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Calculating the concentration of acetic acid in vinegar from pH

DavePhD was kindly helping me with this problem but the site indicated the discussion had gone on too long. Calculating vinegar strength I don't have enough units to start a chat so I am starting a new thread.
The formula above is for determining the concentration or percentage of acetic acid in a vinegar sample when the ph is known. The farmer will get the ph with a ph meter, do the math and calculate the percentage acidity of her vinegar. The Ka, according to the formula DavePhD gave, is .000174 because the pKa for vinegar is 4.76. The molar mass is 60.05196 g/mol. So assuming the ph is 3, could someone write out the "long math" for this? I need someone to walk me through this, step by step to solve for c.
The last time I took algebra was about 50 years ago so, while I remember some things, it is probably better to assume I don't know much :-). Let me know if you need other numbers. Dave says the "M" is just a place holder

Could you come up with a better title for this please?  No one is going to find this during a search.

First off $$\ce{HC_2H_3O_2}\leftrightarrow \ce{H^+} + {\ce{C_2H_3O_2}}^-\\and\\ K_a = \frac{[\ce{H^+}][{\ce{C_2H_3O_2}}^-]}{[\ce{HC_2H_3O_2}]}$$ So let's do some simplifying. The amount of acetic acid left is the orignial molarity minus the amount that's dissociated. The amount of Hydrogen ion and conjugate base left is equal to the amount of that's been dissociated from the original amount of acid. So let's let HA equal the molarity of the acid and x equal to the amount dissociated. Soooooo $$K_a = \frac{[x][x]}{[HA-x]} = \frac{x^2}{HA-x}\\\text{Now let's solve}\\\rightarrow K_a(HA-x) = x^2\\\rightarrow HA -x= \frac{x^2}{K_a}\\\text{ switching the x interms of ph}\\ HA -10^{-ph} = \frac{(10^{-ph})^2}{K_a}\\\text{ plugging in}\\HA- 10^{-3} = \frac{(10^{-3})^2}{.0000174} = .0575$$ This is the molarity of the acid at equilibrium. Add 10^{-ph} for the concentration of the acid before dissociation.

hey @A A Ron! We've got formatting here just for chemical equations in MathJax, just use \ce. Thanks!


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Finding the frequency factor when raising temperature
The question is:

A reaction rate doubles when the temperature increases from $\pu{25^\circ C}$ to $\pu{40^\circ C}$. Calculate $E_\mathrm a$ and the frequency factor.

I found the activation energy to be $\pu{35.8 kJ}$ using the two point form of the Arrhenius equation. What I'm having trouble with is finding the frequency factor. I have two unknowns, $k$ and $A$, and to me it seems like this is impossible to solve without knowing what the rate constant $k$ is. All the examples in the book solves this problem graphically, but apparently you can solve this another way according to my teacher.
The answer given for $A$ is $1.9 \times 10^6$ but what method do you use to solve for this?

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This question has no answer.
The Arrhenius equation is:
$$ k = A e^{-\frac{E_a}{RT}}$$
A linearized form of the Arrhenius equation is
$$ \ln{k} = \ln{A} - \frac{E_a}{R}T^{-1}$$
This equation linearly relates $\ln{k}$ to $T^{-1}$: the intercept is $\ln{A}$ and the slope is $-\frac{E_a}{R}$.
To fully define a line, we need two parameters.  This can be two completely specified points that lie on the line, or any single point on the line plus a slope for the line.  For this problem that would mean either (a) two temperatures and two rates, or (b) one temperature, one rate, and one slope.
Using the information we are given:
$$ \ln{k} = \ln{A} - \frac{E_a}{R}T_1^{-1}$$
$$ \ln{2k} = \ln{2} + \ln{k} = \ln{A} - \frac{E_a}{R}T_2^{-1}$$
Any way we combine those two equations will only yield an equation equivalent to
$$\ln{2} = -\frac{E_a}{R}\left(T_2^{-1} - T_1^{-1} \right)$$
in which $\ln{k}$ and $\ln{A}$ have both cancelled out.  That is because the starting two linear equations have the same coefficients for $\ln{k}$ and $\ln{A}$ in each equation.  Similarly, the two equations $2x=y$ and $2x+2=y+2$  can't be solved for $x$ and $y$.
The problem as stated gives us only a slope, but not even a single point that lies on the line.  The rate could double by going from 1,000,000 $\text{s}^{-1}$ to 2,000,000 $\text{s}^{-1}$ (a very fast reaction!) or by going from 0.1 $\text{yr}^{-1}$ to 0.2 $\text{yr}^{-1}$ (pretty slow).  There is no way to find the intercept of a line when we are given only the slope.  Thus, there is no way to solve for $A$ using the information given.

source: Stack Exchange

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How to take water-dispersed PEDOT:PSS and disperse it in an organic solvent?
I am doing a science fair project on OLEDs. I am testing whether polypyrrole can be used as an emissive layer in a 2-layer OLED. The polypyrrole I am synthesizing is water soluble, and will be spin-coated onto ITO-coated glass substrate. I also need to spin-coat on a hole-injection layer (PEDOT:PSS). I can only seem to find it in what I believe to be water-dispersed form. This will cause obvious problems since both substances are water-soluble. I read on some informal message board that I could drop droplets of water-dispersed PEDOT:PSS into a hot organic solvent with a boiling point higher than water, boil off the water, and be left with organic-solvent dispersed PEDOT:PSS. That makes sense, I am just confused about what organic solvent to use. I believe it needs to have the properties of not dissolving the PEDOT:PSS, ITO, or the polypyrrole, and allow the PEDOT:PSS to become dispersed.
Here is a link to the PEDOT:PSS I am looking at:

Even if you switch to an organic solvent, if the polypyrrole is in water, it will perturb the PEDOT:PSS layer.
Your best bet would be to use a poly-alkyl pyrrole or other conjugated polymer that's soluble in something that doesn't dissolve PEDOT:PSS.
Let's say you switch the solvent to DMF or DMSO and it dissolves the PEDOT:PSS. Then you spin-coat that new mixture onto ITO. Once you drop the aqueous solution of polypyrrole, it will dissolve the PEDOT:PSS.
I think you're looking at this the wrong way. Rather than trying to change the PEDOT:PSS (which will be tricky - PSS is anionic) you should consider changing your polymer. Alternatively, get rid of PEDOT:PSS and use another hole injection layer. (PEDOT:PSS is acidic and will etch the indium out of ITO.)

Funny story.. The original aim of my PhD was to find something more conductive, more transparent and less acidic than PEDOT:PSS.

OK, I will start searching for a method of synthesizing polypyrrole in a different solvent. Thanks!

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What is the ionic equation for reaction of potassium permanganate with potassium iodide?
The half equation for the reduction of the permanganate would be
$$\ce{5e- + 8H+ + MnO4- -> Mn^2+ + 4H2O}$$
But with regards to the iodide, I am very confused. I have seen some sources say that $\ce{I-}$ is oxidised to $\ce{IO3-}$ in this reaction, and others saying that it is oxidised to $\ce{I2}$.

How stable is $\ce{IO3-}$ in the presence of $\ce{I-}$ under acidic conditions?
It would end up being in an equilibrium with very few $\ce{IO3-}$ anions and lots of iodine, due to the following equilibrium:
$$\ce{2 IO3- + 12 H3O+ + 10 I- <=> 6 I2 + 18 H2O}$$

The problem here is that in acidic medium iodide ($\ce{I-}$) is oxidized to iodine ($\ce{I2}$).
But in neutral or faintly alkaline medium, it is oxidized to iodate ($\ce{IO3-}$).
Also in neutral or alkaline medium the ionic half equation of reduction of permanganate would change to:
$$\ce{MnO4- + 2 H2O + 3e-  ->  MnO2 + 4 OH-}$$

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How can the iron in the reaction of iron (ii) chloride and zinc not go to zero oxidation state?
I know that the zinc half equation would be
$\ce{Zn -> Zn^{2+} + 2e-}$.
However I am confused about the half equation for iron. I would think that since iron is completely displaced, its oxidation state goes to zero, so its half equation would be $\ce{Fe^{2+} + 2e- -> Fe}$.
However my teacher said that whenever you have iron, it will always change oxidation states from 2+ to 3+ or vice-versa in a redox reaction, and will not go to zero. For instance, she gave us an example in which there was a reaction between iron(iii) chloride and zinc. I thought that the zinc would completely displace the iron so that it goes from 3+ ion to 0 in the solid iron metal. However my teacher insisted that it goes to 2+, and not to zero.
In the example of iron(ii) chloride, I don't even see how it is possible for it to go to 3+ because then both zinc and iron are oxidised!

Generally, to predict redox reactions, we consider the Standard Reduction Potentials at $25\ ^\circ\mathrm{C}$: The bigger the difference between the potentials of two redox couples, more favorable is the reaction.
$$E^0(\ce{Fe^3+ /Fe^2+})=0.77\ \mathrm{V}$$
$$E^0(\ce{Fe^3+ /Fe})=-0.04\ \mathrm{V}$$
$$E^0(\ce{Fe^2+ /Fe})=-0.44\ \mathrm{V}$$
$$E^0(\ce{Zn^2+ /Zn})=-0.76\ \mathrm{V}$$
So, when $\ce{Fe^3+}$ reacts with $\ce{Zn}$, we get $\ce{Fe^2+}$ (not $\ce{Fe}$) and $\ce{Zn^2+}$.
As for your reaction: when $\ce{Fe^2+}$ reacts with $\ce{Zn}$, we get  $\ce{Fe}$ and $\ce{Zn^2+}$.

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Orientation of halogenation
I was going through my chemistry book (RT Morrison and Boyd) and found a statement:

" the collision frequency of isopropyl and n-propyl w.r.t. Chlorine atoms is the same."

Is it correct because I think the size will matter in this case while both have the same weight?

Welcome! A little bit more context would be nice, like which reaction is this talking about, and what environment.

While the collision frequency may be the same, the reaction rates will differ. This is because not all collisions cause a reaction. The isopropyl radical will be more stericly hindered than the n-propyl radical.

The statement is:

This[collision frequency] must be the same for the two reactions, since both involve collisions of the same particles: a propane molecule and a chlorine atom.

It is clear that n-propyl radical and Isopropyl radical have same carbon structure. In the next step the hydrocarbon radical reacts with a chlorine ($\ce{Cl2}$) and since chlorine is symmetrical the collision frequency cannot depend upon it. It should be easy to judge from the following structures, the very minor difference:

Another reason is: Since the difficult step is the abstraction of a hydrogen and the determination of rates should be done while comparing primary, secondary or tertiary radicals, which have different stability.

This answer, while correct, adds some detail that may occlude the initial point about the probability of collisions.


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Why don't water molecules react alongside aqueous reactions?
To explain my question I'll use an example: We all know that sodium displaces copper in a solution of its sulfate in the reaction:
$$\ce{2Na (s) + CuSO4 (aq) \rightarrow Na2SO4 (aq) + Cu (s)}$$
My question is why don't some of the water molecules present in the aqueous solution of $\ce{CuSO4}$ also react with the sodium (Na) to form sodium hydroxide and hydrogen ($\ce{NaOH + H2}$) along side this displacement reaction. Shouldn't at least a bit of NaOH be formed?
I would appreciate a comprehensive response that a high school sophomore (me) would understand. Thanks in advance.

why don't some of the water molecules present in the aqueous solution of $\ce{CuSO4}$ also react with the sodium (Na) to form sodium hydroxide and hydrogen ($\ce{NaOH + H2}$)

I agree that sodium metal would react with the water.  Where/how did you come up with the equation implying otherwise?
I think you'll enjoy this video: https://www.youtube.com/watch?v=X5yLtqlopTM

"Did you put sodium metal in water?" "No, it was in copper sulfate!"... dissolved in water.


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How is the rate of a reaction defined?
I have learnt for many years that rate of reaction is the rate of change of concentration of a reactant or product with respect to time. I never questioned this before, but recently I have been learning about the rate equation and the order of the reaction with respect to a reactant. Seeing the graphs of how concentration changes with respect to time when you have a different order with respect to a reactant (as shown here)
It seems to me that choosing a different reactant to measure the rate of change of concentration of would give you a different rate of reaction? So how is the rate of a reaction defined?

If in doubt about definitions in chemistry, publications of the IUPAC (International Union of Pure and Applied Chemistry) are the place to look at. Here, the Gold Book gives some definitions for the rate of a reaction and the order of a reaction.

It seems to me that choosing a different reactant to measure the rate of change of concentration of reactant would give you a different rate of reaction?

Suppose a reaction $\rm aA+bB\to cC$, then let rate law comes out to be $\rm r=k[A]^x[B]^y$ then we would write:
$$-\frac1a\frac{\rm \delta [A]}{\rm \delta t}=-\frac1b\frac{\rm \delta [B]}{\rm \delta t}=\rm k[A]^x[B]^y$$

So how is rate of reaction defined?

It is defined in the same way you learned but the rate law is just something we "humans" created to relate the concentration to rate of change of concentration.

Let's take a sample reaction $\ce{\it aA + bB-> pP + qQ}$ with stoichiometric coefficients $a$, $b$, $p$, and $q$ for reactants $A$ and $B$ and products $P$ and $Q$.
The rate of reaction is the rate of change of the extent of reaction.  Extent of reaction is usually given the symbol $\xi$, so reaction rate is $\frac{d\xi}{dt} = \dot{\xi}=r$.
If the reaction rate is $\dot{\xi}$, then the rate of change of concentrations is:
$$\frac{d[A]}{dt}=-a \dot{\xi}$$
$$\frac{d[B]}{dt}=-b \dot{\xi}$$
$$\frac{d[P]}{dt}= p \dot{\xi}$$
$$\frac{d[Q]}{dt}= q \dot{\xi}$$
That is the stoichiometric coefficients link the rate of formation/consumption of the products/reactants, sort of by the definition of stoiciometry.
Now, when talking about the order of a reaction, what we are doing in effect is giving formulas (or partial formulas) for $\dot{\xi}$.  If the reaction is "elementary", then it will be $a$th order with respect to $A$ and $b$th order with respect to $B$.  This doesn't have to be true but often is.  This means that if you change the concentration of A while holding everything else constant, the change in rate will be proportional to the concentration of $A$ to the $a$th power, etc.  In math form that looks like
$$\dot{\xi} \propto [A]^a$$
$$\dot{\xi} \propto [B]^b$$
where $\propto$ means "proportional to", which is saying that the things are equal except for some one unknown constant.  Thus those two $\propto$ "equations" are equivalent to
$$\dot{\xi} = \kappa_1 [A]^a$$
$$\dot{\xi} = \kappa_2 [B]^b$$
It's important to note that $\kappa_1$ could depend on everything except $[A]$, i.e. it could be a function of $[B]$ etc.  Similarly $\kappa_2$ could depend on $[A]$ (but not on $[B]$).  The only way for both equations above to be true is if
$$\dot{\xi} = k [A]^a [B]^b$$
This means that $\kappa_1 = (k [B]^b)$ and $\kappa_2 =  (k [A]^a)$.  Now that we know the reaction order with respect to each reactant, and have written it down in equation form, we can relate it back to the change in concentration of each of the reactants/products:
$$\frac{d[A]}{dt}=-a k [A]^a [B]^b$$
$$\frac{d[B]}{dt}=-b k [A]^a [B]^b$$
$$\frac{d[P]}{dt}= p k [A]^a [B]^b$$
$$\frac{d[Q]}{dt}= q k [A]^a [B]^b$$
These equations clearly show the relationship between the order of a reaction and the rate of change of reagent/product concentrations.

I have learnt for many years that rate of reaction is the rate of change of concentration of a reactant or product with respect to time.

By looking at the four equations above, the rate of reaction will only be equal to the product or reactant concentration change when the stoiciometric coefficient of that product or reactant is 1.  And that's true regardless of what the "order" of the reaction is.  The reaction $\ce{\it aA + bB-> pP + qQ}$ doesn't have to be $a$th order with respect to $a$.  It could have been that $\dot{\xi} \propto [A]^{1.7}$ or even that $\dot{\xi} \propto \frac{[A]^{0.2}}{K_m+[A]}$ or whatever.  But in chemistry, it does often happen that reactions do have orders that depend straightforwardly on stoichiometry, which is why those types of reactions are studied so often.

authors: ÄDñãņ Ãľį Šhëķø


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What does delta letter mean in some ceramics chemical material formula?
Here is a given chemical formula for ceramic material:
La0.6Sr0.4Fe0.8Ni0.2O3-δ

There is a letter δ. What does it stand for? I guess it is something about oxygen vacancies in crystal lattice.

Exactly. In your example, it is related to oxygen vacancy (oxygen deficiency) in the crystal lattice.
It can affect the physical properties of the ceramic material. Take the example of this piezoelectric ceramic material:$\ce{Eu_{0.5}Ba_{0.5}TiO_{3-\delta}} $ Oxygen-vacancy induces antiferromagnetism to ferromagnetism transformation.
In general, $\delta$ denotes deviation from the stoichiometry (exces or deficiency) in a chemical formula


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How does one measure the change in enthalpy of hydration?
I am learning about lattice enthalpies, and my textbook talks about determining the lattice enthalpy from the standard change of enthalpy of solution and the change of enthalpy of hydration of the ions. It says that the changes in enthalpy of solution and hydration can be determined experimentally but that the lattice enthalpy cannot be.
However, I am struggling to think of how one can measure the enthalpy change when one mole of gaseous ions is dissolved in water under standard conditions. I thought one would encounter the same problem as with the lattice enthalpy, and this involves one mole of an ionic lattice being formed from its gaseous ions under standard condition.
I thought that the problem with this was getting the exact amounts of gaseous ions under standard conditions.

You mean the experimental and\or practical way?

Experimental way

Usually, one would measure these changes in enthalpy calorimetricly — or at least it is a nice student excersize in physical chemistry lab courses. I was once tasked to determine the solvation enthalpy of copper(II) sulphate pentahydrate.
The calorimeter consisted of an isolated thermos in which we filled a defined amount of water and a concealed inner container in which we filled a sample of the salt; the apparatus contained a stirrer inside the isolated area. We then let the water temperature equilibrate itself, opened in the inner container from the outside (mechanically) and measured the change in temperature. From $\Delta T$ we could calculate $\Delta q$ and from that $\Delta H_\mathrm{solv}$. All corresponded to the equation
$$\ce{CuSO4 . 5 H2O + H2O -> Cu^2+ (aq) + SO4^2- (aq)} ~~~~~~ \Delta H_\mathrm{solv}(\ce{CuSO4 . 5 H2O})$$
One could then perform a second experiment using anhydrous copper(II) sulphate. According to Hess’ law, the following is true:
$$\ce{CuSO4 + H2O -> Cu^2+ (aq) + SO4^2- (aq)} ~~~~~~ \Delta H_\mathrm{solv}(\ce{CuSO4})$$
$$\ce{CuSO4 + 5 H2O -> CuSO4 . 5 H2O} ~~~~~~ \Delta H_\mathrm{hyd}(\ce{CuSO4})$$
$$\Delta H_\mathrm{solv}(\ce{CuSO4}) = \Delta H_\mathrm{hyd}(\ce{CuSO4}) + \Delta H_\mathrm{solv}(\ce{CuSO4 . 5 H2O})$$
The two $\Delta H_\mathrm{solv}$ can be determined experimentally, thus the $\Delta H_\mathrm{hyd}$ can be calculated from these two values.


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Can a reaction with positive Gibbs free energy change ever occur?
If you have an endothermic reaction with a negative entropy change, is it still possible to induce the reaction in some way despite the fact that the Gibbs free energy change is positive or all temperatures?

Yes, $\Delta G = -RT\ln K$.
If a reaction is endothermic with a negative entropy change, $\Delta G$ is positive.
That $\Delta G$ is positive only means that $K<1$.
$\Delta G$ would need to approach infinity for $K$ to approach zero.
Therefore, at equilibrium, there will always be some products in principle no matter how unfavorable the reaction.
As a practical matter, the reaction would need to be only slightly endothermic and having only a slight negative entropy change to get a significant amount of product.
However, if the small amount of product is continuously removed from the system, as for example by precipitation from a liquid phase, or gas bubbling out of a liquid phase, or by liquid-liquid extraction, one can keep the reaction going forward.

Thank you for your reply! Is K the equilibrium constant in this equation? I have not come across this equation before...

The equation is actually the definition of standard equilibrium constant, so it is a very important equation: http://goldbook.iupac.org/S05915.html

Why does the ozone layer exist although the Gibbs energy for its formation is positive?

@Shub diatomic oxygen in dissociated in to monoatomic oxygen by ultraviolet light, then the monoatomic oxygen reacts with diatomic oxygen to form ozone:  https://csl.noaa.gov/assessments/ozone/2010/twentyquestions/Q2.pdf


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How can 30 ml of water be heated in less than 10 seconds?
How is it possible to heat a tiny amount (30 ml)[1] of water to a high enough temperature to make a coffee, in less than 10 seconds and possibly instantly?
Most heaters that I know of heat water in no less than 90 seconds (induction heater) or 3 minutes which is way too much for my purposes, and a microwave oven takes me one minute and a half to heat.
[1] A single coffee is between 20 - 30 ml and my coffee machine makes 30 ml for each cup.

Good espresso machines do this all the time - a hot heater block with enough thermal mass does just fine.

Thermal conductivity of water would make this not just a function of energy, but also surface area. Too thick and the bottom boils before the top gets up to temperature.  Too thin and it will overshoot your desired temperature too fast to get it off the heating element.

I have a single-cup water dispenser that dispenses around 200-250ml in 30 seconds: presumably it could do the first 30ml much faster with the same technique modified slightly

Troops in Vietnam used tiny amounts of C4 explosive to make coffee in the field. It produced enough energy to heat a cupful extremely fast. http://pomakis.tripod.com/clips/clip23.html - I presume/hope it's not something you have lying around at home, or even in a lab, however.

This part seems to have the energy requirements and surface area needed. It is not expensive either.

10 seconds or less, or it's free! WHAM!! The law of conservation of energy says it isn't free.

@MattThrower Good to clarify they didn't blow up their coffee with the C4, they burnt the C4. C4 burns great and is extremely stable. Just don't run an electric charge through it. And if you have some C4, you probably have better things to do with it. On second thought, if you're a civilian with C4 the best thing (for the rest of us) you can do is burn it to heat your coffee.

@MattThrower Indeed I have access to C4 and other explosives, but I wouldn't feel safe selling a potential bomb to civilians.

@dotancohen That's small enough for my purposes. I think I can find something like it locally.

Tap your radiator. It's cheating because you're not actually heating, it is already hot and tastes like crap. But it's hot water anyway.

Well, let's do some math:
Assuming 30 mL of water is 30 g, and we want to heat our water from 20 °C to 90 °C, the energy required is:
$$\begin{align}E&=C_Pm\Delta K \\
&=\left(4.18 \mathrm{\frac{ J}{gK}}\right)(30\mathrm{\ g})(70\mathrm{\ K})\\
&=8.778\mathrm{\ kJ}\end{align}$$
So how much power do we need to do this in a given time? "Instant" doesn't really mean anything, so let's go with 10 seconds:
$$\begin{align}P&=\frac{E}{t}\\
&=\frac{8778\mathrm{\ J}}{10\mathrm{\ s}}\\
&=877.8 \mathrm{\ W}\end{align}$$
This is not an enormous amount of power, but the trick is that it all has to go into heating the water. A good microwave outputs a fair bit more power than this, but it generally doesn't all get absorbed by such a such a small amount of water in only 10 seconds. Your best bet is probably an electric heating element directly inserted into the liquid, though I don't know if you can get a ~1000 W one small enough to sit in that much water.
As Jon Custer notes, it's not necessary to produce all the heat at once. If you heat some kind of thermal reservoir and flow the liquid past/through it, it reduces the demands on your heat source.

Edit: Also, I just tried this with a 1200 W microwave and it only took 15 seconds. How fast do you really need this coffee?

"How fast do you really need this coffee?" Is this a trick question?

I really don't know which answer to select as best answer as all are great answers and explain many things in different ways. I'll go for the most voted one. About the microwave; I think I have a less powerful and old one. Well, now that I think of it, something in the range of 15-30 seconds is still great for UX. Thank you everyone, hope the answers will help other people like it did for me.

If this is an engineering challenge (i.e. your company manufactures espresso makers), the microwave solution is doable. 30 ml of water will absorb 900W of microwave energy without problems if you can engineer the geometry of the water reservoir and the placement of the microwave tube. A standard microwave oven cannot be optimized like that, and probably is tuned more for a plate of food.

Please bear in mind that heating water using microwaves is an extremely hazardous activity.  Bad steam burns are common when water is heated this way due to pockets of superheated water.  Related to this is the fact that microwaves do not heat water evenly due to their very nature (~3cm wavelength).

@Stephen: 12 centimeters, actually. And since 30 cl is about 3x3x3 cm, a well-designed solution can have a fairly constant heating across the whole container. Also, the dielectric constant of water shifts, so you can tune the system for 293K. Hotter water will then absorb less energy, which means the colder pockets are preferentially heated.

Source: http://en.wikipedia.org/wiki/Nucleate_boiling
As can be seen in the above graph, there is a local heat flux maximum for heating water, when the hot surface contacting the water is 30 degrees C above the boiling point, that is 130 degrees C at atmospheric pressure.
So to minimize time to boiling, contact a 130 degree C surface made of high thermal conductivity material (such as silver, copper or aluminum) with the water.
With heat flux over 100,000 Watts/m^2, you just need to make the surface area of contact large enough to acheive any desired heating time.

(877.8 W) / (100,000 W/m^(2)) = 87.78 cm^(2). 30 mL / 87.78 cm^(2) = 3.4mm water depth (if on a flat plate).

If you want the practical answer, get an instant hot water dispenser.  Practical, HA!  Ok, let's do some science!
First, what kind of coffee?  At 30 ml I'm guessing you're making espresso.  I'm going to go with 92C as the optimal temperature.
Second, I'm going to take your full 10 seconds.  The rest of the process of making espresso takes far more time so whether it's 5 or 10 or 15 seconds doesn't make a real difference to how fast you get your morning hit.
Other answers have pointed out the amount of energy needed to bring your 30 ml up from room temperature, but you can make life easier on yourself and do what a lot of "instant boil" systems do and preheat.  This allows you to use less energy (for the 10 seconds anyway) and bring the water up to temperature slower.  This gives you a wider tolerance between the water being too cold and it being boiled away.  If your water is already at 50 C that will only require about 500 W.
To deal with thermal conductivity, a wider surface area in contact with the heating element would serve to heat the water more quickly and more evenly.  If the water is too thin, it can overheat before it is removed from the element.  Too thick and the bottom may boil before the top reaches temperature.
We need a broad, flat surface that can output at least 500 W and that we can quickly add and remove small amounts of water from.  Sounds like you need an electric skillet!  A very clean one.  Or the precision laboratory version.  Buy one that can heat quickly and evenly.
Knowing when to remove the water is a problem.  Measuring the temperature of 30 ml of water on a hot electric skillet is problematic.  I would suggest trial and error with various skillet settings, times and initial water temperatures.  You can do a lot of experimental rounds in 10 seconds.  This would also allow you to include the cooling effect of whatever vessel you're transferring it into.

According to Michael Dryden's answer, it takes 877.8W to heat 30ml of water by 70C (from 20C to 90C)in 10s.
My electric shower is about 10 times that power (lets call it 8.778kW for convenience.) I don't like to shower at 90C. I am more than happy if my shower heats the water about half that temperature difference, say 35C (from 5C to 40C). So I would set it to deliver 300ml x 2 = 600ml every 10 seconds. That's 3.6L/minute, so my 5 minute shower uses 18L of water. Of course, if I did choose to raise the temperature by 70C in the shower, I would have to restrict it down to 300ml every 10 seconds, making it a continuous version of your experiment, scaled up by a factor of 10.
The last time I took a shower apart, the element was inside a copper vessel of about 300ml volume. So, if you restricted the flow  to 300ml in ten seconds, your residence time in the element vessel would be... ten seconds!
As noted by Schwern, your issue isn't heating it, it's being able to control the temperature accurately. One method that hasn't been mentioned is dunking a calibrated lump of hot metal into it (or pouring it into a hot metal container.) But so far the best method is Michael Dryden's idea of using a microwave.
In industry, batch processes are only used for slow processes like brewing beer. Fast processes (such as those used in oil refining: cracking, reforming, alkylation, etc.) are done continuously. For economic reasons these processes are carried out as fast as possible in the smallest possible equipment, and continuous process is therefore the only practical way of doing them.
If you really need to heat water that rapidly, and the reactants consumed in your experiment are cheap, I suggest you set your experiment up as a continuous process, with a flow of water through it. That way you will be able to control the temperature accurately with an electric heater. A shower will probably be too powerful - try scaling everything down to something like a 25W soldering iron element (make sure it's sealed and 12V DC for safety!)
You can supply your equipment with water from a reservoir at the top and control the flow / residence time with a valve at the outlet at the bottom. If you want to try the same conditions with double the residence time, simply reduce both the water flow and the heater output to 50%.
You may optionally replenish the reservoir with tap water to the point of (near) overflow if you need to mix things into it.
When I did a conversion course from chemistry to chemical engineering, several of the practicals were set up like this. It's definitely the way to go if you want to study fast processes whether they be mixing, reaction kinetics, or whatever.
Regarding Dave's points about critical heat flux, the important thing is to make sure you have sufficient surface area for the volume being heated. I don't see it as a huge issue at your scale though. For 1kW, you need 100cm2 of heating surface for a flux of 100,000W/m2. Many newer kettles provide this (with heating over the whole bottom surface for a neater design) but older models with curly heating elements are above 100,000W/m2, approaching critical heat flux. These electric kettles 'roar' as they come to the boil due to cavitation.
If your reagents are delicate, though, you will need to space your heaters very narrowly, or even use steam injection. I remember one particularly nasty case of electric heaters converting an aqueous polymer solution into volatile aldehydes.

Your shower does not have an output of 8.778kW. Standard electricity cables melt with that amount of power, consumer welding equipment is around 2.5kW. Your shower is not going to beat welding equipment (that would be one hell of a shower).

@Mast: In Europe, a 230V/25A cable is quite common and delivers 5.7 kW. For electric heating, 380V/35A is also available which pushes you well over 13kW. But even in the US, the need to power 10kW+ air conditioners means that high-power supplies are available.

@Mast Any bathroom continuous-flow water heater needs that rating, or more (cf. http://www.amazon.de/b?node=2077608031), like 24kW/100+ Amps. They typically have their own dedicated wiring right to the fuse box, with their own fuse. Oh, and when the landlord decides to install it in an existing apartment building they must renew the house's rising mains as well.

@Mast besides Peter's one from Germany, here's one from the UK where 8.5kW is the lowest power available: http://www.screwfix.com/c/bathrooms-kitchens/showers/cat820272 As Peter says, it needs to be wired in permanently on its own circuit because it's too much current for a wall plug. Regarding welding equipment: Don't underestimate the heat capacity of water. Haven't you ever seen a blacksmith put a large metal object in water to quench it? it cools down pretty quick! I don't know what the rules are in USA, but instant electric showers are rare in Spain because the installed kW is too low

In the Netherlands, a 230V/25A fuse is quite common. But there's usually a 16A fuse behind it, limiting the power usage to 3680W. Wiring directly into the fusebox isn't standard practice as far as I'm aware, but a landlord could do it.

@Mast it all depends on local regs. UK homes have 100A main property fuse installed as standard. My Spanish property on the other hand has only 20A for the whole aparment, because in Spain you pay a premium for any more capacity. As a result, Spanish properties have either instantaneous gas heating, or a storage tank heater of about 50-100L capacity with a 1-2.4kW element (with a plug on it.)

As others have shown the energy requirement is quite reasonable, the problem is delivering it to the water in that timeframe.
I would think a workable approach could be done with a collection of plates with very narrow spaces between them that can hold your 30ml.  Make the plates out of something that's not too good a conductor.  Feed an extreme current through the plates (very low voltage, though--you'll have a big step-down transformer) and it will get hot quickly.  You'll have to figure the energy left in the plates and add it to the amount of power you are dumping through this.
Careful, a second run before the plates have cooled will certainly result in boiling water.

Small problem: your plates supposedly are somewhat-decent electrical conductors. Metal, graphite, etc. Those have lots of free electrons which can conduct electricity, but also heat. So your proposed material is in fact not a thermal insulator.

@MSalters Who said anything about an insulator?

"Not too good a conductor" is an isolator.

@MSalters I mean something with enough resistance to use as a heating element, not something that's an insulator.

In that case, why not use the standard solution - a metal wire coil? The resistance there is obtained by it being thin and long. You want that here, because that length gives you a large surface.

@MSalters A coil won't provide the very thin layers of water I'm picturing.

You could try Induction cooktops. 10 seconds for 30 ml should be enough at 700w or more, if you are using a flat and wide vessel (not all utensils are induction compatible).

How to get instant coffee
There are already some good answers on how you could reasonably get a cup of coffee-temperature water within 10 seconds. However, what if you wanted to get it even quicker than that?
Realistic solution: Use a boiler
If you already have hot water, you won't need to wait for 10 seconds before your cup is ready to go. This may not be the most economical solution to get your morning dose, but if you have a coffee shop and really need to churn out cups during peak times it may be interesting to look into this.
Interesting solution: Use more power
If you don't like the idea of keeping your water hot all the time, you need something to heat it quickly. Of course you could use a huge heating element, but I suppose that it may be even quicker to use something like an explosion. You just need to make sure that no deadly chemicals end up in your cup.
Other answers already indicated that the main problem with using more power, is that it is hard to control the temperature. However, this should be solved with the following steps:

(Over) Heat some water
Measure the temperature (only needed if step 1 cannot produce a fixed temperature reliably)
Mix the right amount of this water/steam with cold water to get a cup ful of water at the desired temperature

Of course it will never be truely instant, but I imagine that you could get it done in a fraction of a second.

Another solution that maybe is not so great for everyday use but very fast, is to enclose water in a metal container, and around it another enclosing where Calcium Chloride [CaCl2] is mixed with water [H2O] (note: not the same water needed to be heated!!) => Ca2+ (aq) + 2Cl-(aq) + Q which should heat the water in less than 10 seconds.

You can easily add $3.9~\mathrm{g}$ steam to $26.1~\mathrm{ml}$ water of $20\mathrm{°C}$ to get $30~\mathrm{ml}$ water of $100\mathrm{°C}$ in less than ten seconds.

Yeah. This is the power of direct steam injection. :o

We need a wide, level surface that could result at the very least 500 W and that we can quickly include and get rid of small amounts of water from. Sounds like you require an good electric skillet! A very clean one. Or the precision laboratory version. Get one that could heat rapidly and also uniformly.


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Solubility of Noble Gases
I read this statement that:

Xenon is the most soluble noble gas in water.

My first doubt is:
Why does the solubility of noble gases in water increase down the group?
Even if this is the trend, why is Radon not considered most soluble?

Wikipedia gives the following solubilities for the noble gases in water at $25^\circ \text C$ in $\mathrm{cm^3~kg^{-1}}$:
$\mathrm{He: 8.61~~~Ne: 10.5~~~ Ar: 33.6~~~ Kr: 59.4 ~~~Xe: 108.1 ~~~Rn: 230}$
Why does solubility of noble gases increase down the group?
The intermolecular interactions of noble gases are in the form of London dispersion forces, which are strongly influenced by the polarizability of the atom which increases as the atomic radius increases. This can be thought of as resulting from the electrons being 'less tightly bound' to the nucleus and the atomic radius increases. Atomic radius increases down the group therefore solubility also increases. For more on polarizability see here.
Why is radon not the most soluble noble gas?
Radon (and other heavy radioactive elements) are often ignored when talking about trends because they are uncommon. Radon is in fact the most soluble noble gas but the half lives of its isotopes are on the order of hours or days so it is not really very useful.

The solubility of radon in water has long been studied, and the Weigel equation in the name of the equation for radon's solubility in water as a function of temperature, at atmospheric pressure.
[Rn(aq)]/[Rn(g)] = 0.105 + 0.405 * exp(-0.0502*T)
Near room temperature, the concentration of radon in the gas phase will be about 4 times the amount in the liquid phase.
Radon is the most soluble noble gas.
See Radon Solubility In Water As A Function Of Salinity And Temperature by Eric B. Lieberman for more information.

The solubility in water is due to Dipole - induced dipole interaction.
The polarizing water molecules distort their electrons with the increase in atomic size.Therefore increase in interaction or Atomic size.
i.e. increase in solubility.

‘The water molecules distort their electrons with the increase in atomic size’ — Water molecules are always the same size. Clarify please.

Also welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. You can [edit] your post at any time with the link in the bottom left. For any questions about the site itself and how it works, visit the [help].


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Pseudo 1st order Kinetic reaction
The pseudo 1st order reaction happens when one of the reactant is in large excess and its concentration doesnt change during the reaction .
Is there any other such situation when a certain higher order reaction turns out to become a pseudo first order kinetic reaction (except the situation where an N$^{th}$ order reaction has N-1 species in large excess) ?

Yes. A 3rd order reaction rate=k[A][B][C]. If both [A] and [B] are in large excess then the rate is solely dependent on [C], pseudo 1st order. The same can be said for any order reaction where only all but one species is in large excess ( and order is of 1 for that species)

@Leeser Ya , but I am asking is there any other situation when this happens ? ( a situation other than "large excess of species ")

Example from biochemistry
The Michaelis-Menten equation for enzyme kinetics is an example.
$v=\frac{k_{cat}E_0S}{K_S+S}$ where $S$ is substrate concentration and $E$ in enzyme concentration.
At low $S$ ($S\ll K_S$), the reaction is 2nd order with $v=\frac{k_{cat}}{K_S}E_0S$, i.e. first order in both enzyme and in substrate.  However at high $S$ ($S\gg K_S$), the reaction is first order only in enzyme with $v=k_{cat}E_0$, and is zero-order in substrate.  The substrate concentration doesn't have to be in molar excess to other reactants.
Example from process control
In chemical engineering, nonlinear systems such as higher-order reactions occuring in a continuous process at steady-state are often modeled using linear (which you can think of as first-order reactions) systems as an approximation for developing process control techniques.  A MATLAB documentation page has some good information.
Another example, somewhat contrived
As a last example, consider an unstable molecule $A$.  It can decompose to $B$ by two entirely separate elementary mechanisms.  The first is $\ce{A + A -> B + B}$.  The second is $\ce{A -> B}$.  At high concentrations of $A$, the second-order kinetics will dominate and the decomposition of $A$ will be second order.  But lower the concentration enough, and the reaction will become first order in $A$ because the rate the bimolecular reaction will be dwarfed by the unimolecular route at low $A$.


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In which all substituents does Inductive effect overpower Mesomeric effect?
In most substituents, the mesomeric effect is considered stronger and more dominating than the inductive effect.
A notable exception are the halogens.
The reason behind this has been well-explained here.
I further had questions about the following:

Since its about the EN, would all other electronegative elements with lone pair show this anomaly?
Are there other substituents with this exceptional behavior?


text:
Acid chlorides and reduction by hydrogenation
An acid chloride being reduced through catalytic hydrogenation is an "unobserved" event in the context of our class, according to my professor. Nonetheless, we are expected to predict that the hydrogenation occurs because acid chlorides are very reactive in general.
Now, this sounds like a case of the right answer with the wrong reason.
Yes, acid chlorides are reactive. We've covered their hydrolysis with water. We've covered their role in Friedel-Crafts acylation. We realize that the $\ce{C-Cl}$ bond in an acid chloride is very weak and that the carbonyl carbon is highly electrophilic. This gives rise to most of the reaction pathways with acid chlorides.
But what does catalytic hydrogenation have to do with the electrophilicity of the carbonyl carbon in an acid chloride? Nothing, in my mind. Hydrogen gas and platinum metal simply work here because the $\ce{C=O}$ part of the acid chloride is flat. I don't think it has anything to do with how reactive the acid chloride is. Right? Hell, even plain jane alkenes can be hydrogenated and they aren't exactly reactive.

Hydrogen gas and platinum metal simply work here because the C=O part
  of the acid chloride is flat.

It's more than just that.  Acid chlorides are significantly more reactive towards catalytic hydrogenation than other carbonyl compounds.  They are reactive enough that a poisoned catalyst must be used along with mild reaction conditions to achieve reduction to the corresponding aldehyde, and have the reaction stop there (Rosenmund reaction).
So the question becomes, "why are acid chlorides so reactive towards catalytic hydrogenation."  It is thought that because the acid chloride $\ce{C-Cl}$ bond is so reactive that the palladium inserts into this bond to form intermediates such the one pictured below.

Other typical carbonyl compounds can't form such an intermediate; consequently acid chlorides are noticeably more reactive towards catalytic hydrogenation. Anhydrides, where the carbonyl is roughly as reactive as an acid chloride's carbonyl towards nucleophilic attack, have not been reported (to my knowledge) to produce aldehydes under Rosenmund conditions supporting this mechanism involving a palladium intermediate.
Historical Note: Prior to the application of metal hydrides to the reduction of carbonyl groups, the Rosenmund reduction was a major synthetic tool for chemists.  It was one of the few methods available for the straightforward transformation of acids, esters, amides, etc, to aldehydes.


text:
Why does the definition of ionisation energy only include gaseous atoms?
I know that in physics there is thermionic emission and the photoelectric effect. These are both method of removing electrons. I think these affects are only for metals due to their de-localized electrons.
So why do they say that you can remove the valence electron of a gaseous atom? Can you not do this with atoms in their solid or liquid state, and does being a metal, non metal or metalloid affect this?

Thermionic emission and the photoelectric effect are applicable to all materials. While these indeed remove an electron from the material, and the energies at which that happens are very interesting, the energies are fundamentally tied to condensed phases with band structures, not to single atomic energy levels.


text:
Does the compound KFO₃ exist?
As the compound potassium chlorate exists, is there something as $\ce{KFO3}$?
All the other halogenates are mentioned somewhere on the web but what about this?
If no, then why?

It doesn't exist - fluorine is more electronegative than oxygen, for starters.

It has been already mentioned in the comments and the other answer, that fluorine is more electronegative than oxygen. Therefore it is highly unlikely that this compound exists.
To illustrate the problem at hand, take a look at similar, known compound with a fluorine oxygen bond: hypofluorous acid, $\ce{HFO}$, which is actually better described as hydroxylfluoride, $\ce{HOF}$. According to Holleman and Wiberg's[1,2] textbook the synthesis of this compound is already difficult. Low pressure fluorine gas is passed over ice at about $-40~^\circ\mathrm{C}$. Obtained is a mixture of $\ce{HF}$, $\ce{O2}$, $\ce{H2O2}$, $\ce{H2O}$ and also $\ce{HOF}$. Fractional freezing until about $-183~^\circ\mathrm{C}$ leads to a white substance, that turns pale yellow at about $-117~^\circ\mathrm{C}$, and boils at about $10-20~^\circ\mathrm{C}$. It decomposes via a radical pathway at about $25~^\circ\mathrm{C}$ and $100~\mathrm{mbar}$ with a half life of about 30 minutes to oxygen and hydrogen fluoride: $$\ce{2HOF -> 2HF + O2}.$$
The electronic structure of the molecule is most likely $$\ce{HO^{\delta+}\bond{-}F^{\delta-}}.$$
Salts, $\ce{M^{+}OF^{-}}$, that are derived from this compound are not known. Covalent compounds, however, are known, i.e. $\ce{O3ClOF}$ or $\ce{O2NOF}$.
It is easy to imagine, that a compound with an even higher oxygen load cannot be stable. Therefore we can conclude that a halogenate like $\ce{HFO3}$, let alone its salt $\ce{M^{+}[FO3]^{-}}$, cannot exist.
Notes and References

Unfortunately I only have access to the German version of this textbook. I tried to excerpt and translate the important parts to the best of my knowledge.
Holleman, A., Wiberg, N., Wiberg, E., et al. (2008). Lehrbuch der Anorganischen Chemie. Berlin, Boston: De Gruyter. Kapitel XII 4.2. Sauerstoffsäure des Fluors. S. 465.  (in german)
Translation of the 101st edition available as Egon Wiberg and Nils Wiberg (2001) Holleman, Wiberg. Inorganic Chemistry. Academic Press.

This compound can't exist. There is no such anion such as $\ce{FO3-}$, why not ?
Because fluorine is more electronegative then oxygen so that anion just can't be formed. Therefore no $\ce{FO3-}$ to combine with $\ce{K+}$ therefore no such compound is present.


text:
Is water a better nucleophile than bromide?
I am a student of biochemistry. My organic chemistry professor was giving a lecture and he talked about halohydrin formation reactions. At one point, he said that "water attacks instead of bromide because water is a better nucleophile than bromide". He repeated that five times or so.
A classmate argued that, because water is the solvent in the reaction and is in greater proportion, it will attack faster. But that still doesn't mean water is a better nucleophile, right? Bromide itself has a negative net charge, while water has no net charge, only 2 free electrons.
The question is, was my professor wrong or is there something I'm missing? Thank you very much.

That is not true. As you noted bromine has a negative charge (and also lower electronegativity than oxygen) and is therefore the better nucleophile (even under these aqueous conditions). Halohydrins are not even usual the result of bromination in water. Special conditions are required to prevent too much $\ce{Br-}$ from attacking the bromonium cation (e.g. certain pH, adding additional reagents as $\ce{AgOH}$ or using $\ce{HOBr}$, see the section Addition Reactions Initiated by Electrophilic Halogen in W. Reusch's online Organic Chemistry book). You would usually expect the dihalide product since $\ce{Br-}$ is already close to the bromonium cation as it forms.

Under many circumstances, I would agree that bromide is a better nucleophile than water. It's negatively charged and, due to its size, bromine is more polarizable than oxygen.
However, in water, the bromide anion is very stable, reducing its nucleophilicity. I suspect that bromide  is still more nucleophilic than water (even considering solvent effect), but combined with the large excess of water, it is more likely that water opens the bromonium ion in the halohydrin formation.

Do you mean stabilised by hydrogen bonding? Will not water will be more extensively hydrogen bonded?

@ReeshabhRanjan Yes, the water molecules are present in a hydrogen-bonded network. However, the concentration effect is indeed overwhelming. The concentration of pure water is around 55.6 M, certainly larger than the [Br-]. Also note that the hydration effect on the Br- may actually be greater as it is in the form of ion-dipole interactions, which are likely stronger than the hydrogen bonding interactions between water molecules.

If the reaction is an alkene reacting with $\ce{Br2}$ in $\ce{H2O}$, the double bond will behave as a nucleophile attacking one of the Br's. Once the Br is attacked it will also keep the carbocation stable until it is attacked by the $\ce{H2O}$ molecule rather than the $\ce{Br-}$. Since there is so much more water in solution the chances of water attacking is more likely than the negative Br attacking.
Once water attacks the carbocation with the lone pair on oxygen it forms a positive charge. Another water molecule will come and pull off a Hydrogen atom forming a Hydroxy. The product will include a Br and Hydroxy group (Halohydrin Formation).


text:
Which is the least stable trihalide of nitrogen?
Which is the least stable trihalide of nitrogen and why?
I have two conflicting theories, one is that fluorine-fluorine electron repulsion will make $\ce{NF3}$ least stable and another is steric reasons of iodine which give $\ce{NI3}$ as the answer.
Which is correct?

Nitrogen triiodide $\ce{NI3}$ is an extremely sensitive explosive that explodes with the slightest touch when dry.
Nitrogen trifluoride $\ce{NF3}$ is so stable that it is a greenhouse gas global warming concern, with lifetime in the atmosphere of hundreds of years.
$\ce{NI3}$ is unstable due to steric strain as you are saying.

I agree with Dave's answer, but would add that iodoform ($\ce{CHI3}$) is stable.  I suspect that $\ce{NI3}$ is unstable due to sterics and an extremely exothermic pathway leading to the formation of the very stable $\ce{N#N}$ triple bond.  A similar decomposition pathway does not exist for iodoform, hence its relative stability.

I could be wrong about the reason, since NCl3 is also unstable.

I believe something's rather different about the fluorine compounds. Even sulfur hexafluoride and carbon tetrafluoride are extremely stable. I guess it's by virtue of their high bond strengths.


text:
Crystal Lattice Defects
Why do defects in crystal lattices of various molecules and compound occur ?
Is it due to the continuous contact or it arises due to impurities and human error during the formation?

There are many reasons for the many different kinds of defects.  Vacancies are part of the thermodynamic equilibrium of 'perfect' crystals because of configurational entropy.  The Simmons-Baluffi experiments are the classic example.

First of all, there are no such thing as a "perfect" crystal.
Defects occur from both human error and the experimental setup.
Defects from human error are primarily due to handling issues, which can never be perfected, especially not if the crystals are solvated.
Experimental defects can occur from either the diffraction experiment or the crystal growth experiment. Impurities can affect crystal growth by displacing the molecule of interest in the lattice and unless you are attempting neutron diffraction, the crystal will be significantly damaged during the data collection.
In my field of protein x-ray crystallography, the typical crystals can have diffraction ranging from nothing at all to ~0.8 Angstrom range (best cases) and as such you can have beautiful crystals that have no diffraction and ugly crystals that diffract very well.
Crystal defects can be perfectly acceptable and are a so common that we usually don't think about them too much.

A perfect crystal can exist only at 0K. Above this temperature, there is always some energy associated with the crystal which causes some particles to leave the crystal as in Schottky or Frenkel defects, or it allows some impurities to enter.

Even at 0K (which is impossible to reach) perfect crystals cannot exist, unless you can crystalize a single isotope of each element present... the isotopes will otherwise be randomly distributed... And you better not be thinking of crystalizing anything with radioactive decay products!


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pKa vs range in choice of an indicator
When choosing the ideal indicator during a titration, there are two considerations to keep in mind:

The pH = pKa of the equivalence point
The indicator has a small range (<±1) over which it can change color

Which factor is more important when deciding over two indicators to see which is best?

The pH = pKa of the equivalence point.
it is the best factor to decided during the titration .
because if the range of the indicator is less , then we can't find the correct volume used in titration.this factor should be considered but not most affective one.


text:
An electric potential is generated when a liquid moves with respect to a solid (streaming potential). Is the reverse true?
Streaming potential is an electrochemical phenomenon that a that relates
electric currents with the relative movement of solid and liquid phases in contact with
each other.
So when a liquid moves past a solid a potential is generated. The equations for the values of the potential and current generated can be found in this Wikipedia article. My question is, is the reverse true? If I applied an electrical potential difference (instead of applying a pressure difference), would a pressure difference be generated?

Your assumption is correct. Electrokinetic pumps use a potential difference across an osmotic membrane, i.e. through narrow channels, the inverse of flow causing a streaming potential difference. There are other aspects of the phenomena such as electrophoresis, capillary osmosis etc.; see:
Electrokinetic Phenomena a
On the Efficiency of Electrokinetic Pumping...
and
short video demo


text:
What is the meaning of "no Hessian Eigenvalue "?
When running GAMESS for finding Hessian Eigenvalues and frequencies ( using Optimization with hessian calculation at the end ) , I have seen there are times that it comes out "without any frequencies".
I think even a non-optimized system should have some vibrational properties and I can not understand what such a " no-frequency state " means physically.

Interesting question. It would be helpful if you quoted some sample output from GAMESS directly in the question.  Also consider adding as much detail as possible about the input into GAMESS that you used too.  It will help you attract higher quality answers.

I second to request relevant parts of both input and output, although, I suspect the problem is quite simple.

What is the meaning of “no Hessian Eigenvalue ”?

The normal modes and frequencies are retrieved from Hessian diagonalization. By diagonalizing it you get the eigen-vectors (describing normal modes) and eigen-values (related to frequencies). If it is not done, the frequencies can not be calculated (and that's why they are not printed)
If the geometry optimization fails there is no point in the calculations above.

I think even a non-optimized system should have some vibrational properties and I can not understand what such a " no-frequency state " means physically.

There is not a no-frequency state here. Trully, each single point calculation does not describe exactly a quantum state of your molecule. Why? Well, it neglect the quantum particle treatment of nucleic. We calculate the quantum state of fermions on some potential focus, and get the corresponding energy for different focus positions. The "movement" of the focus is then treated classically.
The standard procedure is to get the harmonic frequencies, that is, accept the Hook Law for every normal mode. To get the physical insight we can think in a diatomic molecule (that has of course only one vibrational mode).
We can use Taylor series to represent the potential energy $U$ at any distance $r$ around the equilibrium distance $r_e$:
$$U(r) = U(r_e) + U^\prime(r-r_e)+ U^{\prime\prime}(r-r_e)^2+\cdots$$
The heigh over the well is:
$$U(r) - U(r_e) = U^\prime(r-r_e)+ U^{\prime\prime}(r-r_e)^2+\cdots$$
If we asume harmonic behavior, we expect that:
$$U(r) - U(r_e) = \frac{1}{2}k (r-r_e)^2$$
At equilibrium distance: $U^\prime(r-r_e) = 0$, so it is trivial to relate $k$ with $U^{\prime\prime}(r-r_e)$
At others $r$ we can't do that because we don't know $r-r_e$.
In such case, if it has vibrational properties, we don't know much about them. BUT, it is not always true that it has vibrational properties. Imagine a potential energy curve like:

(source: quantavolution.net)
If the molecule is between $B$ and $C$, it will "move" in $A$ direction and no return is expected (no vibration here). It doen't happend in the normal mode of a diatomic molecule, but it is a very common situation in poliatomic species.

I suspect the problem is pretty simple: hessian is not evaluated when geometry optimisation does not converge to a stationary point. It is actually mentioned in GAMESS manual (emphasis mine)

HSSEND = a flag to control automatic hessian evaluation at the end of a successful geometry search. (default=.FALSE.)

And such behaviour is quite standard. The thing is that the conventional vibrational analysis (as descibed in most texts and implemented in most programs) is valid only when the first derivatives of the energy with respect to displacement of the atoms are zero. Or, in other words, the geometry which is used for vibrational analysis must result from a previously converged geometry optimisation run (at the same level of theory, of course).


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Creation and Combustion of Nitrocellulose
I am intending to demonstrate the formation and combustion of nitrocellulose with my high school students, however I am concerned about the potential products of the combustion equation. I have been able to satisfy the safety requirements for the creation which I believe follows this equation:
$$\ce{3HNO3 + C6H10O5 → C6H7(NO2)3O5 + 3H2O}$$
The only equation for the combustion I have found is this one:
$$\ce{2C6H7N3O11 + 9/2 O2→12CO2 + 3N2 + 7H2O}$$
I am concerned about the possibility of $\ce{NO2}$ gas being created instead of or in addition to $\ce{N2}$ gas, my understanding of chemistry if fairly basic so I was hoping someone could confirm whether or not the production of $\ce{NO2}$ is likely.

No, it's not likely, but you should beware of strong heat and possibility of explosion.

You may also get CO out of the reaction. Just keep the demonstration area well away from the students, good cross ventilation, the quantity small, and a fire extinguisher close at hand. I've found the best way to ignite is by electrical spark. Place the guncotton between the electrodes of a jacob's ladder, stand back and turn it on. If your reaction is not fully complete you may have residual cotton burning afterwards and falling down on the counter, so best to place on a fireproof mat.

Thanks for the suggestion, I will look into the jacob's ladder idea.

Your nitration reaction does not include the presence of sulphuric acid, which keeps the formation of water from slowing down the nitration by diluting the remaining nitric acid. Are you aware of this?

Flash paper is commonly used in classroom demonstrations, I think, and it is made of nitrocellulose.  A journal article from 1995 describes rapid creation of flash paper through nitration with nitric acid.
I have never done this procedure myself, but looking through it the biggest safety risk to my eyes is using and handling concentrated nitric and sulfuric acids.  $\ce{H2SO4}$ and $\ce{HNO3}$ are very dangerous especially when concentrated.  Full PPE when handling these is an absolute must.  I'm glad to hear in your question that you are already thinking about safety requirements at this step.
Another important risk is the inherent risk of lighting anything on fire.  Nitrocellulose is an energetic material.  Since it is heavily nitrated, it needs less oxygen gas to completely burn than things like wood, paper, or even gasoline: a lot of oxidizer is already built into the molecule.  This risk is best managed by only ever igniting small quantities of nitrocellulose, and of course by storing any larger quantities of the nitrocellulose far away from sources of ignition.  Full PPE is a good idea too.  Perform the combustion in a well-ventilated area.
I'm less worried about the particular risk incomplete combustion forming dangerous levels of $\ce{NO_{x}}$, especially if you limit the combustion to small sheets of nitrocellulose at a time.  High levels of $\ce{NO2}$ are unlikely to form via incomplete combustion; if large amounts did form, since $\ce{NO2}$ is a strong oxidizer, it would react with remaining nitrocellulose to less toxic, more reduced nitrogen oxides or more likely to nitrogen gas.
Small amounts of $\ce{NO2}$ do form from high-temperature combustion of anything in air, but if you use small amounts of flash paper, exposure of your students to $\ce{NO2}$  is likely to be background sources in the atmosphere will vastly exceed anything produced by your demonstration.
All that said, don't take too much safety advice from the internet.  Talking to experienced chemists in person who have done this demonstration is probably the single best thing you can do to stay safe.
Update: I should explicitly say that although I have never made flash paper, I have burned small pieces of it on numerous occasions, never with any problems or even noticeable amounts of smoke or fumes.


text:
How is streaming potential measured?
This paper describes the measurement of streaming potential using a potential divider configuration. However it is unclear where the potential difference is created. From the diagram it appears that it is created across the length of the flow channel, so at one side there is a low potential and the other a high potential. However this doesn't seem to agree with the description of streaming potential. From other sources it appears that the potential difference is across the outside of the flow channel near the surfaces, to the inner middle of the flow channel (at the point furthest away from the surface). Which is correct?
EDIT: there isn't a tag for electro-chemistry, so I have just put this under battery chemistry. However this is not necessarily for a battery.

Any chance you could share the paper or the figure you are mentioning?

It should appear as a hyper-link in the first two words of the question. If not, here:
http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1548283&tag=1

Yes, but I don't have access to the paper. That is why I was asking.

Ah okay i'll add the figure to the question. Just give me 30mins, im in a meeting.


text:
Confused on normallity
So normality is 'equivalents' over liters solution right? I'm having a hard time understanding exactly what an equivalent is. At first from one thing I read i thought it was just the H released from acids in the reactions but then something else started talking about ti with compounds without H, and in double/single displacement reactions. From part of what I read it sounds like a single would have 1 eq and 2 for double, but what found to read was pretty brief so I'm not fully sure. Can someone please explain these a bit better?

Normality = Equivalents of solute/liters of solution
Equivalents are determined by the number of electrons exchanged (or protons in water solutions, or yet other ions in different solvents). For example, if HCl is used to neutralize an alkali, then one mole of HCl will neutralize one mole of NaOH, but it takes two moles of HCl to neutralize one mole of Ca(OH)2 or three moles of Al(OH)3, because they have, respectively, two and three active OH- groups.
There are some pretty clear explanations in more detail at Molarity, Molality and Normality and at Normality.
And then there's Mel Brooks Abby Normal...


text:
Is there any abbreviation for hydrates?
I'm getting tired of writing out the full formulas for hydrates, adding $\ce{.$x$\,H2O}$ to all formulas. I'm wondering if there's standard shorthand for these.

Well, instead of $\ce{.$x$,H2O}$ one can use $\ce{(H2O)_x}$. Going cryptic, instead of 3 symbols for water one can try to use alchemical "" symbol instead, e.g. like $\ce{CaSO4^{5}}$ :)

There is, as far as I know, no shorthand for that, but you might be able to omit it completely in a first order approximation. The closest might be adding $\ce{(aq)}$ when working in aqueous environment.
Adding $\ce{.$x$\,H2O}$ to the formula usually indicates, that it is a solid and that there are $x$ water molecules enclosed in the crystal structure. It depends on what you are looking at, which kind of reaction, what you might be able to omit.
For example, when you are describing a drying process, it is not possible:
$$\ce{\overset{blue}{CuSO4.5H2O} ->[\Delta] \overset{white}{CuSO4} + 5H2O}$$
When you are working in aqueous solution it is usually not important, as there is much more water around it, hence
$$\ce{\overset{blue}{CuSO4.5H2O} + Na2S -> \overset{black}{CuS} v + 2Na+ + SO4^{2-} + 5H2O}$$
is essentially the same as
$$\ce{CuSO4~(aq)  + Na2S~(aq) ->CuS~(s) v + 2Na+ ~(aq) + SO4^{2-}~(s)}$$
If you are describing a redox reaction, it might be the same, hence the following should be fine
$$\ce{CuSO4~(aq) + Zn~(s) -> Cu~(s) v + Zn^{2+}~(aq) + SO4^{2-}~(aq)}$$
Often you can omit the $\ce{(aq)}$ or $\ce{(s)}$ all together.
So depending on what kind of chemistry you are talking about, some information can be implied and therefore be omitted. If you add an example of what kind of reactions/ substances you are working with, I can expand my answer to take that into account.


text:
Why does the electron configuration for some elements not follow the diagonal rule?
I'm doing a high-school assignment, and I came across a question that I didn't quite understand.

Explain how the electron configurations for the following elements do not follow the diagonal rule:

Gold
Curium
Thorium
Molybdenum
Palladium

From what I understand, all of these elements do follow the diagonal rule. How exactly do they not follow this rule?

Look them up—they're not what you think. For example, gold is [Xe] 4f14 5d10 6s1, not [Xe] 4f14 5d9 6s2.

The reason for these exceptions (from the diagonal rule) is that some elements are more stable with fewer electrons in some subshells and more electrons in others:

Electronic configuration of molybdenum: $[\ce{Kr}] \ce{4d^{5} 5s^1}$, instead of  $[\ce{Kr}] \ce{5s^2 4d^{4}}$ according to the diagonal rule, because a half-full $\ce{4d}$ subshell and a half full $\ce{5s}$ subshell are more stable than $\ce{4d}$ filled with four electrons and a full $\ce{5s}$ subshell.

Electronic configuration of Gold: $[\ce{Xe}] \ce{4f^14 5d^{10} 6s^1}$, instead of  $[\ce{Xe}] \ce{4f^14 5d^{9} 6s^2}$ according to the diagonal rule, because a full $\ce{5d}$ and half full $\ce{6s}$ subshell is more stable than $\ce{5d}$ filled with 9 electrons and a full $\ce{6s}$ subshell.

Electronic configuration of Palladium: $[\ce{Kr}] \ce{4d^{10}}$, instead of  $[\ce{Kr}]\ce{ 5s^2 4d^{8}}$ according to the diagonal rule, because a full $\ce{4d}$ orbital is more stable than $\ce{4d}$ filled with eight electrons.

Electronic configuration of Curium: $[\ce{Rn}] \ce{7s^2 5f^7 6d^{1}}$, instead of  $[\ce{Rn}] \ce{7s^2 5f^8} $ according to the diagonal rule, because a half full $\ce{5f}$ orbital is more stable than $\ce{5f}$ filled with eight electrons.


text:
Hydrogen fuel cell - why do the H+ ions move through the electrolyte
Consider the following hydrogen fuel cell

http://butane.chem.uiuc.edu/pshapley/Enlist/Labs/FuelCellLab/FuelCell.html
At the anode, hydrogen is oxidised (losing electrons). My first question is this: why is the anode negative if it is attracting electrons?
My second question refers to the fact that once the hydrogen atoms are iodised, the H+ ions move through the electrolyte towards the oxygen ions at the cathode, reacting to form water.
If the anode is negative, why would positive H+ ions move away from it?

Here is a better picture: https://phys.org/news/2010-08-catalyst-platinum-nanoparticles-conk-out-free.html
The anode is labeled negative because that is where the electrons enter the wire. The cathode is labeled positive because that is where the electrons leave the wire (so it is from the perspective of the consumer of electricity).

At the anode, hydrogen is oxidised (losing electrons). My first question is this: why is the anode negative if it is attracting electrons?

It is not attracting electrons, it is attracting hydrogen atoms. The electrode binds the hydrogen and releases the electrons. The electrons are then able to flow out through the wire.

My second question refers to the fact that once the hydrogen atoms are iodised, the H+ ions move through the electrolyte towards the oxygen ions at the cathode, reacting to form water. If the anode is negative, why would positive H+ ions move away from it?

Once the electrons have returned to the cathode, the cathode is more negative than the anode. The positive ions move to the cathode and the negative ions move to the anode.

In a galvanic cell, the anode is positively charged.  The cathode is negatively charged.  A good mnemonic I learned a while ago is that anions go to the anode, and cations go to the cathode.  This is because opposite electric charges attract.  So negatively charged anions are attracted to the positively charged anode.  And positive cations are attracted to the negative cathode.
The electron itself was originally discovered as the component particles of cathode rays from cathode ray tubes.  That electrons come out of the cathode is a hint about cathode charge: the buildup of electrons on the cathode eventually results in the emission of an electron beam.  If electrons are building up, the cathode must be negative.

in the diagram in my question, the anode is negative.

Can you add a link or reference to where you got your diagram from in the question?

The writeup at wikipedia is pretty good.  Check it out: http://en.wikipedia.org/wiki/Anode .  Reading it made me realize that I need to add a few qualifiers to my answer which I will do now.

Reference added. I still dont get why the anode is negative in the diagram.

Is the diagram could be right if it were a "reverse" fuel cell where electrical energy and water is converted to hydrogen.  This is a an electrolytic cell (the reverse of a galvanic cell).  In this mode the anode is negative.

Also, have you considered whether your diagram is wrong?

haha quite a bold statement! Yes i considered it, but a quick google image search reveals that most of the hydrogen fuel cell diagrams all have a negative anode.

You should check the standard reduction potentials of each reaction in the cathode and anode. In a fuel cell (which is a type of galvanic cell), the hydrogen oxidation reaction occurs on the anode at 0V vs RHE while the oxygen reduction reaction on the cathode occurs at around 1.23V vs RHE. As you can see, the anode reaction occurs at a negative potential compared to the cathode reaction. That is why the anode is marked negative and the cathode is positive.

At the anode, hydrogen is oxidised (losing electrons). My first
question is this: why is the anode negative if it is attracting
electrons?

The anode is negative because it is accumulating electrons from the oxidation of hydrogen atoms.

My second question refers to the fact that once the hydrogen atoms are
iodised, the H+ ions move through the electrolyte towards the oxygen
ions at the cathode, reacting to form water. If the anode is negative,
why would positive H+ ions move away from it?

As there is an accumulation of electrons at the anode, electrons repel each other and are attracted towards the cathode where there is a deficiency of electrons. At the same time, the cathode is becoming the "meeting place" for the incoming electrons and provided oxygen atoms which attract the hydrogen ions to drive the oxygen reduction reaction that forms water. So H+ ions are attracted to the cathode due to the increasing number of electrons in the cathode.

Anode as such is positive and cathode is negative.Hydrogen gets oxidized at the anode by giving an electron,so the anode becomes rich in electrons. The flow of electrons is from anode to cathode. Hence anode which is rich in electrons, compared to cathode, is represented as negative. Cathode being negative attracts the hydrogen ions, which gets reduced when it receives electrons from anode via an external circuit.

"Hence anode which is rich in electrons, compared to cathode, is represented as negative." - does this not contradict your first line? sigh I am so confused


text:
Can I make a buffer from two different weak acids or bases?
Can I make a buffer using two different acids/bases? For example, adding 1 mole of acetic acid and 1 mole of ammonia?
The only information I can ever find is that I can make a buffer using only an acid (or base) and its conjugate base (or conjugate acid). So, carbonic acid and carbonate, $\ce{H2CO3}$ and $\ce{HCO3-}$.
Why do I need a conjugate acid-base pair and not two random weak acid-base pairs to make a buffer?

In your specific case, a combination of acetic acid $\ce{HC2H3O2}$ and ammonium $\ce{NH4+}$ would not work, since both species are acids.

@BenNorris you, sir, are a pedant. I think Nova meant ammonia NH3. Not everyone here is a native English speaker.

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

Yes, having multiple weak acids and/or bases having different pKa values would give buffering capacity in more than one pH range, or over a greater pK range.
There is a term "universal buffer" which refers to a buffer having multiple pKas spaced thoughout the pH ~2-12 range to give some buffering capacity over the enter pH range.
As explained in A Neutral Buffered Standard for Hydrogen Ion Work and Accurate
Titrations Which Can be Prepared in One Minute J. Am. Chem. Soc., 1932, 54, pp 1911–1912, ammonium aceate has the property that if it is added to pure water it produces a buffered solution of pH 7.00, accurate to a hundreth of a pH unit.

In theory you can, but it depends on the way you prepare the buffer.
A buffer solution is prepared by mixing a weak acid (or base) with a strong salt of that acid
For example you can prepare the following buffer in two ways:
$\ce{HOAc + NaOAc}$
1:
Mixing directly a solution of the salt with the acid
2:
Adding (let's say dropwise) a solution of $\ce{NaOH}$ to the acid. In this case $\ce{NaOAc}$ will be formed in situ.
What you cannot do!
Prepare a buffer by mixing $\ce{HOAc}$ with $\ce{NaCl}$ for example
On the other hand you can mix two different buffers to get a new one with a different pH range like in the case of universal buffers.

The $\ce{OAc}$ shorthand is typically used with $\ce{O}$ preceeding the $\ce{Ac}$, unless the connection continues to the right, in which case $\ce{AcONa}$ would be preferred.

Please refrain from readding clutter once it has been edited out.

@Jan I don't see why I can't include that line. Any explanation?

In short: the help centre which includes a note about ‘taglines’. And hope this helps is one of the most generic taglines out there. It does not much to answer the question and is thus perceived as noise and clutter. Check out this tongue-in-cheek meta post and a similar but more earnest meta-post.

Of course it doesn't answer anything, I agree, it was just to be friendly. That's why it is at the end of the answer, to separate it from the rest of the content I'm aware that it's irrelevant and probably unnecessary, but is not disturbing anything there. Also thanks for the other corrections my native language is spanish and I'm triying to do my best.

@Marange Hi.  Thanks for offering the answer.  Please see http://meta.stackexchange.com/questions/2950/should-hi-thanks-taglines-and-salutations-be-removed-from-posts as to why someone was trying to be helpful and remove it.  I'm going to leave the tagline in the post for now, but know that this behavior is discouraged for the reasons outlined in that post.  Thanks and welcome!

@ jonsca ♦ Ok I get it and removed it, no problems with it.


text:
Why are oxime geometrical isomers stable?
231st page of Clayden OC 2nd edition:

Unlike the geometrical isomers of alkenes, however, those of an imine
  usually interconvert quite rapidly at room temperature. The
  geometrical isomers of oximes on the other hand are stable and can
  even be separated.

Why are oxime geometrical isomers much more stable than the geometrical isomers of immines?

I assume that by "stable enough to be separated", you mean that the two isomers do not equilibrate. Could you provide a reference that this is true? I can think of examples of imines that should be unlikely to equilibrate, except, for example in the presence of aqueous acid (but then oximes should equilibrate under those conditions, too).

It is the same reasoning that is used to predict relative barrier heights in amines.
In the example pictured below, the imine starts with the nitrogen lone pair in "more or less" an $\ce{sp^2}$ orbital.  In the transition state for interconversion, the lone pair is in a $\ce{p}$ orbital.

The term "more or less" is important.  If we start with a molecule that has the nitrogen lone pair in an orbital that has a little bit more p character than $\ce{sp^2}$ (say $\ce{sp^{2.2}}$ for example), then it is already closer to the $\ce{p}$ hybridized transition state and the activation energy required to reach the transition state will be reduced.  On the other hand, If we start with a molecule that has the nitrogen lone pair in an orbital that has a little bit less p character (more $\ce{s}$ character) than $\ce{sp^2}$ (say $\ce{sp^{1.8}}$ for example), then it is further from the $\ce{p}$ hybridized transition state and the activation energy required to reach the transition state will be greater.
In the case of oximes ($\ce{X=OH}$) or other molecules where the substituent on nitrogen is electronegative (electron withdrawing), the electron density in the $\ce{N-X}$ bond is shifted towards the electronegative atom.  Consequently there is less electron density in the nitrogen portion of this bond.  As a result the nitrogen rehybridizes. Since $\ce{s}$ orbitals are lower in energy than $\ce{p}$ orbitals, orbitals with increased $\ce{s}$ character are better able to stabilize electrons. Since there is less electron density in the orbital nitrogen is using for this bond, $\ce{s}$ character is removed from the orbital.  This "extra" $\ce{s}$ character is used to stabilize the orbital containing the lone pair of electrons.  The lone pair orbital now has more $\ce{s}$ character and is further away from the $\ce{p}$ hybridized transition state - the barrier to reach this transition state increases.  If our substituent is an alkyl group (X=R, an imine), the carbon is less electronegative then nitrogen, there will be increased electron density in the nitrogen orbital used to form the $\ce{N-R}$ bond.  The nitrogen will hybridize such that there is more $\ce{s}$ character in this orbital so as to better stabilize these electrons.  As a consequence, there will less $\ce{s}$ character (more $\ce{p}$ character) in the lone pair orbital.  The lone pair orbital is therefore closer to the $\ce{p}$ hybridized transition state and less energy will be required to reach the transition state.
As a result of this reasoning $\ce{NF3}$ has a higher inversion barrier then $\ce{NH3}$ and oximes have a higher barrier to inversion than imines.

That seems like Bent's rule.

Yes, that's right.

It's a matter of energetic barrier between the $E$ and $Z$ stereoismers. It's  sufficiently low for imines. So, it's relatively difficult to isolate $E$ or $Z$ stereoismer.
In the case of oximes, the energetic barrier of interconversion between the two stereoisomers is much higher and allows the isolation of the two stereoisomers.
For more details, please see http://chemweb.bham.ac.uk/~coxlr/Teaching/1st_Year/Carbonyl_Group/Carbonyl_pdfs/lecture%207%20%28student%20HO%29.pdf


text:
Intramolecular hydrazide condensation
Say I have acetohydrazide ($\ce{CH3CONHNH2}$).
Is it possible, that when I add an acid, an intramolecular condensation reaction occurs, forming the cyclic compound 3-methyl-1H-diazirine:

Furthermore, in general, the mechanism of such a reaction is nucleophilic substitution, right?

Nope, no chance. Three-atomic rings are extremely strained, even more so with double bonds in the ring, so they form only in reactions with extremely high-energetic particles, such as carbenes.

But if we, for a second, discount the strain factor, such a reaction will be possible right?

Your starting material is a hydrazide. Where is the semicarbazone?

@DhananjayGupta The usual synthesis of imidazole (see wiki for the structure) involves condensation of o-phenylenediamine with formic acid. But you can't dismiss strain factor here.

To avoid the strain, it would be much more likely for an intermolecular condensation to occur.

The concept of 'discount the strain factor' is nonsensical.  Furthermore, hydrazides are relatively resistant to nucleophilic addition-elimination

Your reaction product is a double bond isomer of a diazirine.
Halogene-substituted derivatives of the latter are typically prepared via oxidation of amidines with hypochlorites (Graham oxidation).
Heating your hydrazide in an aqueous solution of a mineral acid will only turn it to the corresponding carboxylic acid.


text:
The meaning of different parts in the naming of glucose
There are two main types of glucose, $\alpha$-D-(+)-glucose and $\beta$-D-(+)-glucose.
Can anyone please explain what is $\alpha$,D or (+)?

So (maybe by permutation and combination), how many types of glucose are there?

A highly related question: http://chemistry.stackexchange.com/questions/7308/whats-the-difference-between-alpha-glucose-and-beta-glucose

@CurtF. It would be better to consult meta before introducing a new tag to a whole lot of questions. I personally did not see any benefit in the new tag. The community might disagree with me on this, so please go ahead and post to meta about it.

Sorry I did not realize the protocol.  I probably won't write a post on meta because I did not maintain my own separate records of the posts tagged, and without those posts to display, I can't make a good argument for the tag.  I don't feel like re-searching through old questions to find the ones I already found.  I guess we won't have a "carbohydrates" tag here after all, at least not any time soon.

Glucose can exist in the open form or cyclic form.

In the open form, there are only two types, D-glucose (illustrated in Fischer projection above) and L-glucose which is the mirror image of the above.
Experimentally, D-glucose rotates plane polarize light clockwise, hence the symbol "+", and L-glucose rotates plane polarize light counterclockwise, hence the minus symbol.  For more information on the meaning of + versus - see: http://en.wikipedia.org/wiki/Dextrorotation_and_levorotation
In the cyclic form, there is an additional chiral center, so there is alpha (where 3 OH groups are on the same side of the ring) vs beta (where 2 OH groups are on each side of the ring) as ilustrated in the drawing of the question.
Each of alpha and beta have a D and an L form, so there are 4 types in the cyclic form.
So in summary, there are only two types in the open form, but 4 types in the cyclic form.

So if "D" and "(+)" have the same meaning why do we have to write both of them? Or is there something like "D-(-)" ?

No, it just happens to be that way in this case.  For glucose, D experimentally is + and L experimentally is -, but for other molecules it can be reversed.  Read more about this here: http://en.wikipedia.org/wiki/Chirality_(chemistry)

It's worth noting that in aqueous solution the forms interconvert fairly rapidly.  So for D-glucose, or D-(+)-glucose if you prefer, three species are interconverting: the open chain D-glucose, the cyclic $\beta$-D-glucopyranose, and the cyclic $\alpha$-D-glucopyranose.  (Glucopyranose is a name that expicitly indicates the six-membered ring closed form of glucose.)


text:
How can Nernst equation apply in this redox reaction?

$$\ce{CuSO4 + Zn -> ZnSO4 + Cu}$$
What is the cell potential of the voltaic cell with copper and zinc electrodes if the system is at 50 degrees and the solutions are both at 1 mol? How can it be equal to the standard potential from the nernst equation? $$E_\text{cell} = E^\circ – \frac{RT}{nF}\ln Q$$

If the solutions are at 1 mol concentrations each then $Q = 1$ and $\ln Q = 0$ resulting in $E_\text{cell} = E^\circ$.
But the above basically shows that concentration affects cell voltage but temperature doesn't have an effect on cell potential – which isn't the case. How can I calculate how different temperatures affect the cell voltage?

The flaw in your reasoning is that you are believing that $E^\circ$ is a constant, when really it is a function of temperature.
See Standard Electrode Potentials and Temperature Coefficients in Water at 298.15K for a linear approximation of the temperature dependence of $E^\circ$.


text:
What is the temperature of heat generated from LPG gas?
I'm working on a project and I require around 300-400 degree Celsius of temperature. As of now I'm taking the normal LPG gas that is available in our house for cooking purposes for my project, but I'm not sure whether it can reach 300-400 degree Celsius or not? I can definitely use multiple burners but how much heat/temperature does 1 burner generate?
Maybe I'm wrong because heat is counted in calorific value, but then how do we know the temperature of that calorific value? Or is there some relation which says that x amount of LPG would give x degree Celsius or maybe x calorific value? But yet again, how can I know the temperature?

LPG is mostly propane and butane both of which have adiabatic flame temperature near 1970 degrees C when burned in air.  See Flame Temperatures of some Common Gases.
So 300-400 degrees C is certainly reachable.

That looks promising, but here comes one contradiction (I may be wrong), when you heat water on your stove, it takes around a minute or so in order for evaporation to start which takes place at 100C. Now if that can take around a minute,  then how can it be 1970C flame temperature? Or is the rest of the heat getting used up by the pan in which water is being heated?

Because the time it takes for a material being heated to reach a certain temperature, is a function of the amount of material, the heat capacity of the material and the rate of heat transfer to the material (including both heating mechanisms and cooling mechanism; for example the water could be getting heated from below and cold from above).

I wonder that is there any way to find out that "time" that heat would take for its transfer in that material?

in what material?

The material that I'll be using would be Stainless Steel. Or is there any formula to find "time" in any material? Thanks a lot again!

OK, the heat capacity of stainless steel is 500 Joule per kg per degree C.  But that is only one piece of the puzzle.  If I put a stainless steel needle into a gas stove flame to sterilize it to get a splinter out of my skin, the tip of the needle glows orange within one second, meaning it is over 1500 degrees C.  You will need to consider amount of material, surface area, and the temperature all surface are exposed to, and a heat transfer coeffient.  Read about newton's law of cooling: http://en.wikipedia.org/wiki/Convective_heat_transfer

Really, thanks a lot! BTW I'm working on a project, so if it doesn't bother you then can you share your mail address or facebook's username with me?


text:
Are there any magnetic particles that are safe to inhale?
In medicine, there is an (old?) experiment where a subject is asked to inhale radioactive Xenon gas, and radiation counters are placed at different positions near the lungs. Using this experiment, it has been demonstrated that the (alveolar?) volume of the lungs is highest in the bottom parts, and smallest in the top parts.
I don't know if this experiment is still used today, since inhaling radioactive gas has the obvious radiation related risks.
I was wondering if there are some magnetic particles that would be safe to inhale? I know that for example helium does not diffuse into blood, and it does not permanently bind into anything. However, I think it does not have any suitable magnetic properties. Let's assume the detector is very very sensitive to any magnetic fields (so, a super-conducting quantum interference device).

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. || Not bad, not bad! We're starting to get good questions from everyone, basically.

By "magnetic" I assume you mean ferromagnetic.  Ferromagnetism is generally viewed as a property of bulk matter.  It has long been debated whether a gas could display "magnetic" properties.  A few years ago there was a claim that a magnetic gas had been produced.  I think the debate is still going on.

Current state of art in safe imaging of biological objects is NMR-tomography. It relies on magnetism of nucleus, specifically on interaction of radiowave, nuclear spin and magnetic field. It allows very detailed imaging, for example a detailed structure of inteconnections in a living brain.

Ferromagnetic nanoparticles do exist, and are more or less safe to inhale in small amounts (though they ARE a health risk, just like any mineral dust). On the contrary, carefully controlled inhalation of selected radioactive gas in low concentration is realtively (relatively!!!) safe. Humans do have radioactivity tolerance above zero and can tolerate minor expositions without negative effects. Nobody cares about plane flights, though they does expose passengers to above-average radioactive background.

$\ce{O_2}$ is paramagnetic.

source: http://www.chem1.com/acad/webtext/chembond/CB-images/liquidoxygeninmagnet.jpg
The stable Xenon isotope Xe-129 has been used in MRI studies:
Hyperpolarized Xenon MR Imaging of the Brain

A radically new magnetic resonance imaging (MRI) technique, using hyperpolarized 3He and 129Xe, is being developed to produce high-contrast images of important body tissues that have resisted conventional MRI techniques. Hyperpolarization of the noble gas nuclei is achieved by spin exchange with rubidium atoms that have been optically pumped into one electronic spin state with the use of circularly polarized light from a high-power diode laser system. Hyperpolarization can increase the detectability of 3He and 129Xe up to a hundred thousand times. This technology is showing dramatic results for diagnostic imaging of the lungs, brain, and other parts of the body. Most notably, we have achieved high-resolution dynamic images of the lungs and airways using 3He, and measured cerebral perfusion in the brain using 129Xe.

See video of lung being imaged: http://www.spl.harvard.edu/archive/HypX/about.html

Does NMR/MRI count here? If so, then any NMR active nucleus should be considered 'magnetic'?

As @DavePhD states, hyperpolarized xenon is used as an MRI [NMR for the non-PC] contrast medium and is not radioactive. For a study on the lungs, see http://ultra.bu.edu/projects.asp?project=mri.


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How can the electronic structure of pentaaquanitrosyliron be explained?
The magnetic moment of $\ce{[Fe(H2O)5NO]^2+}$ is 3.87 Bohr magneton, which implies that there are 3 unpaired electrons. Since iron is in $+1$ oxidation state (due to charge transfer with nitrosyl group), it's electronic configuration is $[\ce{Ar}]\ \mathrm{3d^7}$. Having 3 unpaired electron is only possible if the 4th and 5th electron enter the higher energy orbital $\mathrm{e_g}$ (crystal field theory). The $\mathrm{t_{2g}}$ or the lower energy orbital has 5 electrons. Why do the 4th and 5th electron enter $\mathrm{e_g}$ despite a strong field ligand (nitrosyl group) being present?

As explained in Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of
Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction
Revisited Inorg. Chem. 2002, 41, 4-10, the $\ce{Fe}$ is really 3+ oxidation state.
The $\ce{Fe^3+}$ is spin 5/2 and antiferromagnetically couples to spin 1 $\ce{NO-}$ to yield a S=3/2 state.

The presence (or absence) of a single strong field ligand is rarely enough to tip the entire complex’ balance from high to low spin (or vice-versa). I have often come across questions here that attemt to argue ‘but $\ce{X}$ is a strong field ligand, why is complex $\ce{Y}$ not low-spin?’, and the answer is always that the entire picture must be considered — merely looking at just one ligand is never enough.
To add to the confusion, some complexes are known in both the high and low spin state, indicating that there is a tipping point that can be crossed but which also needs to be reached; and if you only barely reach it you might tip this way or the other. So do not take the presence of a single cyanido or carbonyl ligand and immediately assume that the complex should be low-spin.
Finally, the actual electronic structure of pentaaquanitrosyliron $\ce{[Fe(H2O)5(NO)]^2+}$ must be considered. This complex is a very tricky one because iron is very redox active and nitrosyl is what is known as a very non-innocent (promiscuous?) ligand: it can occur as all three of $\ce{NO+, NO^.}$ and $\ce{NO-}$. To add to the confusion, these can also bond in a linear or bent fashion — and one cannot simply look at a crystal structure and infer the charge of nitrosyl from its bent or linear configuration. And lastly, $\ce{NO-}$ which is isoelectronic to $\ce{O2}$, can adopt both a singlet and a triplet configuration. Depending on which nitrosyl variant you decide to use, the oxidation state of iron will be $\mathrm{+I}$, $\mathrm{+II}$ or $\mathrm{+III}$ in the same order.
For IUPAC nomenclature purposes — which are based on observing a certain configuration and deriving a systematic name without knowledge of the intrinsic chemistry — nitrosyl is always considered a neutral three-electron donor ($\ce{NO^.}$; donating the radical and a lone pair) leading to a formal IUPAC oxidation state of $\mathrm{+II}$. The physical reality was hard to determine. The spin is known to be $\frac32$ which is consistent with a number of different states. The currect state of science is to assume iron(III) ($S = \frac52$) and a singlet nitrosyl anion ($S=1$; hyponitrite?). These then couple antiferromagnetically to give an overall spin state of $S=\frac32$ as DavePhD’s answer already indicates.
What seems to be a very simple problem is, in fact, rather complicated! Indeed, the notation $\ce{\{FeNO\}^7}$ has been proposed for nitrosyliron complexes where the superscripted $7$ denotes the number of electrons in iron’s d orbitals and the ligand’s coordinating orbitals. This allows one to write a structure without having to imply any given actual electronic configuration.