes, there are two reactants.
No, all reactants are in excess.
No, there is only one reactant.
propane
oxygen
carbon dioxide
water
Not enough information to identify which is the limiting reactant.

Yes, one reactant is naturally limiting.
Yes, there are two reactants.
No, all reactants are in excess.
propane
oxygen
Not enough information to identify which is the limiting reactant.

It seems that you would need information about how much of each reactant you have when conducting this. edit: "Excess" is the key word here; so it hints that Oxygen is NOT the limiting reagent. Considering how combustion reactions are structured, the answer seems to be the other reactant here.

This was actually a 3-part question, but I figured out the first 2 parts. (a) Potassium chlorate decomposes to form potassium chloride and oxygen. (No, there is only one reactant.) (b) Silver nitrate and hydrochloric acid react to produce silver chloride and nitric acid. (Yes, there are two reactant, but not enough information to tell). I wasn't given any masses for either a or b, yet it was still possible to determine the answer. Using this, can you determine an answer for c, or should I saw there isn't enough information?

Because it says excess, can I assume that oxygen isn't the limiting reactant? And what does the term "naturally limiting mean?

Oh snap. I totally missed "excess". Whenever you see that it basically means there is an unlimited amount of that reactant. My apologies!
As for "naturally limiting reactant", I'd take it to just be a syntax modification to deal with the fact that you do in fact have excess Oxygen, I've never heard of a reactant being "naturally limiting" before.

Reading the question multiple times again, for "naturally limiting reactant", it is probably a corresponding answer to selecting one of the chemicals also listed. i.e. Yes, one reactant is naturally limiting (because there is excess Oxygen), and then you specify that the reactant would be propane.

My homework is online and allows for 3 tries before I miss it completely. I used my first try on putting Propane would be the limiting reactant. I used my second my saying Propane because it was naturally limiting (I didn't understand what naturally limiting meant though)


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What is the reason a grain of salt alters the perceived bitterness of coffee by such a high degree?
Coffee served hot can be quite bitter (presumably high pH).
Several people I have known will add a single grain of salt to the coffee - which reduces the perceived bitterness.
My question is: What is the reason a grain of salt alters the perceived bitterness of coffee by such a high degree? Can you explain the pH change that is occurring during this reaction?

The addition of even a moderately large amount of salt to pure water does not significantly make the solution any more alkaline or acidic. Coffee or tea aren't as simple as pure water, but I imagine the effect is about the same. I don't think bitterness has to do with the pH at all; the bitterness in coffee and tea come from dissolved alkaloids, of which caffeine is one. Alkaloids often produce basic solutions, but you can certainly make basic solutions without them. Whether such solutions are bitter, I don't know, as I haven't really felt the urge to taste $0.1\ M\ \ce{NaOH}$!

That's exactly my point - why salt and not baking soda? What is the reaction that reduces the bitterness?

Is simply a dissolution no reaction occur and as said Niculau the pH change, if occur, is absolutely negligible.

Salt in general affect taste lowering the bitterness of the food enhancing the positive sensory attributes of foods. In fact the reason of this behavior is the sodium ion $Na^+$ (even litium ion works) and his interaction with the receptors in taste buds. So sodium acetate therethically works as well, Gery Beuchamp has studied many of this salt, quoting from NCBI:

various sodium-containing ingredients have been known to reduce the
  bitterness of certain compounds found in foods, including quinine
  hydrochloride, caffeine, magnesium sulfate, and potassium chloride

This is an article about the suppression of bitterness by sodium. From here I'm afraid is more a question of physiology.
I read this the first time here:
This, Herve. Molecular gastronomy: exploring the science of flavor. Columbia University Press, 2006 p. 94


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Why does confinement of quantum dots result in discrete energy levels?
I'm doing a project with lead (II) sulfide quantum dots because they have optical transitions in the near infrared. I understand the simple "particle in a box" approach where confinement strengthens exponentially when the box shrinks.
What I don't get is how this confinement results in discrete energy levels? Can someone tell me how this shrinking of molecular size and confinement of the electron hole pair (exciton Bohr radius) results in discrete energy levels? Is it the electron shells being changed?

As a rule of dumb, you can have a continuous spectrum if electrons are delocalized. This happens (i) for extended systems (like solids) or (ii) for unbound electrons. In both cases the electrons move in an infinite space, and due to position momentum commutator rules and the fact that momentum is related to energy, one can have a continuum spectrum.
Another way of looking into it is that in a confined space you can use a discrete basis set to describe your wave-functions. Since the eigenstate of the hamiltonian are a possible basis set, then the eigen-states are discrete.
Quantum dots are an example of a way to localize electrons and thus to get discrete energy levels for the bound part of the spectrum. Same happens in a box (where all electrons are bound by a potential with an infinite barrier).
See also the discussion in this post.


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Determining order of a reaction without isolation
I'm writing a computer program to calculate the order of a chemical reaction with respect to each reactant. When I do this by hand, it's easy to isolate which reactant concentration stays relatively the same while the others change. However, when you don't have an isolation method available, is there a non-guess and check way to determine rate orders?
For example, with the sample data of a 1st order reaction with respect to A and B, and second order overall
+--------+-----+-----+------+
| Trial# | [A] | [B] | Rate |
+--------+-----+-----+------+
|      1 |   1 |   1 |   10 |
|      2 |   2 |   3 |   60 |
|      3 |   3 |   2 |   60 |
+--------+-----+-----+------+

There isn't a trial with two constant values, so I'm not sure how you would programmatically determine the orders with respect to each of the reactants without manually trying every possibility until one fit (I realize that this wouldn't take too much computing time to do, but it won't scale well for multiple trials, and I also would prefer a non-brute force solution if possible). If there is no way to determine the rate with respect to each reactant, is there a way to determine the overall rate? This would greatly reduce the number of possibilities of orders when faced with larger rates

I think this is more of a programming/maths problem than specific to chemistry... Have you tried asking on Math.SE?

Although this is indeed a bit more a math problem than chemistry, I think it is useful to answer it, because this is a commonly encountered problem in reaction kinetics.
First of all, Yes it is possible to get the rates with respect to each reactant separately, as long as you have sufficient experimental data. With sufficient I mean: $1+\text{number of reactants}$. This is because you don't only need the order for each reactant, but also the rate constant for the total reaction
The rate equation
In general a rate equation will have the following structure:
$$ r = k [A]^\alpha [B]^\beta[C]^\gamma...\text{etc}$$
where with etc I mean that you can continue for as many reactants that you have. In your example you have 2 reactants so your rate equation in general will look like this:
$$ r = k [A]^\alpha [B]^\beta $$
To find the reaction order with respect to each reactant we need experimental data on $r$ for given $[A]%$ and $[B]$ like you provided.
The fitting procedure
You will have to write a short program that takes an initial guess for $k$, $\alpha$ and $\beta$ and checks how good or bad that guess is by comparing the calculated rates $R$ with the experimental rates $r$ for each set of $[A]$ and $[B]$. The normal way to make this comparison is based on least-squares:
$$S = \sum_{i=1}^N (r_i-R_i)^2 $$
So you sum up the squared differences between the reaction rate you found experimentally and the calculated rates for all $N$ experimental datasets.
To get the reaction orders and rate constant you want $S$ to be minimal, preferably 0 (although that normally doesn't happen with experimental data due to small measurement errors). Normally $S$ is not immediately optimal on your first guess for $k$, $\alpha$ and $\beta$, so you have to 'guess again'. You could indeed really guess again, but there are much smarter ways of determining the new guess, which will make sure your new guess is a better one than the old one. A common technique for this is called Newton's method. With the new guess you again check the value of $S$. You then repeat the cycle until you cannot find a lower value for $S$, which means that you have found the values of $k$, $\alpha$ and $\beta$ that are optimal for this experimental dataset and thus should be the correct reaction orders for the reaction.


text:
Radial expectation value for particle in a cubic box

A hydrogen atom is in a cubic box with side lengths equal to $\require{mediawiki-texvc}\pu{100 \AA{}}$. For what value of $n$ (principal quantum number) will the expectation value of the radius be equal to one-half the box size?

Welcome to Chemistry.SE. Please have a look at our Homework policy.

\[
\left< r_{nl} \right> = n^2\cdot a_0\ \left\{ 1+ \frac{1}{2} \left[1-\frac{l(l+1)}{n^2}\right] \right\}
\]
should be close to what is applicable to solve to problem in a simple way. If the value is a bit off, blame it on Niels Bohr or the cabinetmaker who built the box.


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How alkanes can be prepared by aldehydes and ketones?
Preparation of alkanes by aldehydes and ketones, its chemical reaction and mechanism.

Please add additional details into the question.  What information have you found thus far? (hint: take a look at your textbook)

I need the chemical reaction

Please see our homework policy.

Look here

You want to deoxygenate the carbonyls under conditions where you do not get condensation reactions.  Do you want to pull the oxygen or dimerize to the olefins then hydrogenate?
Wolff-Kishner or Clemmensen reductions, thioacetal then Raney nickel and hydrogen.  McMurry reaction for dimerization.  Given polymethylhydrosiloxane,
Synlett 276 (2009)
Org. Lett. 13, 584 (2011)


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What are the products of the dissociation of sodium bicarbonate in water? What is the relative pH of the solution?
I had a recent question on a test that asked what the products would be if sodium hydrogen carbonate were dissolved in water. I had a few candidate answers

$\displaystyle\ce{NaHCO3 -> Na+ + HCO3-}$, but that one doesn't involve water at all
$\displaystyle\ce{NaHCO3 + H2O -> Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?
$\displaystyle\ce{NaHCO3 + H2O -> Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.

Part two of the question asked whether the solution would be acidic, basic, or neutral. What is the proper reaction for this?

once NaHCO3 is dissolved, the dissolution products are the Na+ ion and the HCO3- ion . According to wikipedia the HCO3- ion reacts with water to produce H2CO3 and a OH- ion, so the solution will be basic. The H2CO3 will decompose into water and carbon dioxide. So I guess that the most likely dissolution reaction will be the 3rd that you proposed

It's not as intuitive as it seems and your questions are all legitimate questions, but sometimes even good arguments can't be used as evidence in chemistry.

$\ce{NaHCO3 -> Na+ + HCO3-}$, but that one doesn't involve water at all

Look at this reaction:
$$\ce{NaCl <=>Na+ +Cl-}$$
Even this reaction doesn't "involve" water in the schematics but is right, we assume that is a dissociation of a salt in water.

$\ce{NaHCO3 + H2O -> Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?

Yes, but in fact this is an equilibrium:
$$\ce{H2CO3 <=> H2O + CO2}$$
And then you have to take into account another important equilibrium many time ignored:
$$\ce{CO2 (aq) <=> CO2 (g)}$$
So the amount of $\ce{CO2}$ present in the solution depends on by the type of system and the pressure exerted. Like a Coca-Cola bottle: if the system is closed, the $\ce{CO2}$ remains dissolved otherwise the equilibrium shift to the right and the $\ce{CO2}$ "escapes" until the rate between the $\ce{CO2}$ molecule dissolved and the $\ce{CO2}$ molecule escaped forms the solution is the same.

$\ce{NaHCO3 + H2O -> Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.

Okay, it is not a net ionic equation but the stoichiometry is still right. Maybe water is used to indicate that we are talking about a reaction that takes place in water.

Which reaction is not right?
All the reactions here are right, I think maybe they put some formal inaccuracies to help you, but in fact they are all part of series of equilibria that normally occur and can be described as:
$$\begin{multline}
\ce{NaHCO3 + H2O <=>[A] Na+ + HCO3- + H2O <=>[B] Na+ + H2CO3 + OH-} \\
\ce{ <=>[C] Na+ + OH- + H2O + (CO2(aq)<=>[C_2]CO2 ^ (g))}\end{multline}$$
The question should be more specific asking for the ultimate products of the reaction because all these are products of the dissociation of sodium hydrogen carbonate and because are equilibria is probable that all the species are present in different amounts when the reaction reach the equilibrium. However, the ultimate products are described by the third answer. And this equation can give you a hint about the acidity of the solution. When $\ce{CO2}$ escape from the solution (equilibrium $\mathrm{C_2}$ goes to the right) the concentration of the manly acid species goes down but the hydroxides from the equilibrium $\mathrm{B}$ are still present so at the end the solution in an open system with an atmosphere with a low partial pressure of $\ce{CO2}$ will have a higher pH, will become mildly alkaline.

First, ionize the salt as you have shown. NaOH is a very strong base so Na+ must be a very weak acid. It is solvated. Being a singly-charged, rather large cation, it stops there. If it were $\ce{Al^{3+}}$, coordinated water would be acidic by charge withdrawal from that brutally small trication.
Bicarbonate is the salt of the first ionization of weak carbonic acid. It will then be a modestly weak base by hydrolysis, as you have shown (reversible). Bicarbonate solutions are weakly basic. If you add acid (even weak acetic acid) they fizz. Boiling the solution will drive out carbon dioxide.
Carbonate is the salt of the second ionization of weak carbonic acid. It is a strong base.

So what would be the net ionic equation?


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How does voltage and resistance affect an electrolysis?
At school we've learned about calculating the voltage needed to perform an electrolysis. The current is how many electrons pass per time. The formula is given $$I = \frac UR$$
This means that given an electrolysis with $U=\pu{4.0 V}$ we get $$I = \frac{\pu{4.0 V}}{R}$$
If I am correct, what would decide how fast the electrolysis will go, is the resistance. Is the resistance fixed, or can I do anything to lower it? How would increasing the voltage affect the electrolysis? Would it just make it go faster (because $I$ is increased)?

Voltage determines the kind of chemistry at each electrode.  Gold is purified by plating the anode to the cathode at minimum voltage for the half-reaction, so impurities either do not dissolve and do not plate out.  Copper electrorefining electrode mud contains valuable trace elements in concentrated from.
Current determines the amount of chemistry, ~96,500 coulombs/mole of electrons.
Cell internal resistance converts input energy, power multipleid by time, $\ce{I^2Rt}$, into useless heat.  You then want small electrode spacing and good electrolyte stirring.  You then need enough voltage to do the chemistry, plus more to compensate for resistance losses, but not so much voltage that undesirable chemistries are enabled.  Does cell resistance increase (ohmic) or decrease with temperature?

Thank you for a good answer! So you say that the voltage should be as high as possible without enabling undesired reactions to increase the current? Additionally have a small distance between the electrodes to reduce  the resistance?

How many ampere should I get with 5 cm spacing, $3 V$ and a saline NaCl-solution in water? Is it safe to connect a car battery charger?

"I2Rt, into useless heat" of only minor importance: While all waste heat is lost energy in that it does not contribute to actual electrolysis, a hot electrolyte  is more conductive then a cold one, thus retained heat will increases the rate of electrolysis. Since rate of electrolysis is effected by surface area, strength of electrolyte and heat,  then to a degree, even waste heat can be an important contributing factor of a production unit.

Regarding your question: "Is the resistance fixed, or can I do anything to lower it?"
The energy used in electrolysis produces both product and waste heat. Naturally, you want to maintain sufficient production while limiting wasted energy.
A few important formulas:

Amps = volts / resistance
Watts = amps * volts
Joules = watts * seconds.

Things that effect resistance, and their effects on product versus waste heat and rate of production: Adjusting each parameter individually:

Voltage: Per unit of energy supplied, in general, a lower voltage produces more product and less waste heat from electrolysis. Below a certain threshold voltage no electrolysis will occur.  Although less power is consumed to produce the same product at a low voltage at a higher efficiency, the process proceeds slower because with a lower voltage, there is less current and power is consumed more slowly. Alternativly one can increase the voltage to drive more current and produce their product faster, but this happens at an expence to efficency.
Electrolyte type and concentration: Different electrolytes have different capacities to conduct and
Its concentration above or below an ideal will negatively effect the electrolyte's capactiy to conduct electricity.
Temperature: Increasing the temperature of electrolyte lowers the resistance of an electrolyte. Additionally,  some of the energy used to break up water's molecular bonds is provided by the thermal energy of molecules bouncing off each other. But, the effect really only becomes appreciable with high temperature electrolysis, like say 800c.
Surface Area: Surface area and current are proportional. All other parameters being equale, if you double the surface area of your electrodes, you effectivly half the resistance and double the current.
Spacing: Decreasing the spacing between electrodes decreases resistance.


text:
Is there a danger in electrolyzing NaCl?
To make $\ce{NaOH}$ one could electrolyze a solution of $\ce{NaCl}$ in water ($\ce{H2O}$) It would go like this:
$$\ce{2H2O_{(l)} + 2Cl-_{(aq)} + 2Na+_{(aq)} -> H2_{(g)} + Cl2_{(g)} + Na+_{(aq)} + 2OH-_{(aq)}}$$
After calculating a bit, I found that even just a few minutes of electrolyzing this at $10\ \mathrm{A}$ could be deadly.
Can anybody confirm this?

If the cell is mixed, you would make bleach, sodium hypochlorite and some free chlorine (and probably chew up your anode). Since the solution is not very basic, hypochlorite would probably disproportionate to chlorite, and that possibly to chlorate.
One mole of electrons is one faraday, about 96 500 coulombs. One ampere is one coulomb per second. Five minutes passes $(5\ \mathrm{min})(60\ \mathrm{s/min})(10\ \mathrm{C/s}) = 3000\ \mathrm{C}$ or $0.031\ \mathrm{F}$ in a reaction that requires two moles input for each mole of chlorine. $0.01554\ \mathrm{mol}$ chlorine or $350\ \mathrm{ml}$ at STP, tops. It wouldn’t be good for you.

Thank you again for your answer. I found this on a Web page explaining how to make NaOH, but no mention of the dangers.

You buy it - lye.  If you make it, a diaphragm cell to rpevent mixing of compartments.