text:
How to judge if a reaction is exothermic or endothermic
Mix $\ce{HCl}$ and $\ce{NaOH}$ together (reactants are at the same temperature) in a calorimeter. The temperature increases, and the book says it's an exothermic reaction. I don't get it. The thermometer is in the solution which means it measures the temperature of the solution. It's the system. The temperature of the system increases, this means that heat flows in. How could this be an exothermic reaction?

The purpose of the calorimeter is to isolate the system and prevent heat from entering or exiting the system. If the temperature increases, then the heat must come from what is inside the calorimeter.

I just confuse about what the system is in this reaction.

if we just put a thermometer in a container with opening to carry out this reaction, still the number on thermometer will increase and the container, which is the surroundings will feel hot because the heat flowed from the system to surroundings. But what is the system? When we measure the temp of solution, it also increases.

Is the solution surroundings or system?

The system is the solution, colorimeter, thermometer and anything inside the container (colorimeter). If it is open on top, the heat can transfer to the outside. If the solution temperature increases above room temperature, the heat from the room could not have entered the system. The system must have produced the heat.

Does this answer your question? How can decreasing in temperature indicate an endothermic reaction?

This is one of the most commonly misunderstood principles in thermochemistry - if you can wrap your mind around it, you are most of the way there!
The good news is, there is an easy way to do that. In any physics problem (thermochemistry is really physics) you have to be very careful with how you define the system and the surroundings. This is important for two reasons.

It lets you write energy balance equations across the system boundaries (so we can use math to solve the problem)
By choosing the right system boundaries, we can make the math much, much easier.

Since thermochemistry at this level always involves the same types of problems, chemists have done the hard part for you and already selected the best system boundaries.
The system is whatever is in the reaction equation.
That's it. The entire rest of the universe is the "surroundings."
Looking back at your question:

The thermometer is in the solution which means it measures the temp of the solution. It's the system. The temp of a system increases means heat flows in, how could this be an exothermic reaction?

If the reaction molecules are the only thing in the system, then that means the water, the calorimeter, the themometer, you, me, the room - everything else - counts as surroundings. Now, the water did heat up. We know this because the thermometer absorbed some heat and showed a temperature increase. Since they are all part of the surroundings, the heat must have come from the system. Since heat left the system, the sign of $q_{reaction}$ is negative, and the reaction is exothermic.
As a teacher, I have found that getting system boundaries and the signs right is one of the biggest problems students have. If you focus on getting that concept down early - and really understanding why they are that way, you are well on your way.

HCl + NaOH give NaCl plus water.  Add up the free energies of formation of the species on each side and compare. Binding energies are negative.  The more strongly bound side is created with an exotherm.  The outside is irrelevant.

I kinda confuse what's the system. But now I found out that the calorimeter is isolated. So the solution is system, and the heat the reaction released turned out heating the solution up, it that what happened?

Yes, you are now correct. Heat did not flow into the system; heat was liberated by the system components. Hydronium ion reacted with hydroxide ion to form bonds. Bond forming or bond "improvement" is exothermic.

Also I wouldn't say that $\ce{NaCl}$ was created unless we know we were mixing liquid $\ce{HCl}$ with molten $\ce{NaOH}$ or if we were otherwise somehow working with pure $\ce{HCl}$ and pure $\ce{NaOH}$. Throw water into the mix and what you get isn't $\ce{NaCl}$ but mostly solvated  $\ce{Na^+}$ and $\ce{Cl^-}$ ions.


text:
Rate Constant Units and Eyring Equation
Rearranging the Eyring equation leads to the following:
$$\Delta^\ddagger S^\circ = R \ln{\frac{k \times h}{{k_\text{B}}{T}}}+\frac{\Delta^\ddagger H^\circ}{T}$$
where $k$ is the rate constant, $h$ is the Planck constant, $R$ is the universal gas constant, $k_\text{B}$ is the Boltzmann constant, $T$ is temperature, $H$ is enthalpy, and $S$ is entropy. The entropy term should ultimately be in units of $\mathrm{\frac{J}{mol\ K}}$. The right term of the equation will clearly yield the correct units. For the left term, the universal gas constant can be expressed in $\mathrm{\frac{J}{mol\ K}}$, which means that the log term should be dimensionless. For the bottom part of the fraction, the units evaluate to $\mathrm{J}$. For the top part, the Planck constant has units of $\mathrm{J\cdot s}$. To get the top part of the fraction to evaluate to $\mathrm{J}$, the rate constant would have to be in units of $\mathrm{s^{-1}}$. For first order reactions, this is indeed the case. For reactions of higher order, this will never work out to the correct units.
This makes me think that the Eyring equation is simply using the unitless values of these variables. This still leads to some ambiguity about which value of the rate constant to use. Take for instance a second-order reaction in which the rate constant is in units of $\mathrm{M^{-1}\ s^{-1}}$. If we rearrange the units to SI units ($\mathrm{m^3\ mol^{-1}\ s^{-1}}$), the value of the rate constant changes. So is the rate constant $k$ in the Eyring equation simply the value of the rate constant when that rate constant is expressed in certain units? And if so, are the concentration units always expressed in molarity? I kept missing a homework problem because I was converting $k$ from molarity to SI units, which makes me believe this is indeed the case.

Using Transition State Theory (TST) one has to be aware of at least some of the many flaws of it. And may also often lead to wrong assumptions. However, it is a wonderful tool to describe basic kinetics in a theoretical manner.
The most important/ fundamental assumptions are

separation of electron and nuclear movement (comparable to Born-Oppenheimer approximation)
(all possible) transition states (TS) are in equilibrium with the reactants  $$\ce{A + B <=> [AB]^{\ddagger}}\tag1$$
combined Boltzmann statistic for reactants and TS
All molecules that reach the transition state from the reaction coordinate of the reactants will leave the TS to the reaction coordinate of the product(s) - and vice versa
\begin{align}
\ce{A + B <=> &[AB]^{\ddagger}->P}\tag{2a}\\
\ce{P <=> &[AB]^{\ddagger}-> A + B}\tag{2b}
\end{align}
at the TS the transition vector can be separated from all other motions and hence may be treated classical as translation

In other words TST reduces a very complicated multidimensional problem to a one-dimensional translation. That implies that all treated molecules must be connected on a potential energy hypersurface with one vector. (Borrowing from the amazing wikipedia)

Hence reactions that are described with TST are of first order by default. One would have to make various adjustments to treat higher orders (and I would assume that not matching units would be of minor concern).

You are correct in your analysis that the rate constant is first order with units $\ce s^{-1}$. The problem you have is that the Eyring equation you use is not complete as I try to show below after some introduction. (The derivation I use can be found in several texts, but I outline that in Steinfeld, Fransisco & Hase, 'Chemical Kinetics & Dynamics')
The idea behind Transition State Theory (TST) is that the reactants are assumed to be in thermal equilibrium with the transition state. This is a small region at the top of the potential energy barrier that separates reactants and products. A contour plot of an A + BC reaction ($\ce{F + H2}$) is shown in the first figure (together with a single reaction trajectory).

The transition state is labelled with a dot and lies at a saddle point, which is to say that the potential is convex along the reaction path and concave in a perpendicular direction to this, see second figure.
Following the reaction path from right to left (contour first figure) we see that on collision the reactants climb the (rather small) barrier and cross to products. The product is formed with some considerable vibrational energy. The profile along the reaction path (or reaction coordinate) is the familiar potential energy hill between reactants and products.

There are a number of assumptions in deriving TST. The two basic ones are
(a) that nuclear and electron motion can each be separated (similarly to the Born-Oppenheimer approximation use to solve the Schrodinger equation), and it is further assumed that the energy for translation, vibration and rotational motion can be treated additively),
(b) the reactant molecule’s energy is distributed among its states, i.e. translation, vibration, rotation according to the Maxwell-Boltzmann distribution.
Other assumptions are necessary for the theory and are that:
(c) the transition state is crossed only once, i.e. it is a point of no return,
(d) The motion along the reaction coordinate at the transition state can be separated from other motions and treated classically as a translation.
(e) even in the absence of equilibrium between reactants and product molecules the energy in the modes of the transition state is distributed according to Maxwell-Boltzmann. This assumption is sometimes called the ‘quasi-equilibrium hypothesis’.
To calculate the rate constant $k$, the speed of crossing the transition state and the number of molecules doing so has to be calculated. Let the reaction be a general second order reaction - $\ce{A + B -> C}$ with rate equation:
$$\frac{\mathrm d[A]}{\mathrm dt}=-k\ce{[A][B]}$$
In the reaction $\ce{A + B -> X^\ddagger -> C}$, where $\ce{X^\ddagger}$ is the transition state, suppose that there is a small region of length $\delta$ at the top of the barrier that the reactants must all pass through.
Imagine this as a very thin dividing surface, reactants on one side and products on the other. At equilibrium the rate of forwards and backwards motion through this region must be the same, consequently if the concentration is $N^\ddagger$ then:
$$N^\ddagger = K^\ddagger\ce{[A][B]}$$
Now suppose that all the products are removed from equilibrium and as these are half of the total $N^\ddagger$ the number going forwards to products
$$N_\mathrm f^\ddagger = K^\ddagger\frac{[A][B]}{2}$$
which is the quasi equilibrium hypothesis.
The rate of reaction to reactants to products is given as:
$$\frac{\mathrm dN}{\mathrm dt} = \frac{\delta _{N^\ddagger}}{\delta_t}$$
Here, the number of transition states with velocity $v$ to $v+\delta v$ in one direction is:
$$\delta_{N^\ddagger} = \frac{N^\ddagger}{2}$$
The average time to cross the dividing surface at average velocity $v_\mathrm a$ is:
$$\delta_t = \frac{\delta}{v_\mathrm a}$$
where $\delta$ is the thickness of the dividing surface and by substituting gives
$$\frac{\mathrm dN}{\mathrm dt} = \frac{N^\ddagger}{2}\cdot\frac{v_\mathrm a }{ \delta }$$
The average velocity in one direction can be calculated in the standard way$^1$ assuming equilibrium distribution of velocities and is:
$$v_\mathrm a = \left(\frac{2k_\mathrm BT}{\pi\mu}\right)^{1/2}$$
where $\mu$ is the reduced mass. Substituting again gives:
$$\frac{\mathrm dN}{\mathrm dt}=\frac{N^\ddagger}{2\delta}\cdot\left(\frac{2k_\mathrm BT}{\pi\mu}\right)^{1/2}$$
The equilibrium constant is given by:
$$K^\ddagger= \frac{N^\ddagger}{\ce{[A][B]}} = \frac{Z_\mathrm{tot}^\ddagger}{Z_\mathrm A Z_\mathrm B}e^{-E_0/k_\mathrm BT} $$
Where $Z_\mathrm{tot}^\ddagger$, $Z_\mathrm A$ and $Z_\mathrm B$ are the respective partition functions per unit volume and are derived using statistical mechanics.
The $Z_\mathrm {tot}^\ddagger$ partition function is that of the combined species $\ce{AB}$ at the transition state which by the condition (d) above, is separated into two terms - one term for the reaction coordinate($Z_\mathrm r$) and $Z^\ddagger$ for all other $3N-1$ degrees of freedom.
$$Z_\mathrm{tot}^\ddagger = Z_rZ^\ddagger$$
The translational partition function in one dimension is
$$Z_r=\left(2\pi\mu k_BT\right)^{1/2}\cdot \frac{\delta}{h}$$
Thus substituting and simplifying produces:
$$\frac{\mathrm dN}{\mathrm dt}= \frac{k_\mathrm BT}{h}\frac{Q^\ddagger}{Q_\mathrm AQ_\mathrm B}e^{-E_0/k_\mathrm BT}\ce{[A][B]}$$
from which we get the second order rate constant to be:
$$k= \frac{k_BT}{h}\frac{Q^\ddagger}{Q_\mathrm AQ_\mathrm B}e^{-E_0/k_\mathrm BT}$$
The constants $\delta$ and $\mu$, which were used in the derivation, cancel out.  This constant $(k_\mathrm BT/h = \pu{6.25 \times 10^{12} s-1})$ is often called the frequency factor.
The rate constant will have units of $\pu {m3 s-1}$
All that now remains is to calculate the partition functions.
The parameters needed have to be guessed for the transition state as they are generally unknown, no transition state ever having being identified spectroscopically. The values needed are  vibrational frequencies, moments of inertia reactants and shape (linear or bent). The  general partition function is split into a product of terms, $Z_\mathrm {trans} \cdot Z_\mathrm {rot} \cdot Z_\mathrm {vib} \cdot Z_\mathrm {elec}$. This is assumption (a) above, separating electronic and nuclear motions.
In the case of atom plus diatom reaction $\ce{A + BC -> AB + C}$ of which many have been measured - $\ce{O + H2, Cl + H2, H + H2, F + H2}$ etc. - the rate constant equation becomes the monster equation:
$$k=\frac{k_\mathrm BT}{h}\left[\frac{Z^\ddagger}{Z_\mathrm A Z_\mathrm {BC}}\right]_\mathrm {vib}\left[\frac{Z^\ddagger}{Z_\mathrm A Z_\mathrm {BC}}\right]_\mathrm {rot}\left[\frac{Z^\ddagger/V}{Z_\mathrm A/V \cdot Z_\mathrm {BC}/V}\right]_\mathrm {trans}e^{-E_0/k_\mathrm BT}$$
Finally, the whole rate constant may need to be multiplied by a symmetry or statistical factor, for example - in $\ce{F + H2}$ the rate constant needs an addition factor of $2$.
The rate constant has units $\pu{m3 s-1}$ which when multiplied by the Avogadro number becomes $\pu{dm3 mol-1 s-1}$, normal $2^\mathrm {nd}$ order units.
Notes:
(A) The volume appears because we need to convert the thermodynamic equilibrium constant $k$ (ratio of partition functions) into $k_c$, the chemical one  which uses concentrations.
(B) The equations for partition functions are well known, are listed on many phys. chem. textbooks, e.g. McQuarrie & Simon ‘Physical Chemistry’ chapter 1,  but the Wikipedia page is rather technical so they are listed below$^{2}$.
(C) It is important to note that partition functions have widely differing values, translation $\frac{Z_\mathrm{trans}^\ddagger}{V} \approx \pu{10^{11}m-3}$; rotation $Z_\mathrm{rot}^\ddagger \approx 10^4 $, vibration $Z_\mathrm{vib}^\ddagger \approx 10$ and electronic $Z_\mathrm{elec}^\ddagger \approx 1$ are all dimensionless. (Thus, the translational partition function divided by $V$ has dimensions of $\pu{m-3}$).
Limitations:
(A)As the transition state properties are not known, reactions between complicated molecules are unlikely to give accurate values.
(B) Molecules in solution are limited by diffusion to a maximum value determined by diffusion constants and hence solution viscosity.
(C) More recent treatments by Kramers discuss how, in solution, the rate constant depends on friction of the reaction coordinate (see Daune ‘Molecular Biophysics).
Footnotes
(1) The average of a quantity $x$ with distribution $p(x)$ is calculated with the integral ratio
$$\frac{\int \exp(x)dx}{\int p(x)dx}$$
(The denominator is the normalizing term, equivalent to the total number when making a simple average of any set of objects, i.e. $\sum (x/N)$)
In the case in point, the average velocity distribution is $p(v) = \exp(-\mu v^2/(2k_BT))$ and the integration limits are 0 to $\infty$.
(2) The partition function is defined as the ‘sum over states’,
$$Z=\sum_i g_i \exp(-E_i/(k_\mathrm BT))$$
where $g_i$ is the degeneracy of the energy levels $E_i$.
When the energy levels are close (as in translation) the summation is replaced with an integral.
The partition functions are
$$Z_\mathrm {trans} = \frac{(2\pi \mu k_\mathrm BT)^{3/2}V}{h^3}$$
For a generally shaped molecule, with moments of inertia $I_\mathrm a$ etc. about the three principal axes, and symmetry factor $\sigma$ (which is the number of ways the molecule can be rotated back to the original):
$$Z_\mathrm {rot} = \left(\frac{8\pi ^2k_BT}{h^2}\right)^{3/2}\cdot\frac{\left(\pi I_aI_bI_c\right)^{1/2}}{\sigma}$$
For each vibrational mode.
$$Z_\mathrm {vib} = \frac{1}{1-e^{-h\nu/k_\mathrm BT}}$$
These partition functions are multiplied together for many vibrational modes; $\nu$ is the vibrational frequency.
The electronic partition function is the multiplicity of the state, most commonly 1.

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text:
How much heat is gained/released based off of equation?
Using this equation,
$\ce{CaCl2(s) -> Ca^{2+}(aq) + 2Cl^{-}(aq)}$
and $\Delta H = - 325.0 \; kJ $
If you have 15.0 grams of the solid, how much heat is released or gained in kilojoules?

Are you sure you have the right units for $\Delta H$?

@ldc3 Ya, it was a given value for the problem.

Suppose you dissolve a gram of calcium chloride and then a  kilogram, both in excess water. They cannot each and both be the value you posted. If you have the correct units, the path to a correct answer is obvious.  Open your textbook.  The first time you do it there is a crackling sound.  Google it, "enthalpy of solution".  BTW, the value is for infinite dilution.

$\Delta H$ is usually in $kJ/mol$. That is why I am asking. Either the question has a mistake or maybe it is $\Delta H^°$

@LDC3 I updated the post with a picture of the question. It very may well be kJ/mol but it isn't stated.

@UncleAl I updated the post with a picture of the question.

Yeah, I've seen these sorts of questions in my lecture notes where they say ΔH=(some no)kJ; they make the assumption you're working in moles.

I'm going to assume that the question is presenting the given change in enthalpy as a molar enthalpy change (so as to say that -325.0 kJ are evolved per mole of CaCl2).
Under that assumption, answering the question is then a matter of using the a molar fraction of CaCl2 for the amount you have present.
CaCl2 has a molar mass of 110.98 g/mol so with 15.0 g of CaCl2, you'd have roughly 0.1352 mol of CaCl2. 0.1352 mol * -325.0 kJ/mol = -43.94 kJ. The negative here means that the CaCl2 will release 43.94 kJ of heat.

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text:
Unexplained peak cluster in proton NMR trace?
I am working on a straight-forward NMR graph as part of my test prep. I have a pretty limited understanding of NMR, but I thoroughly read the chapter leading up to this practice question and am having difficulty explaining all of the peaks that I am seeing. Let me show the question and my understanding:
Here is the chemical and the trace that results.

My understanding of this problem is that there are four groups of equivalent hydrogens, and I numbered them 1-4. The 1st group has 3 equivalent hydrogens, with two adjacent non-equivalent hydrogens (group #2). Using the n+1 rule, the H group #1 should create 2+1=3 peaks in a cluster. Same goes for H group #4, since there are two adjacent hydrogens, so another 3 peaks. When considering H-group #2, the three hydrogens from H-group #1 should be 3+1=4 peaks. Same goes for H-group #3, which has the three hydrogens from H-group #4 so 3+1=4.
So, I am expecting to see 2 groups of 3 peaks and 2 groups of 4 peaks. I see the two groups of 4 peaks, but not the 2 groups of three peaks. Instead, I see 1 group of 6 peaks.
Am I supposed to believe that the 2 groups of 3 are just located so near it appears to be 6? Or am I missing something?
Thanks for any help!

AS the answers correctly say, the two methyl peaks overlap. It is worth having a play with NMR simulation tools (many structure drawing packages like ChemDoodle now include NMR simulators). These can help develop a good intuition about what to expect from spectra.

Sounds like a good idea, thanks for the recommendation!

I did a simulation of the peaks (I got the shifts wrong) for 60 MHz and 400 mHz. As you can see, the peaks overlap on the weaker magnet.

Wow, thanks for the graphical comparison, that really helps me visualize what is going on

You are correct. The chemical shift of methyl protons 1 and 4 are close to one another.  You still see the two expected triplets, they just have similar chemical shifts.  That's not too surprising since they are both methyl groups attached to a methylene group.  We could tell which triplet belongs to which methyl carbon by measuring and comparing the 2 sets of CH2-CH3 coupling constants.


text:
Electrolysis of aqueous lead nitrate
When going through one of my chemistry textbooks, I saw that the electrolysis of aqueous lead nitrate led to oxygen being formed at the anode and lead being formed at the cathode.
However, in school, I was taught that only metals less reactive than hydrogen will form at the cathode.  This electrolysis contradicts what I learned, as lead is more reactive than hydrogen, and should remain as ions in solution.
Therefore, my question is: What forms at the cathode during electrolysis of aqueous lead nitrate and why?

Firstly, the reactivity being referred to is not reactivity in the sense of the reactivity series of metals, but rather in the sense of standard electrode potentials (standard reduction potentials).
This comes into play when setting up an electrolytic cell in determining the reaction that will take place at the cathode/anode. In this case, the reaction that requires the least amount of work (or voltage in this case, since voltage is essentially a measure of work) will take place. So, assuming that this is done aqueously, if the reduction/oxidation reaction to take place requires more voltage than the reduction/oxidation of the water, then the water will simply be reduced/oxidized instead. (I think this is where the reactivity of hydrogen comes into play in how you were taught).
So to answer your question more directly, it is actually lead that forms at the cathode.
Edit: To add some detail, the reduction potential of H2O is -0.8277 V, while the lead(II) ion is -0.13. In this case, less energy means closer to zero, the negative being essentially irrelevant. Hope this helped.

The sentence learnt at school is not perfectly correct. It should not be :
In school, I was taught that only metals less reactive than hydrogen will form at the cathode.
It should have been modified by adding "at comparable ionic concentration", to give :
In school, I was taught that only metals less reactive than hydrogen at comparable ionic concentration will form at the cathode.
As an example, in $1 M$ solution of $\ce{Pb^{2+}}$ or $\ce{Sn^{2+}}$ or $\ce{Ni^{2+}}$, containing also $\ce{H+}$ with the same concentration $\ce{[H+]} = 1 M$, the metals $\ce{Pb, Sn, Ni}$ will not be deposited by electrolysis. $\ce{H2}$ will be deposited before the metals, as their redox potentials are too negative, being respectively $-0.13 V, -0.14 V$, and $-0.23 V$, and because of the redox potential of $\ce{H2/H+} = 0 V$
But in non-acidic solution, the situation is reversed. In the same $1 M$ solutions of $\ce{Pb^{2+}, Sn^{2+}, Ni^{2+}}$ but with $\ce{[H+]}$ = $\ce{10^{-7}}M$, the redox potential of $\ce{H2/H+}$ becomes very negative, being = $-0.41 V$. The metals are then deposited by electrolysis before $\ce{H2}$, because at $p$H $7$, all three redox potentials are less negative than $-0.41 V$

What was stated is correct but there is an exception. For the metals, zinc to lead in the reactivity series( zn, Fe, Pb and what's in between) what is formed at the cathode can be affected by concentration. If the solution is concentrated, the metal forms at the cathode and if the solution is dilute, hydrogen forms. Hope this helps.


text:
Possible methods for copper (II) reduction to copper (I)
How can I make $\ce{Cu2O}$ from copper?
I am seeing if electrolysis is a good route (in the context of kitchen style chemistry) to cuprous oxide, instead of the traditional heating copper and removing the cupric oxide, etc.
With Epsom salt electrolyte, it is very easy to make copper II hydroxide (cloudy ammonia soluble, according to the wiki). This can be heated to $\ce{CuO}$ easily enough, but I have no idea if this can be reduced.
Can $\ce{CuO}$ be reduced easily? Or is there a simple way to produce (from copper) a copper (I) salt, from which a precipitate of $\ce{Cu2O}$ may be formed? What reducing agents could be used?

using the more sensible electrolyte NaOH gives better results and ammonia is indeed a good solvent - the sky blue hydroxide forms a deep blue solution like copper sulfate. Now the challenge remains to reduce this solution, as with Benedict's reagent or Fehling's solution. Any thoughts?

i bought glucose and hesitantly boiled the ammonia solution after adding an excess - it create a very fine orange suspension. i have no idea how to cause it to settle though, as heating will produce the black oxide (as the last water evaporates, possibly?).

Copper (I) has a half-filled orbital, so transforming this would be relatively easy,

Six years after the original post, I wanted to do the same thing (get some Cu2O), and Benedict's reagent seems to be the way to go, as far as kitchen chemistry is concerned.
I made 100 mL Benedict's according to wikipedia (only I used citric acid and washing soda, as I did not have trisodium citrate at hand, but the math is simple...). It's not a high-throughput technique, only 2 grams of copper sulfate (sorry, my kitchen scale only has 1 gram resolution), I reduced it with about 2 grams of glucose (smarties candy) by boiling in microwave oven. All of the Cu 2+ gets reduced to Cu2O.
The brick-

There are several ways to make copper(I) oxide from copper:-

Heat it

$$\ce{4Cu + O2 ->[\Delta] 2Cu2O}$$
Copper react with oxygen to produce copper(I) oxide. This reaction takes place at a temperature of over 200°C.

Heat it with nitrous or nitric  oxide

$$\ce{2Cu + N2O->[\Delta] Cu2O + N2}$$
$$\ce{4Cu + 2NO->[\Delta] 2Cu2O + N2}$$
Copper react with nitrous and nitric oxide to produce copper(I) oxide and nitrogen. Both reaction takes place at a temperature of 500-600°C.

Heat it with copper(II) oxide

$$\ce{Cu + CuO->[\Delta] Cu2O}$$
Copper react with copper oxide to produce copper(I) oxide. This reaction takes place at a temperature of 1000-1200°C.

If you made copper (II) oxide, just heat it

$$\ce{4CuO ->[\Delta] 2Cu2O + O2}$$
The thermal decomposition of copper oxide to produce copper(I) oxide and oxygen. This reaction takes place at a temperature of 1026-1100°C.
Since you are doing this at home, method 2,3,4 cannot be used as it requires laboratory grade heater but method 1 can be used since it require 200°C which is achievable.
But the best method(according to the comment) I would recommend is the Fehling's test. All you need is Fehling's solution and glucose which can be purchased from sigma-aldrich at affordable price.
$$\ce{RCHO + (2Cu^2+ + 5OH^{-})->Cu2O + RCOO^- + 3H2O}$$


text:
Reactions involving trimethylaluminium Al2Me6
I need to write balanced equations describing the following reactions:

one mole of $\ce{Al2Me6}$ with two moles of water
excess of $\ce{Al2Me6}$ with silicon dioxide
excess of $\ce{Al2Me6}$ with tin(IV) chloride

My answer so far:

a) $\ce{2H2O + Al2Me6 -> Al2Me4(H2O)2}$?
b) $\ce{SiO2 + Al2Me6 -> SiMe4 + Al2Me2O2}$?
c) $\ce{SnCl4 + Al2Me6 -> SnMe4 + Al2Me2Cl4}$?

Organoaluminium compounds are very reactive. If a substance is capable of attacking one of the methyl groups in trimethylaluminium, then the process likely doesn't stop until all carbon-aluminium bonds are broken. Can you figure out the answer now?

Balance your equations.  Draw structures to see what is happening.  Why is trimethylaluminum dimeric?  For (a), what happens when you mix a Bronsted acid with a Bronsted base given a potent oxophile?

I'm voting to close this question as off-topic because it's an AMIRITE question unlikely to be of help to future visitors.

For a): Many organometallic compounds are easily hydrolyzed to the respective hydrocarbons and metal (hydr)oxides. In this case, the products are $\ce{CH4}$ and more likely $\ce{Al2O3}$, because the formation of $\ce{Al(OH)3}$ requires more moles of water.
Reactions b) and c) should be transmetallations where the methyl groups are transferred from aluminium to the other metal. The excess of $\ce{Al2Me6}$ helps to shift the equilibrium of the reaction to the product side. $\ce{SiMe4}$ and $\ce{SnMe4}$ would be the correct products which you have identified, but aluminium is also converted into its respective oxide/chloride.


text:
Reaction of CpFe(CO)2Cl with phenyllithium

The reaction of $\ce{(C5H5)Fe(CO)2Cl}$ with an equimolar amount of $\ce{Li+[C6H5]-}$ gives a product. Draw its structure and apply the 18 electron rule.

In a question I was given the molecular weight and percentages of each element in the compound. I worked out that there is $\ce{1 Fe}$, $\ce{13 C}$, $\ce{10 H}$ and $\ce{2 O}$.
The product seems to be an iron atom surrounded with ligands, $\ce{2 CO}$ ligands, $\ce{1 C5H5}$ and $\ce{1 [C6H5]-}$ ion.
The 18 electron rule expands to 8 valence electrons from $\ce{Fe}$, 2 electrons from each $\ce{CO}$ ligand, 5 electrons from $\ce{C5H5}$ and 1 electron from the $\ce{[C6H5]-}$ anion: $8+4+5+1 = 18$. But what would the structure look like?  Especially how is the $\ce{[C6H5]-}$ anion connected?

http://voh.chem.ucla.edu/vohtar/fall06/classes/277/pdf/18-Electron%20Rule.pdf
Summing ligand electron contribution.
DOI:10.1006/jmsp.2001.8470
http://www.ccp14.ac.uk/ccp/web-mirrors/farrugia/~louis/chem2x/Organomet-2.pdf
 Document p. 34.  Note error in drawing the graphic.
Direct displacement gives $\ce{CpFe(CO)2Ph}$, with the phenyl being a simple anion sigma-bonded to the metal. If there are two carbons plus two oxygens in the product, implying the CO ligands are intact, your are done.  If some fancy stiff happened with those ligands and the phenyl anion, you must work it out or look it up.


text:
Why does no Robinson Annulation occur after the Michael addition?

Under basic conditions a Michael addition reaction can be used to convert compound A into product C. Explain why no Robinson annulation cyclic product is observed?
Compound A

Product C

I have worked out that the other compound involved in the Michael addition reaction would be compound B (below), but I don't know why no Robinson annulation reaction could occur after the Michael addition reaction?
Compound B

As illustrated in the following figure, the Robinson annulation reaction involves two steps.  The first step involves the Michael addition of a ketone enolate (the enolate derived from cyclohexanone in the example pictured below) to an $\alpha,\beta$-unsaturated ketone (methyl vinyl ketone below).  In a second step the methyl group adjacent to the carbonyl is deprotonated and undergoes an intramolecular aldol condensation to form the final product. This last step is very favorable since the reaction occurs intramolecularly and involves a favorable 6-membered transition state to form a stable 6-membered ring.

Often in Robinson annulations methyl vinyl ketone is used. It is important that  the $\alpha,\beta$-unsaturated ketone have an alkyl group attached to the carbonyl so that the enolate required in the second step can be formed.

Explain why no Robinson annulation cyclic product is observed?

In your example, the phenyl vinyl ketone has no alkyl group attached to the carbonyl carbon - only a phenyl group. Hence, the enolizable hydrogens present in your Product C can only react with the other carbonyl through a highly-strained 4-membered transition state preventing any ring formation through an intramolecular aldol from taking place.

Either of the enolates at either acidic $\alpha$ position in C could form, but the intramolecular aldol would produce an unfavorable four-membered ring (angle strain).

Well, technically it could. There is nothing stopping the Michael addition product from being enolised and when enolised there is little to stop the enolate attacking a different carbonyl.
When I say ‘little’, that was a little lie, though. The main inhibiting factor is the Baeyer strain on three- and four-membered rings. These rings are rather strained (try building them with a molecular model kit!) and thus do not form easily. Especially, a nucleophilic attack onto the carbonyl carbon would be very difficult if not impossible to perform if it generates a four-membered ring. You can also try that with a molecular model kit, and remember that the nucleophile has to attack the carbonyl group in the Bürgi-Dunitz angle of approximately $107^\circ$.
In a regular Robinson annulation, as Ron described, the Michael acceptor system can be enolised on the ‘far’ side of the ketone — the one where your molecule has a phenyl group. It is then this enolate that attacks the other carbonyl group to form a six-membered ring — one that does not have Baeyer strain and is generally the most thermodynamically stable ring system. Your system cannot do this as it is lacking α-hydrogens on the ‘far’ side of the Michael acceptor.


text:
How can the 18 electron rule be used to explain that Fe(η5-C5H5)2 doesn't react with hydrogen but Ni(η5-C5H5)2 does?
Fe(η5-C5H5)2 does not react with hydrogen, but its nickel analogue, Ni(η5-C5H5)2 is readily hydrogenated to give Ni(η5-C5H5)(η3-C5H7). Use the 18 electron rule to explain this behaviour.
My answer so far...
Fe(η5-C5H5)2 18 electron rule = valence electrons from Fe is 8, electrons from C5H5 is 5 therefore 8 + 5 + 5 = 18 so it is stable.
Ni(η5-C5H5)2 18 electron rule = valence electrons from Ni is 10, electrons from C5H5 is 5 therefore 10 + 5 + 5 = 20 so Ni is not in most stable form
Ni(η5-C5H5)(η3-C5H7) 18 electron rule = valence electrons from  Ni is 10, electrons from C5H5 is 5.....
how many electrons from (η3-C5H7) and how does this all explain why the iron complex doesn't react with hydrogen but the nickel one does? It is to do with the fact it has two too many electrons (20e) so it bonds with two protons from hydrogen? Also what does this symbol η mean?

Hydrogenation of NiCp2 takes place on the ligand, not the metal, to reduce donor electron count, giving a low-energy closed-shell 18 electron count.
http://en.wikipedia.org/wiki/Nickelocene
http://en.wikipedia.org/wiki/Ferrocene
http://en.wikipedia.org/wiki/Hapticity
http://www.chem.mun.ca/homes/cmkhome/Organometallics_The%20Basics.pdf
   p. 9, 10.  "η" is "hapticity," the number of ligand atoms contributing pi-electron orbitals to the metal center.
$\ce{Cp^-}$ is aromatic (draw the Frost diagram) when coordinated $\ce{η^5}$ on-face, or  $\ce{η^1}$ non-aromatic as a simple anion coordinated on-vertex.  Saturating one localized double bond leaves cyclic delocalized allyl anion coordinated $\ce{η^3}$.
In the case of $\ce{TiCp4}$, two Cp anions are face pentahapto, two are vertex monohapto.  If we take a proton or C-13 NMR, will we see discrete protons or carbons for the $\ce{η^1}$-Cp, or will allowed 1,5-shifts scramble all five protons?  All five carbons?  Will it be temperature dependent if it happens?  If it happens, will H and C have different kinetics?


text:
Finding the enthalpy of formation of magnesium oxide using Hess' Law
I have the following reactions:
$$\begin{align}
\ce{Mg(s) + 2HCl(aq) &-> MgCl2(aq) + H2(g)}\\
\ce{MgO(s) + 2HCl(aq) &-> MgCl2(aq) + H2O(aq)}\\
\ce{H2(g) + 1/2O2(g) &-> H2O(l)}\\
\ce{Mg(s) + 1/2O2(g) &-> MgO(s)}
\end{align}$$
I also have $\Delta H$ in $\mathrm{kJ/mol}$ for the first three reactions.
Out of this information, I need to find $\Delta H^\circ_\mathrm{f}{\left(\ce{MgO(s)}\right)}$.
I’m pretty sure I have to use Hess’ law but I don’t know how to apply it here. I’ve tried using the second reaction as my “main” reaction and then I can use the third and fourth reaction as formation reactions for two of the compounds of the main reaction but after that I’m not sure how to proceed.

You are trying to find out the enthalpy of formation.
The enthalpy of formation is the energy change when one mole of a substance is formed from its constituent elements in their standard state.
The last equation matches this, so that should be your 'main' or 'target' equation. From there you should be able to use Hess' law to calculate the enthalpy change of the formation of magnesium oxide.

Out of the four equations that you have, you need to find $\Delta H^\circ_\mathrm f$ of $\ce{MgO}$, or rather, the $\Delta H$ value of  the reaction
$$\ce{Mg(s) + 1/2O2(g) -> MgO(s)} \tag{1} \label{1}$$
Why is this? This is because of the definition of heat of formation is as follows:

Standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states. (emphasis mine)

In the above equation, the elements are in their standard states and they are used to form one mole of $\ce{MgO}$.
Onto the solution:
We need to apply Hess's law of constant heat summation, which states that:

The change of enthalpy in a chemical reaction (i.e. the heat of reaction at constant pressure) is independent of the pathway between the initial and final states.
In other words, if a chemical change takes place by several different routes, the overall enthalpy change is the same, regardless of the route by which the chemical change occurs (provided the initial and final condition are the same).

Therefore if we can write any equation in terms of two or more other reactions, the change in enthalpy of the first reaction can be written as a linear combination of enthalpies of said reactions.
Here the other equations for which we know the enthalpies are:
\begin{align}
\ce{Mg(s) + 2HCl(aq) &-> MgCl2(aq) + H2(g)} \tag{2} \label{2}\\
\ce{MgO(s) + 2HCl(aq) &-> MgCl2(aq) + H2O(l)} \tag{3} \label{3}\\
\ce{H2(g) + 1/2O2(g) &-> H2O(l)} \tag{4} \label{4}
\end{align}
Now, in order to proceed we need to write ($\ref{1}$) in terms of a linear combination of ($\ref{2}$), $(\ref{3})$ and $(\ref{4})$.
Finding the possible combination using observation (Method 1)
We need to notice that in our final reaction equation ($\ref{1}$), we need to have an $\ce{MgO}$ on the right and the only equation where an $\ce{MgO}$ molecule exists is in $(\ref{3})$, similarly, we observe that only $(\ref{2})$ has an $\ce{Mg}$ molecule and therefore, we can say that the final combination needs to have $(1) = (2) - (3) + x(4)$.
Substituting the equations and solving for $x$, we get x = 1, (which could have also been guessed using the above mentioned method).
Therefore our final linear combination is as follows:
$$\Delta H^\circ_\mathrm f = \Delta H^\circ_2 + \Delta H^\circ_4 - \Delta H^\circ_3$$
Finding the answer through applying math (Method 2)
We write
$$(1) = x(2) + y(3) + z(4)$$
After this, we compare the left hand side and right hand side which should satisfy the conservation of mass.
Therefore:
\begin{align}
\ce{Mg(s) + 1/2O2(g) &-> MgO(s)} \tag{LHS}\\
\ce{xMg + yMgO + (2x + 2y)HCl + zH2 + z/2O2 &-> (x + y)MgCl2 + (y+z)H2O + xH2} \tag{RHS}
\end{align}
Comparing co-efficient on both sides, we get:
\begin{align}
x &= 1 \\
y &= -1 \\
x + y &= 0 \\
z &= x \\
z &= 1\\
\end{align}
Either way, we get the same combination for Hess' Law and that is:
$$\Delta H^\circ_\mathrm f = \Delta H^\circ_2 + \Delta H^\circ_4 - \Delta H^\circ_3$$

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


text:
How can I carry out a titration to identify mercury in wastewater?
First, I would like to do a some form of extraction to only get the mercury, then followed by a titration to identify the concentration or the presence of mercury.

http://www.rdmag.com/news/2013/03/high-sensitivity-detection-method-found-mercury-water

Is there a specific reason you want to determine the mercury content by titration? Atomic absorption spectroscopy, among other spectroscopic techniques, are likely simpler (so long as you have the instrument, of course), more precise, and more sensitive.

I'm confused about your question - how will you titrate to determine concentration/presence after you extract the mercury?

The potentiometric titration of mercury with dithiooxamide has been successfully applied after preliminary separation of mercury by reduction (Comprehensive Water Analysis, Volume 2, pp.186-187, Elsevier).
Selective identification of mercury in waste water can be accomplished by fluorescence enhancement chemosensors (SENSOR LETTERS 2010, 8, 1–6). In this recent paper, the ﬂuorescent emission intensity of the probe is enhanced upon binding to mercury ions. The ﬂuorescence enhancement is attributed to a 1:1 complex formation between the probe and Hg(II) ion. The chemosensor can be applied for the quantiﬁcation of mercury; it shows a high selectivity toward Hg(II) ions in comparison with common metal ions.


text:
Major product formed when 1,1,1-trichloropropane is treated with aqueous KOH

What are (is) the product formed when 1,1,1-trichloropropane is treated with aqueous potassium hydroxide?

I think the major product should be propionic acid, but I'm not sure.

Figure: 1,1,1-trichloropropane
${\color{green}{\checkmark}}$ The above compound involves $\color{red}{\ce{C-H, C-C, C-Cl}}$ bonds, whose bond energies are $413, 348, 328 kj/mol $ respectively. So, initially $\color{orange}{\ce{C-Cl}}$ bond must break.
${\color{green}{\checkmark}}$ It may be noted that generally alcoholic $\color{blue}{\ce{KOH}}$ causes elimination (dehydrohalogenation reaction) in the molecule of haloalkane, while aqueous $\color{blue}{\ce{KOH}}$ leads to substitution. As in your reaction, aqueous $\color{blue}{\ce{KOH}}$ is used, substitution reaction is most preferable.
${\color{green}{\checkmark}}$ As primary carbon and strong $\color{orange}{\ce{C-Cl}}$ (compared to other $\ce{C-Br, C-I}$ bond) is involved in 1,1,1-trichloropropane, $S_{N2}$ mechanism is preferable.
From the above thought process, the most preferable reaction would be:
$$\ce{Cl_3C-CH_2-CH_3 + 3KOH(aq) ->[\text{heat}] (OH)3C-CH_2-CH_3 + 3KCl} $$
${\color{green}{\checkmark}}$ It may be noted that more than one $\color{blue}{OH}$ group cannot be present on the same carbon atom. In such a case, the compound will be extremely unstable and will readily lose a water molecule to form more stable aldehyde, ketone or carboxylic acid.
$$\ce{(OH)_3C-CH_2-CH_3 -> (OH)CO-CH_2-CH_3 + H_2O}$$
You are right, propionic acid along with water molecule is going to be the product.

$\ce{OH^-}$ could displace $\ce{Cl3C^-}$ to give ethanol plus chloroform plus tetrachloroethylene (via cage effect on dichorocarbene).
$\ce{OH^-}$ could eliminate HCl to give 1,1-dichloro-1-propene, and on to 1-chloropropyne, and on to....
$\ce{OH^-}$ could displace $\ce{Cl^-}$ to give the acid chloride hemicarbonyl chloride, then on to propionic acid.
Solvent, concentration, and temperature are important for biasing reaction paths.

The techinfo sheet for 1,1,1-trichloroethane says that major product of hydrolysis is 2,2-dichloroethylene, so I'd say that for 1,1,1-trichloropropane it would be 1,1-dichloropropene-1.


text:
Carbon monoxide detection without a detector
I have just been waken up by an alarm and found out that my neighbour did something terrible with his lighter. Luckily, there was no fire, only a plenty of smoke, and our flats were hermetically isolated, so that smoke didn't go to my flat. But, most probably because of my suspiciousness, I noticed some smell of burnt organic, which usually accompanies CO. I switched on ventilation and went out - just in case. There were no CO detectors in the house - only smoke detectors - my bad.
Now the adventure is over (I hope so, at least) and I have got an (im)practical question:
Are there any ways to detect CO using only "kitchen chemistry"?

Surf a civilian aircraft supply.  They sell a little absorbant white tab (chemical spot detector) containing $\ce{PdCl2}$.  When exposed to CO it goes black.
http://www.aircraftspruce.com/menus/ps/carbonmonoxidedetectors.html
Canary.  Goldfish won't work - they can survive on anaerobic respiration,
http://aquaticcommons.org/3663/1/14P089.pdf
   who needs powdered alcohol?

I have seen more than one website recommend a canary. If you'd prefer not to have a canary in the house I highly recommend investing in a CO detector... they save lives!

If you would still not like to have a CO detector; I highly recommend knowing what the symptoms of CO poisoning are, if you experience these symptoms and believe they maybe caused by CO, get out of the house and call for help immediately! Flu-like symptoms without the fever, dizziness and headache. Here are two links, I hope you find their information helpful!

Good luck!

http://www.nytimes.com/2004/01/18/realestate/your-home-detecting-carbon-monoxide.html

http://www.health.state.mn.us/divs/eh/indoorair/co/


text:
Multiplication of the reaction and Equilibrium constant
$$\begin{align}\ce{2A + 2B &-> 2C }\\ \ce{A + B &-> C}\end{align}$$
Why does the equilibrium constant change?
And why the rate of the first reaction square the rate of the second? I understand it mathematically, but logically and scientifically I can't understand it at all.

A simple(?) thought experiment:
A+B involves a 2-body interaction (i.e. 2 molecules are interacting at one common location).
2A+2B involves a 4-body interaction. That's (MUCH) less than half as likely to occur than 2x 2-body interactions. So there's not much reason to believe that the distribution of reactants and products (i.e. the equilibrium constant) will be in a 1:2 ratio because you doubled the number the particles involved in the reaction, right?

If you consider collisions at hard sphere model, even three bodies hitting each other is highly unlikely. Four however is virtually almost impossible. So the basic question is more likely to be, if there are two of A and B necessary. Upscaling reactions and therefore the change in the kinetics is most likely connected to concentrations.

Thanks @Martin I had meant to make those changes earlier. :)

Related question: https://chemistry.stackexchange.com/questions/6641/why-does-equilibrium-constant-change-when-the-reaction-is-multiplied-by-a-number

Rate will not be square in most of the cases and who said so? The equilibrium constant scientifically changes because Gibb's free energy is extensive quantity and it is related to equilibrium constant through relation that you might know $\Delta G=\Delta G^\circ +\mathcal{R}T\ln Q $. Where $Q$ is reaction quotient which at equilibrium becomes equilibrium constant and then it gives $\Delta G^\circ = -\mathcal{R}T\ln K$

I don't know what Gibb's free energy energy is, what I mean by my question is what happens to the equilibrium constant on doubling the stoichiometry of the equation at constant concentrations???

It will become squared of earlier.Rate doesn't but equilibrium constant does become squared unless all substances are either pure solid or liquid.


text:
Bond angles in PDB
people I've been trying to adjust the bond angles in a protein data bank (PDB) file, but I'm a novice when it comes to computing so I would like to ask your assistance. I have a file for metamizole, but as has been pointed out by a user on Wikipedia the 90 degree bond angle in it isn't consistent with the trigonal planar geometry of the nitrogen atom in the molecule. Here is my PDB file:
COMPND    UNNAMED
AUTHOR    GENERATED BY OPEN BABEL 2.3.2
HETATM    1  C   LIG     1      -0.971  -5.437  -1.069  1.00  0.00           C
HETATM    2  N   LIG     1      -1.931  -4.484  -0.575  1.00  0.00           N
HETATM    3  C   LIG     1      -3.342  -4.804  -0.629  1.00  0.00           C
HETATM    4  S   LIG     1      -4.562  -3.540  -0.090  1.00  0.00           S
HETATM    5  O   LIG     1      -4.430  -2.149  -1.074  1.00  0.00           O
HETATM    6  O   LIG     1      -4.326  -3.106   1.433  1.00  0.00           O
HETATM    7  O   LIG     1      -6.044  -4.127  -0.251  1.00  0.00           O
HETATM    8  C   LIG     1      -1.479  -3.307  -0.058  1.00  0.00           C
HETATM    9  C   LIG     1      -1.218  -3.011   1.277  1.00  0.00           C
HETATM   10  C   LIG     1      -1.353  -3.951   2.413  1.00  0.00           C
HETATM   11  N   LIG     1      -0.863  -1.703   1.337  1.00  0.00           N
HETATM   12  C   LIG     1      -0.731  -0.968   2.567  1.00  0.00           C
HETATM   13  N   LIG     1      -0.775  -1.218   0.118  1.00  0.00           N
HETATM   14  C   LIG     1      -1.185  -2.152  -0.766  1.00  0.00           C
HETATM   15  O   LIG     1      -1.262  -2.037  -2.012  1.00  0.00           O
HETATM   16  C   LIG     1      -0.309   0.020  -0.235  1.00  0.00           C
HETATM   17  C   LIG     1      -1.201   0.998  -0.733  1.00  0.00           C
HETATM   18  C   LIG     1      -0.737   2.291  -1.040  1.00  0.00           C
HETATM   19  C   LIG     1       0.619   2.617  -0.856  1.00  0.00           C
HETATM   20  C   LIG     1       1.520   1.640  -0.390  1.00  0.00           C
HETATM   21  C   LIG     1       1.059   0.346  -0.078  1.00  0.00           C
HETATM   22  H   LIG     1      -1.483  -6.301  -1.438  1.00  0.00           H
HETATM   23  H   LIG     1      -0.312  -5.724  -0.276  1.00  0.00           H
HETATM   24  H   LIG     1      -0.403  -4.993  -1.861  1.00  0.00           H
HETATM   25  H   LIG     1      -3.492  -5.667  -0.014  1.00  0.00           H
HETATM   26  H   LIG     1      -3.539  -4.923  -1.674  1.00  0.00           H
HETATM   27  H   LIG     1      -3.538  -1.778  -0.991  1.00  0.00           H
HETATM   28  H   LIG     1      -1.646  -4.913   2.048  1.00  0.00           H
HETATM   29  H   LIG     1      -2.094  -3.586   3.092  1.00  0.00           H
HETATM   30  H   LIG     1      -0.414  -4.034   2.920  1.00  0.00           H
HETATM   31  H   LIG     1      -0.853  -1.635   3.395  1.00  0.00           H
HETATM   32  H   LIG     1      -1.481  -0.206   2.610  1.00  0.00           H
HETATM   33  H   LIG     1       0.238  -0.518   2.611  1.00  0.00           H
HETATM   34  H   LIG     1      -2.196   0.763  -0.872  1.00  0.00           H
HETATM   35  H   LIG     1      -1.394   3.001  -1.399  1.00  0.00           H
HETATM   36  H   LIG     1       0.954   3.571  -1.064  1.00  0.00           H
HETATM   37  H   LIG     1       2.519   1.873  -0.276  1.00  0.00           H
HETATM   38  H   LIG     1       1.721  -0.367   0.265  1.00  0.00           H
CONECT    1    2   22   23   24
CONECT    1
CONECT    2    1    3    8
CONECT    3    2    4   25   26
CONECT    3
CONECT    4    3    5    6    7
CONECT    4
CONECT    5    4   27
CONECT    6    4
CONECT    7    4
CONECT    8    2    9   14
CONECT    9    8   10   11
CONECT   10    9   28   29   30
CONECT   10
CONECT   11    9   12   13
CONECT   12   11   31   32   33
CONECT   12
CONECT   13   11   14   16
CONECT   14   13    8   15
CONECT   15   14
CONECT   16   13   17   21
CONECT   17   16   18   34
CONECT   18   17   19   35
CONECT   19   18   20   36
CONECT   20   19   21   37
CONECT   21   20   16   38
CONECT   22    1
CONECT   23    1
CONECT   24    1
CONECT   25    3
CONECT   26    3
CONECT   27    5
CONECT   28   10
CONECT   29   10
CONECT   30   10
CONECT   31   12
CONECT   32   12
CONECT   33   12
CONECT   34   17
CONECT   35   18
CONECT   36   19
CONECT   37   20
CONECT   38   21
MASTER        0    0    0    0    0    0    0    0   38    0   38    0
END

If you can also give me a general method of correcting this problem it will be much appreciated. Here is the gif file I get out of this PDB file from Jmol.

This is what I got from optimisation.

My eyesight must be poor. I don't see any 90° bond angle in the figure.

Good point! I'll say that and see what response this guy has. I think he might have an inflamed ego, thinking he's the chem God. Thanks, Chem isn't my strong suit so I'm kind of letting this guy walk all over me.

He said it should be tetrahedral, not trigonal planar.

I made the molecule in ACD ChemSketch and it made the nitrogen almost planer. The C=C-N-C(-SO2CH3) torsion angle is 22.34° and -178.68° to the other carbon. So, your configuration should be OK. If he still thinks it should be tetrahedral, ask him for the molecular modeling calculation.

Thank you! I agree, this guy is pretty strict when it comes to chem drawing; he's almost blocked me for my images.

The reason that it is almost planer is that the lone pair on the nitrogen has formed a $\pi$ bond with the 5-member ring (i.e. conjugation). This stabilizes the molecule; the non-planer nitrogen configuration requires more energy.

@LDC3 The reply he gave was too long for me to give it here see https://commons.wikimedia.org/wiki/User_talk:Fuse809#Nitrogen_geometry_in_File:Metazimole3Dan.gif

The crystal structure of metamizole can be found in Acta Cryst. (1979). B35, 612-615. It is also available via the Cambridge Crystallographic Data Centre (CCDC) database, entering the DOI: 10.1107/S0567740879004271.

There are essentially two conformations. The phenyl ring is rotated of plane of the pyrazolone moiety by about $42.5/44.5^\circ$. The nitrogen(4) bound to the phenyl moiety is slightly out of plane of the surrounding atoms ($-0.17/-0.18~\mathring{\mathrm{A}}$), so is the neighbouring nitrogen(5) ($0.39~\mathring{\mathrm{A}}$). The methylaminomethane group is also pyramidalised [nitrogen(9)], as can be seen by the dihedral angles with respect to the pyrazolone ring,  $\ce{C(3)-C(2)-N(9)-C(12)}$ and $\ce{C(3)-C(2)-N(9)-C(12)}$, being $89.3/104.6^\circ$ and $-43/-42.2^\circ$.
Here is a stick model screenshot of this structure, hydrogen atoms are not included in the files. (grey: carbon; red: oxygen; blue-ish: nitrogen; yellow: sulfur; purple: sodium)

A calculation (BP86/cc-pVTZ in gas phase at $O~\mathrm{K}$) based on one of the modifications in the crystal structure reveals the same picture.

And for completeness the xyz coordinates of the optimised compound.
  38
BVP86/cc-PVDZ scf done: -1369.314862
 C     0.634952    10.049072     4.900129
 C    -0.038364     9.052812     5.625610
 C    -0.843603     9.404086     6.721280
 C    -0.967888    10.746974     7.078721
 C    -0.317360    11.745954     6.347432
 C     0.482122    11.389915     5.258069
 N     0.088512     7.682548     5.283073
 N     0.524862     7.332867     3.979484
 C     1.016111     6.033439     4.052906
 C     1.037214     5.613638     5.356904
 C     0.456058     6.652593     6.201473
 C    -0.418749     7.686282     2.908454
 C     1.415171     5.281438     2.830834
 N     1.587824     4.410206     5.852441
 C     0.695266     3.536091     6.539780
 S     0.466632     1.963130     5.547604
 O     1.708338     1.174668     5.559744
 O     0.313685     6.699922     7.422867
 C     2.964673     4.465074     6.365693
 O    -0.210611     2.273661     4.290449
 O    -0.634313     1.159020     6.505778
 H    -0.141454     0.406631     6.889481
 H    -1.345054     8.622837     7.287776
 H    -1.589369    11.013825     7.934553
 H    -0.429995    12.793897     6.626878
 H     1.005549    12.158525     4.687515
 H     1.289838     9.765864     4.076298
 H    -1.364716     7.127508     2.998793
 H     0.053315     7.462436     1.945681
 H    -0.625790     8.761093     2.953974
 H     0.547989     5.057675     2.191339
 H     2.145811     5.841566     2.228931
 H     1.854760     4.324136     3.132627
 H    -0.315695     3.946934     6.648080
 H     1.067111     3.196537     7.517244
 H     3.601787     4.973633     5.631670
 H     3.344097     3.442734     6.490780
 H     3.030898     5.004553     7.327684

Upon request, I will also provide a pdb file. There are probably a million ways for easy conversion of *.xyz files to *.pdb files. For example on linux with openbabel via
PROMPT> babel <filename>.xyz -opdb <filename>.pdb

Much more easier, because with graphical user interface: Avogadro (because it has openbabel built in).

open the *.xyz via file - open or ctrl+o
save the file as *.pdb via file - save as or ctrl+shift+s
make sure you select as filetype "common molecule formats" and enter the ending

And as a result you will get this beautiful file:
COMPND    METAMIZOLE scf done: -1369.314862
AUTHOR    GENERATED BY OPEN BABEL 2.3.0
HETATM    1  C   LIG     1       0.635  10.049   4.900  1.00  0.00           C
HETATM    2  C   LIG     1      -0.038   9.053   5.626  1.00  0.00           C
HETATM    3  C   LIG     1      -0.844   9.404   6.721  1.00  0.00           C
HETATM    4  C   LIG     1      -0.968  10.747   7.079  1.00  0.00           C
HETATM    5  C   LIG     1      -0.317  11.746   6.347  1.00  0.00           C
HETATM    6  C   LIG     1       0.482  11.390   5.258  1.00  0.00           C
HETATM    7  N   LIG     1       0.089   7.683   5.283  1.00  0.00           N
HETATM    8  N   LIG     1       0.525   7.333   3.979  1.00  0.00           N
HETATM    9  C   LIG     1       1.016   6.033   4.053  1.00  0.00           C
HETATM   10  C   LIG     1       1.037   5.614   5.357  1.00  0.00           C
HETATM   11  C   LIG     1       0.456   6.653   6.201  1.00  0.00           C
HETATM   12  C   LIG     1      -0.419   7.686   2.908  1.00  0.00           C
HETATM   13  C   LIG     1       1.415   5.281   2.831  1.00  0.00           C
HETATM   14  N   LIG     1       1.588   4.410   5.852  1.00  0.00           N
HETATM   15  C   LIG     1       0.695   3.536   6.540  1.00  0.00           C
HETATM   16  S   LIG     1       0.467   1.963   5.548  1.00  0.00           S
HETATM   17  O   LIG     1       1.708   1.175   5.560  1.00  0.00           O
HETATM   18  O   LIG     1       0.314   6.700   7.423  1.00  0.00           O
HETATM   19  C   LIG     1       2.965   4.465   6.366  1.00  0.00           C
HETATM   20  O   LIG     1      -0.211   2.274   4.290  1.00  0.00           O
HETATM   21  O   LIG     1      -0.634   1.159   6.506  1.00  0.00           O
HETATM   22  H   LIG     1      -0.141   0.407   6.889  1.00  0.00           H
HETATM   23  H   LIG     1      -1.345   8.623   7.288  1.00  0.00           H
HETATM   24  H   LIG     1      -1.589  11.014   7.935  1.00  0.00           H
HETATM   25  H   LIG     1      -0.430  12.794   6.627  1.00  0.00           H
HETATM   26  H   LIG     1       1.006  12.159   4.688  1.00  0.00           H
HETATM   27  H   LIG     1       1.290   9.766   4.076  1.00  0.00           H
HETATM   28  H   LIG     1      -1.365   7.128   2.999  1.00  0.00           H
HETATM   29  H   LIG     1       0.053   7.462   1.946  1.00  0.00           H
HETATM   30  H   LIG     1      -0.626   8.761   2.954  1.00  0.00           H
HETATM   31  H   LIG     1       0.548   5.058   2.191  1.00  0.00           H
HETATM   32  H   LIG     1       2.146   5.842   2.229  1.00  0.00           H
HETATM   33  H   LIG     1       1.855   4.324   3.133  1.00  0.00           H
HETATM   34  H   LIG     1      -0.316   3.947   6.648  1.00  0.00           H
HETATM   35  H   LIG     1       1.067   3.197   7.517  1.00  0.00           H
HETATM   36  H   LIG     1       3.602   4.974   5.632  1.00  0.00           H
HETATM   37  H   LIG     1       3.344   3.443   6.491  1.00  0.00           H
HETATM   38  H   LIG     1       3.031   5.005   7.328  1.00  0.00           H
CONECT    1   27    6    2
CONECT    2    1    7    3
CONECT    3    2    4   23
CONECT    4    5    3   24
CONECT    5    6   25    4
CONECT    6   26    1    5
CONECT    7    8    2   11
CONECT    8   12    9    7
CONECT    9   13    8   10
CONECT   10    9   14   11
CONECT   11    7   10   18
CONECT   12   29   30   28    8
CONECT   12
CONECT   13   31   32   33    9
CONECT   13
CONECT   14   10   19   15
CONECT   15   16   14   34   35
CONECT   15
CONECT   16   20   17   21   15
CONECT   16
CONECT   17   16
CONECT   18   11
CONECT   19   36   14   37   38
CONECT   19
CONECT   20   16
CONECT   21   16   22
CONECT   22   21
CONECT   23    3
CONECT   24    4
CONECT   25    5
CONECT   26    6
CONECT   27    1
CONECT   28   12
CONECT   29   12
CONECT   30   12
CONECT   31   13
CONECT   32   13
CONECT   33   13
CONECT   34   15
CONECT   35   15
CONECT   36   19
CONECT   37   19
CONECT   38   19
MASTER        0    0    0    0    0    0    0    0   38    0   38    0
END

If you do not see any double bonds (or $\pi$ bonds as they should correctly be referred as) then you messed up your programs settings. (This is probably valid for any common molecular geometry file)

There's no double bonds in this PDB file. Could you please fix this?

This is no PDB file, these are xyz atomic coordinates. They have no representation of bonds whatsoever. I consider the use of double bonds in the depiction of 3D structures of molecules as too incorrect, as it will never be close to the actual bonding picture (except for ethene maybe). I do not have any experience with PDB, so I cannot fix it.

OK, this question is about PDB files so I'm going to need a valid PDB file for me to accept your answer. Although if you know how to import these coordinates into avogadro, I could probably use it to create the PDB file I desire.

I have updated the answer with your desired pdb. I included also a short tutorial on how you can do it yourself. Just a note on the display of double bonds. There will be double bonds between sulfur and oxygen, which are wrong. There might be double bonds in the phenyl ring, which is absolutely not correct, since it is a delocalised electron system.

Your interpretation of the remark on wikipedia is wrong.
It seems that you provided a pdb file of the structure as drawn and without any geometry optimization. As a result, N2, N3, and N13 are trigonal planar in your file, while there should in fact be pyramidalisation (there are lone pairs!).
Moreover, "the guy" isn't writing about bond angles, but about the dihedral angle formed by C1-N2-C8-C9, which is indeed close to 90° in your structure.
The criticism is justified.
Your should run a geometry optimization on the structure before you submit it.
In order to do that, open your pdb file in Avogadro, select MMFF94 as your force field and run a geometry optimization pressing AltCtrlO. If necessary, fiddle with the dihedral in question and repeat the optimization.

How could I do this?

@BrentonHorne Info on geometry optimization added to answer.

How do I select the force field?

@BrentonHorne Menu: Erweiterungen->Molekül-Mechanik->Kraftfeld einrichten. My Avogadro is in German.

Nvm I found out myself; but the amine still has a ~109-120 degree bond angle.

(See my question I just edited it to include the new structure of metamizole)

@BrentonHorne In the molecule nitrogen with planar trigonal geometry is connected with $sp^2$ carbon wich is clearly not electron-enriched. While it is unclear if nitrogen should be $sp^2$ in this particular case, it is common for nitrogen in such conditions to be (almost) $sp^2$. However, all substituents of both the carbon and the nitrogen and the carbon and the nitrogen must be in one plane in the case.

I performed the optimisation with Avogadro and it gave me the same structure as I showed in my question.

The nitrogen is trigonal planer with the lone pair orbitals aligned with the pi-bond orbitals on the 5-member ring. Also the benzene ring needs to be planer with the 5-member ring.


text:
How could alcohol be made into a powder?
In the news recently is 'Palcohol', powdered alcohol, that you add to water or existing liquids to make it alcohol. Presumably this would dissolve and break apart into alcohol and {some other compound}.
What could bond to ethyl alcohol in order to:
1) Create a solid, and
2) Not be disgusting as a side product?
Most things have some associated taste, $\ce{NaCl}$ is certainly out. If the 'other' components somehow combined to $\ce{H2O}$ or a gas like $\ce{O2}$ or $\ce{CO_2}$ that might remove any residual taste.
What would be in the list of available combinations with ethyl alcohol that are safe to consume, gaseous, or don't otherwise add weird tastes to the liquid?

Cyclodextrins, which comprise a class of cyclic carbohydrates, use their properties of being hydrophobic within their ring(s), yet hydrophilic outside. Alcohol can be put inside these rings. Then, the entire molecule is still capable of dissolving in water, releasing the encapsulated alcohol molecules. (src)
This popsci link has one way of making edible powdered alcohol. A nice casual step-by-step tutorial, actually...
Still you want to maintain a balance between maximizing alcohol (or do you?) and having the powder "feeling wet" due to high absorption of alcohol.
Also this may be a relevant Wikipedia article.


text:
Mistake in Wikipedia article on fugacity?
The Wikipedia Article on Fugacity contains statement

The contribution of nonideality to the chemical potential of a real gas is equal to RT ln $\phi$

Here $\phi$ is fugacity coefficient.I think it is wrong it should be fugacity coefficient multiplied by partial pressure of gas divided by standard pressure.So for the example of nitrogen gas at 100 atm it's chemical potential should be μ = μ$_{id}$ + RT ln 97.03 and not μ = μ$_{id}$ + RT ln 0.9703 as given in Wikipedia Article.
I am confused, who is wrong me or wikipedia ?

I think you have misinterpreted the passage on Wikipedia a little. Of course, you are right that the chemical potential is given by the formula
\begin{align}
 \mu_{i} (p_{i}, T) &= \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \log a_{i} \\
 &= \mu_{i}^{0} + RT \ln \left(\phi_{i} \frac{p_{i}}{p^0} \right)
\end{align}
where $R$ is the universal gas constant, $T$ is the temperature, and $\phi_{i}$, $p_{i}$, $a_{i}$, and $\mu_{i}$ are the fugacity coefficient, the partial pressure, the activity and the chemical potential of the $i$th component in the system, respectively.
But the passage you are quoting says that the contribution of nonideality to the chemical potential of a real gas is equal to $RT \ln \phi$ and in the equation
\begin{align}
 \mu = \mu_{\text{id}} + RT \ln \left(\phi \right)
\end{align}
$\mu_{\text{id}}$ is not the same as $\mu^{0}$. Rather it is meant in the following way: For an ideal gas the fugacity coefficient is $1$ by definition and so the chemical potential of an ideal gas, $\mu_{\text{id}}$, is given by
\begin{align}
 \mu_{\text{id}} = \mu^{\text{id}}_{i} (p_{i}, T) = \mu_{i}^{0} + RT \ln \left(\frac{p_{i}}{p^0} \right)
\end{align}
If you use the mathematical property of logarithms that $\log(a \cdot b) = \log(a) + \log(b)$ you can rewrite the equation for the chemical potential of real gases to
\begin{align}
 \mu_{i} (p_{i}, T) &= \underbrace{\mu_{i}^{0} + RT \ln \left(\frac{p_{i}}{p^0} \right)}_{= \, \mu_{\text{id}}} + RT \ln \left(\phi_{i} \right) \\
 &= \mu_{\text{id}} + RT \ln \left(\phi_{i} \right)
\end{align}
and so you see, that the contribution of nonideality to the chemical potential of a real gas is indeed equal to $RT \ln \phi$ and you get to the equation from the Wikipedia article.

Thanks i didn't look at $\mu_{id}$.I am ashamed of such silly mistake.

I can't vote up answer due to lack of reputation

@VishvajeetPatil If you keep participating in this site, I'm sure you'll get the chance :)

@VishvajeetPatil Mistakes like that happen very easily, nothing to be ashamed of. It shows you that it is important to always define the symbols in your equations. If the Wikipedia article had spelled out what $\mu_{\text{id}}$ was explicitly in the first place your confusion couldn't have arisen.

I have another question about chemical potential can you solve that Question


text:
Explain isotopic uranium/perchloric acid digestion
What is happening in this video? Can you explain this technically as well as in layman's terms?
I'll summarize it. I see a technician moving fuming mixtures of presumably uranium and organic compounds on and off hot plates. Perchloric acid is mentioned, and a large bottle is present.

If that is a perchlorate digestion, the technician needs more training.  The hood door must be lower despite the Official marker.  His cotton lab coat must NOT be exposed to vapors!  Sleeves should be inside long gloves.  He has a bottle of perchloric acid next to a running hot plate!  I'd have a 5-gallon bucket of water nearby if the gloves ignite.  As perchlorate digestions occasionally explode, and he's doing a lot of them, I'd wear goggles under the face shield and wear Dyneema or Spectra body armor under the lab coat.  One might have some concern for hands (longer pair of tongs) and neck, too.
He's apparently cooking off a sample of organics-contaminated uranium whatever for trace contamination or isotopic analysis.  Everything oxidizable will be gone, leaving uranium oxides.
That is NOT marked as a perchlorate fume hood!  Everything organic downwind of those fumes is a potential high explosive, and certainly a violently flammable material including its own oxidizer.  There will be massive corrosion of all metals.  You don't want to be standing, living, downwind of the exhaust stack.  OTOH, the operation is low cost/sample, and anybody off the street is a potential lab tech.


text:
Reaction involved in Combustion of Nitrocellulose-Nitroglycerine(NCNG) solid propellant?
I'm working on solid propellant based microthrusters and I am using NCNG as the solid propellant. I want to know the reaction taking place and the mass fraction of the products obtained on combustion of NCNG propellant.

Nitrocellulose is complex stuff. It is generally considered to be the fully nitrated version of cellulose, in which every available alcohol functional group is converted to a nitrate:
$$\ce{[C6H10O5]n +3HNO3 -> (C6H7O2(ONO2)3]n +3H2O}$$
In principal it probably contains some small amount of partially and/or non-nitrated units.
If we consider nitrocellulose to be fully nitrated, then the combustion reaction is:
$$\ce{ 2C6H7N3O11 + 9/2O2 -> 12CO2 + 3N2 + 7H2O}$$
For nitroglycerin $(\ce{C3H5N3O9})$:
$$\ce{2C3H5N3O9  -> 6CO2 + 3N2 + 5H2O + 1/2O2}$$
It will depend on the ratio of nitrocellulose to nitroglycerin what the overall $\ce{CO2}:
\ce{N2}:\ce{H2O}$ mole ratio is. I'll leave you to figure out the mass ratios using the molecular masses of the products.

Energetic materials can emit a significant fraction of their carbon as CO.  Take advantage of excess local oxygen with more fuel, or push to CO2 with more oxidizer.  The lower the average molecular weight of the exhaust, the higher the Isp, all other things being equal.  US hydrogen/LOX engines run heavy on hydrogen.
re Cordite and Ballistite, grain size, and internal channel geometry if any.
The Mechanism of the Burning of Double-Base Propellants
DOI: 10.1021/j150480a015
http://www.google.com/patents/US4072546
http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA425264
Solid Propellant Chemistry, Combustion, and Motor Interior Ballistics, Volume 185. Vigor Yang, Thomas B. Brill, Wu-Zhen Ren, Paul Zarchanm, 2000,. p. 288 ff.
Double-base propellants (DB) give minimal smoke with medium-high performance, Isp ~ 235 s. Adding aluminum gives Isp ~ 250 s with visible smoke.  DB have 50-60% nitrocellulose and 30-40% nitroglycerin content.  High nitroglycerin for extrusion and casting, low for pressing.  DBs are boosted by adding dispersed micronized ammonium perchlorate.  Both nitrocellulose and nitroglycerin decompose to afford nitric acid.  This must be neutralized during long term storage, certainly if perchlorate is in there.  Beware of detonation under pressure, hence phlegmatizing agents.
double-base propellant energetics


text:
Heating CrCl3(bpy)3 — isomerism and ligand exchange

$\ce{CrCl3(bpy)3}$ (A) when heated forms (B) which has the formula $\ce{CrCl3(bpy)2}$ and has one free $\ce{Cl-}$ per mole of (B). B can react with ethylene diamine ($\ce{en}$) to form (C) which has the formula $\ce{CrCl3(bpy)2(en)}$.
(B) can also be converted into (D) which also has the formula $\ce{CrCl3(bpy)2}$ but cannot react with $\ce{en}$.
Identify compounds A–D and draw their structures

I know that B is cis-$\ce{[CrCl2(bpy)2]Cl}$
C is $\ce{[Cr(bpy)2(en)]Cl3}$. The $\ce{en\bond{->}metal}$ bonds are where the $\ce{Cl-}$ ions in B are.
D is the trans isomer of B.
What I dont understand is how you're meant to know what A is. I assume it's $\ce{[Cr(bpy)3]Cl3}$ as a later part of the question asks about relative molar conductivity of A and B.
Why would heating that cause a $\ce{bpy}$ to leave the complex and two chlorides to replace it? Or is it $\ce{[CrCl2(bpy)2]Cl(bpy)}$ in which case, why does heating it not form the more stable complex I've stated above?

It is Cr(III) not Co(III).  Cr(III) has reasonable substitution kinetics. Your structure for A is good. Heating drives off a mole of volatile bipy, then two chlorides inner sphere coordinate to take its place.  The third chloride remains outer sphere and readily exchangeable. Bidentate en comes in to displace two inner sphere chlorides outer sphere again.

This does not answer the question.

If you are given only the options $\ce{Cr(bpy)3Cl3}$, then it is mandatory for the bipyridine ligand to actually be attached to the chromium centre. If it were not, as in your suggestion $\ce{[Cr(bpy)2Cl2]Cl . bpy}$, this additional bipyridine can dissociate away as there is nothing keeping it close to the complex. On the other hand, isolateable substances are always neutral, thus if your complex cation is $\ce{[Cr(bpy)3]^3+}$, then the three chloride anions will remain in the same crystal structure and you can isolate a complex salt of the type $\ce{[Cr(bpy)3]Cl3}$. This molecular structure can be proven by being able to precipitate three moles of chloride per mole of complex, e.g. with $\ce{AgNO3}$.

Why would heating that cause a $\mathrm{bpy}$ to leave the complex and two chlorides to replace it? Or is it $\ce{[CrCl2(bpy)2]Cl(bpy)}$ in which case, why does heating it not form the more stable complex I've stated above?

This somehow assumes that $\ce{[Cr(bpy)3]Cl3}$ is the most stable of these complexes. However, the chemical literature thinks different. Burstall and Nyholm first reported the synthesis of $\ce{[Cr(bpy)2Cl2]Cl}$ from anhydrous chromium(III) chloride using three (!) equivalents of $\ce{bpy}$ and heating to reflux.[1]
On the other hand, traditional methods to generate $\ce{[Cr(bpy)3]^3+}$ salts — most commonly perchlorates $\ce{[Cr(bpy)3](ClO4)3}$ — typically started from a corresponding $\ce{[Cr(bpy)3]^2+}$ and employed oxidation.[1] (This was apparantly first published in 1932 by Barbieri and Tettamanzi in an obscure journal that I can’t find.)[2] Relatively recently, a new method to synthesise the $\ce{[Cr(bpy)3](ClO4)3}$ complexes from $\ce{CrCl3 . 6 H2O}$ was published — which, however, builds on completely removing any chloride from the solution by using silver(I) salts as in equation (1).[3]
$$\begin{multline}\ce{CrCl3 . 6 H2O + 3 [Ag(bpy)2](ClO4) ->\\ [Cr(bpy)3](ClO4)3 + 3 AgCl v + 3 bpy}\tag{1}\end{multline}$$
Seemingly, having chloride ions in solution around the $\ce{[Cr(bpy)3]^3+}$ ion can’t be trusted.
Partially due to SciFinder being down for maintenance as of now, I was unable to find any references for the synthesis or properties of $\ce{[Cr(bpy)3]Cl3}$, but given the Burstall-Nyholm method I would not assume the cation to be more stable than $\ce{[Cr(bpy)2Cl2]+}$.
A likely explanation for the lower stability could be the charge balance. The dichloridobis(bipyridine)chromium(III) has an overall charge of $1-$, which is generally okay. In tris(bipyridine)chromium(III), there is a very small $\ce{Cr^3+}$ ion in the centre of a complex surrounded by otherwise neutral ligands. Any method to relieve the charge could well be favourable; amoung them reduction of the metal or ligand exchange with anionic ligands. Remember that charge balance is a thing: $\ce{[CoCl4]-}$ forms when cobalt(III) reacts with chloride ions mainly due to charge balance reasons.

References:
[1]: F. H. Burstall, R. S. Nyholm, J. Chem. Soc. 1952, 3570. DOI: 10.1039/JR9520003570.
[2]: Barbieri, Tettamanzi, Atti R. Accad. Lincei. 1932, 15 877. Reference as cited in [1].
[3]: W. Kharmawphlang, S. Choudhury, A. K. Deb, S. Goswami, Inorg. Chem. 1995, 34, 3826. DOI: 10.1021/ic00118a035.

Nicholas, Cr(III) is generally 6-coordinate and both bipy and en are bidentate.  So 3 bipy in (A) connect to the metal to form the cation, leaving 3 $\ce{Cl-}$ in the outer sphere.  Then in (B) one bipy leaves and its two positions are occupied by $\ce{Cl}$, and as you say the configuration must be cis (both because a bidentate ligand departed and because another can easily come in).  Isomerization of (B) to (D), the trans isomer, probably occurs slowly, and (D) cannot accept en because the $\ce{Cl}$ sites are opposite instead of adjacent.

This does not answer the question.


text:
Generally available fluid with good heat transport properties
I have designed a container which is heated up and has to heat up some tube-coils of water inside of it. This container is filled with a fluid that can help store and transfer the heat received from the container to the tube-coils.
My plan was to find some material that I can easily buy and which is not very expensive to fill the $636cm^3$ container.
It has to have a high coefficient of heat transfer, high heat capacity and an operating temperature of at least the range of 10-400 degrees Celsius. I will also need to calculate how much the volume changes when it goes from around -35 to 500 degrees Celsius.
I also think a high density and if the material is non-toxic might be a plus.
Does anybody know any fluid that could be used for this or somewhere where I could find such information?

400 C is beyond long term use of any organic heat transfer fluid.  Dow Corning boasts 427 C for one of its silicone fluids.  Don't bet on that.  Inorganic eutectic salts will long term decompose into corrosive volatiles.  Best bet for a red hot closed system is fusible alloys of indium, gallium, and bismuth - beware of metal boiling points.  Open systems will air oxidize.   Accommodate sealed inert gas fill (argon) with a metal bellows add-on, preferably on a pipe extension away from the hot zone.  Fill composition must be compatible with container composition.  Mercury is out for toxicity and boiling point, 357 C.  Thermal coefficient of linear expansion, solid vs. liquid alloy, must be looked up.  The increment is cubed for volume change.
500 C (red heat) will change the metallurgy of stainless steels over time.  They will harden (carbide precipitation), among other things.  Don't try to drill a hole or tap a thread in stainless that has been soaked at red heat.

Have you tried Wax?
'Paraffin wax is a white or colorless soft solid derivable from petroleum, coal or oil shale, that consists of a mixture of hydrocarbon molecules containing between twenty and forty carbon atoms. It is solid at room temperature and begins to melt above approximately 37 °C (99 °F);[1] its boiling point is >370 °C (698 °F).[2] Common applications for paraffin wax include lubrication, electrical insulation, and candles. [3] It is distinct from kerosene, another petroleum product that is sometimes called paraffin.
Paraffin candles are odorless, and bluish-white in color. Paraffin wax was first created in the 1850s, and marked a major advancement in candlemaking technology, as it burned more cleanly and reliably than tallow candles, and was cheaper to produce. [4]
In chemistry, paraffin is used synonymously with alkane, indicating hydrocarbons with the general formula CnH2n+2. The name is derived from Latin parum ("barely") + affinis, meaning "lacking affinity" or "lacking reactivity", referring to paraffin's unreactive nature.[5]"

Welcome to Chemistry.SE! Could you please site the source you took this from?


text:
About the nomenclature: manganate or manganite?
I am a bit confused about the correct name of $\mathrm{LaMnO}_3$. Is it lanthanium manganate or lanthanum manganite?
I was assuming that since $\mathrm{SrTiO}_3$ is called strontium titanate, $\mathrm{LaMnO}_3$ would then be called lanthanium manganate, but I got confused when I heard a collegue referring to it as manganite and after reading a wikiepdia article on $\mathrm{La}_x\mathrm{Sr}_{1-x}\mathrm{MnO}_3$ where they also used the term manganite.
Which is the correct form, or are they simply interchangeable?

Let's go down the ladder.  $\ce{KMnO4}$ is permanganate, Mn(VII).  $\ce{K2MnO4}$ is manganate, Mn(VI).  $\ce{K4MnO4}$ is manganite, Mn(IV).  $\ce{LaMnO3}$ is called both manganate and manganite, but Mn(III). In a proper world it should be lower than that manganite.  They are pedagogically both wrong if it is La(III).
http://dictionary.sensagent.com/potassium%20manganite/en-en/
Manganese oxidation state nomenclature is a real world mess. Drawing nomenclature parallels to Cl(VII) and downward is not valid.  Drawing nomenclature parallels to titanium is poor, for Ti(IV) is the highest common oxidation state.  $\ce{SrO + TiO2}$ gives $\ce{SrTiO3}$, properly called titanate, for it is Ti(IV) on both sides.
Organic had the same problem. The highest valence carbocation was tricoordinate, hence a carbonium ion.  Magic Acid gave $\ce{CH5^+}$.  Carbonium ions were renamed carbenium ions to general confusion.  Then, $\ce{CH6^{2+}}$.  Organic dropped the ball with direction of optical rotation versus geometric left and right.  It was flat out stupid designating E ("trans") and Z ("cis") olefins.  The designation should have been abstract rather than language-originated.

The last paragraph of your answer does not seem to have anything to do with with the admittedly clunky nomenclature of the metal oxyanions.

Also, your statement about optical isomerism is confusing, given that the direction of rotatation of plane-polarized light is generally unrelated to the absolute spatial geometry of the chiral molecule

Do better: http://www.mazepath.com/uncleal/norbors.gif  http://www.mazepath.com/uncleal/norone.png P3(1)21 quartz and P3(2)21 berlinite are both levorotatory despite being opposite geometric hands, J. Appl. Crystallogr. 19, 108 (1986). CIP notation flips within unchanged geometry (L-(S)-alanine, L-(R)-cysteine). Achiral PhCOCH3 impurity changes resolved PhCH(OH)CH3 optical rotation, J. Org Chem. 38(10), 1870 (1973).


text:
How does the process of protein denaturatation work?
I know it's the proteins that somehow change their properties at some temparature.
But what kind of change, in terms of molecule-level structure?

protein folding is biophysics, which is a separate department at many universities.  Isn't biophysics ok here?

@jinawee I would think the biologic effects of the changes are irrelevant in regards to the question, as are the chemical effects too - what remains is physics.

@jinawee http://meta.physics.stackexchange.com/q/4403/

Let's see what others think. But I guess it could be more specific, like: What is the physical process of denaturatation?

This is related to unfolding of structures of proteins... This is certainly chemistry... In fact I remember seeing a type of this question in my chem book..

http://en.wikipedia.org/wiki/Denaturation_(biochemistry)#How_denaturation_occurs_at_levels_of_protein_structure

@jinawee Ok, makes sense, I'll change to that title!

@Awesome Just for future reference--rather than input on this question--it is possible for questions to be on-topic on more than one Stack Exchange site. In particular there are substantial areas of overlap between physics and both chemistry and astronomy. In those the poster can put it where (s)he likes, but for new posters you might appraise them of the existence of the other site.

Basically, the egg-white (just another protein structure) is made of long chains of amino acids. They're held in shape by weak bonds. When you heat the structure, the energy you put in, is enough to break those bonds, thereby destroy the structure. This process has a name for it.
Once the temperature is high enough, new covalent bonds can form between the amino acids, and the amino acids in some other protein and they become entangled into a much more complicated structure, which is completely different from the real one. As you cool, you get this finished structure as cooked egg...

Ah, thanks, now I start to see how it's chemistry, I saw it as related to a phase change... I'll ask for migrating it - Oh, I'll better take out of the title the explicit "physical" before


text:
Calculation of the enthalpy change of reactions
First, I’m a high school student and I wonder why every textbook just tells you how to set up those equations and cancel them to get the entropy change but never tells you why you can cancel the same species on both sides.
Second, for example,
$\ce{2B + 3H2O -> B2H6 + 3/2O2} \tag{  $\Delta H= 762~\mathrm{kJ}$}$
$\ce{H2 + 1/2O2 -> H2O} \tag{  $\Delta H=-286~\mathrm{kJ}$}$
Here the textbook tells us we should add up two reactions in order to get what we want (I didn’t post the whole question, just go with this)
Now, why do we have to add the two whole reactions up instead of just change the $\ce{H2O}$ on the left side into $\ce{H2 + 1/2O2}$?  What I understand about this $\ce{H2 + 1/2 O2 ->H2O}$ is if we reverse it, we can say if $\ce{H2O}$ takes in $286~\mathrm{kJ}$ of energy, it will turn into $\ce{H2}$ and $\ce{1/2 O2}$. So why we can’t just add $286~\mathrm{kJ}$ to the equation? Why we still need to add the whole thing $\ce{H2 + 1/2 O2 -> H2O}$ to the other equation?

You always need to add the equations since the break of bonds or the formation of bonds leads to an exchange in energy.
Here is a simple case
$$\begin{alignat}{2}
\ce{2 B + 3 H2O &-> B2H6 + 3/2 O2}\quad&&(\Delta H = 762\ \mathrm{kJ})\\
\ce{H2 + 1/2 O2 &-> H2O}\quad&&(\Delta H = −286\ \mathrm{kJ}) \times 3\\
\hline
\ce{3 H2 + $\boldsymbol{\ce{3/2 O2}}$ &-> $\boldsymbol{\ce{3H2O}}$}\quad&&(\Delta H = −858\ \mathrm{kJ})\\
\ce{2 B + $\boldsymbol{\ce{3 H2O}}$ &-> B2H6 + $\boldsymbol{\ce{3/2 O2}}$}\quad&&(\Delta H = 762\ \mathrm{kJ})
\end{alignat}$$
Notice the ones bold get canceled since the product is the reactant in the other equation.
$\Delta H_\mathrm{r}$ = sum of the $\Delta H$s so it’s equal to $-96\ \mathrm{kJ}$.
And in the second equation you were talking about one $\ce{H2O}$ molecule however in the first you are talking about three.


text:
at constant volume why the change of enthalpy doesn't equal the change of heat
At constant pressure $p=\text{const.}$, change of internal energy equals heat subtract work(PV). So, $q=\Delta E - pV$. So is the the change of enthalpy. So we say at constant pressure, the change of enthalpy equals the change of heat. But when the volume is constant $V = \text{const.}$, no work is done. $\Delta E= q$, $\Delta H = \Delta q$. So why not in this situation too?

At constant volume, the pressure is not necessarily constant. The pressure will usually increase when the system is heated. This makes the $PV$ term not constant. For this reason, you get:
$$\Delta H = \Delta (E + PV) $$
$$\Delta H = \Delta E + \Delta (PV) $$
$$\Delta H = \Delta E + V\Delta P $$
Now heat is just $\Delta E$, because there is no work. So,
$$\Delta H = q+ V \Delta P $$


text:
Predicting the pH of a weak acid and weak base solution
I know that it is possible to predict whether combinations of acids and bases will be acidic, basic, or neutral:

weak acid (WA) and strong base (SB) reacts basic
strong acid (SA) and and weak base (WB) reacts acidic
SA and SB reacts neutral

How does a WA react with a WB? I've been told that it isn't neutral. How do you predict what it is and why is it not neutral?

Pretty cryptic indeed... At that point I'd say compare the Ka and Kb of both and then you can have an idea on which of the molecules is "stronger"

Note: The expected $\text{pH}$ values refer to 1:1 mixtures as suggested in the question

One major feature of weak acids and bases is that they do not dissociate completely in water, producing less free oxonium or hydroxide ions than strong acids and bases do (Brønsted). For example in Sodiumacetate, a weak base, there is an equilibrium where the reactions left to right and right to left occur at the same speed, hence cancelling each other.
\begin{aligned}\ce{
   2NaOAc_{(aq)} + H2O &<=> 2Na+_{(aq)} + {}^{-}OAc_{(aq)} + HOAc_{(aq)} + {}^{-}OH_{(aq)}} &(1)
\end{aligned}
The same also applies to weak bases like ammonium hydroxide $\ce{NH4OH}$. Note, that ammonium itself is not a Brønsted base as it lacks to donate hydroxide ions.
\begin{aligned}\ce{
   NH4OH_{(aq)} &<=> NH4+_{(aq)} + {}^{-}OH_{(aq)}} &(2)
\end{aligned}
Weak acids can be expanded in the same way, example benzoic acid:
\begin{aligned}\ce{
   Ph-COOH_{(aq)} + 2H2O&<=> Ph-COO^{-}_{(aq)} + H+3O_{(aq)}} &(3)
\end{aligned}
In the Lewis definition, weak acids are not able to complex the hydroxide ions of water's autoprotolysis.
\begin{aligned}\ce{
   2H2O &<=> H+3O_{(aq)} + {}^{-}OH_{(aq)} &\\
   BF3_{(aq)} + 3H2O &<=> [HOBF3]^{-}_{(aq)} + 2H+3O_{(aq)} + {}^{-}OH_{(aq)} &(4)
}\end{aligned}
These equations are also in equilibrium, meaning all species are present in aqueous solution.
In the Lewis definition ammonia $\ce{NH3}$ is a base and it reacts analogous to the above.
\begin{aligned}\ce{
   2NH3_{(aq)} + 2 H2O &<=> NH4OH_{(aq)} + NH4+_{(aq)} + {}^{-}OH_{(aq)}} &(5)
\end{aligned}
Now the rate of dissociation is strongly dependent on the weakest bond. In $(1)$ that is $\ce{Na-O}$ and successively $\ce{H-O}$. Overall the reaction will produce excess hydroxide ions, hence the solution will have a $\text{pH}>7~(@25~^\circ{}\!C)$.
The same applies analogously to $(2)$ and $(5)$
In $(3)$ the opposite is the case. The reaction will produce excess oxonium ions, hence having a $\text{pH}<7~(@25~^\circ{}\!C)$. The same applies to $(4)$.
So when adding a weak base to a weak acid, one has to evaluate the whole equilibrium:
\begin{aligned}\ce{
   NH4OH_{(aq)} + HOAc_{(aq)} + H2O
   <=>&\\ [(NH4)(OAC)]_{(aq)} + NH4+_{(aq)} &+ {}^{-}OH_{(aq)} +  {}^{-}OAc_{(aq)} + H+3O_{(aq)}
} &(6)
\end{aligned}
It is therefore very much dependent on the chemical properties of all compounds involved in the solution. These systems are also very well known as buffer solutions.
So a good start would be comparing $\text{p}K_a$ and $\text{p}K_b$ values of the involved components, giving you a hint whether it will be more acidic or more basic. But you also have to pay attention to the amount of substance in particular in these cases. Dilution also plays an important role in this kind of chemistry.


text:
How can I reduce Crystallization temperature of seed oil to be used in winter as biodiesel?
How can I reduce Crystallisation temperature of seed oil to be used in winter as biodiesel?
The glycerol has been removed from the oil, and white spirit is used to reduce its viscosity.

Trans fat oils don't crystallise usually.

Waxing and pour point are typical specs for fuel. How cold is your winter? If it is very cold, and your tank is above ground, there is little to be done. Insulate the tank (metal-backed rock wool) and heat it.  Use a more unsaturated feedstock with no saturated components. That lowers its pour point and removes waxing.  Did you dry your fuel?  Unsaturated fuels oxidize and deposit varnish.  "Biofuel" is a good substrate for microbial growth.
Enviro-whinerism, sustainability, social consciousness, are Luddite lies.  They assume thermodynamics is amenable to legislation and good feelings, and weather is always mild and cooperative.  None of that obtains in the real world.
http://en.wikipedia.org/wiki/Pour_point
http://www.infineum.com/Pages/pourpoint.aspx
http://www.aftonchemical.com/Solutions/LubricantComponents/Pages/PourPointDepressants.aspx
http://www.globalp.com/products/product.cfm?productID=32
Chemicals

That second paragraph seems to be a little off topic doesn't it?


text:
Where does the characteristic peak for phtalic acid diethyl ester at m/z 149 originate?
Which fragmentation is possible at this m/z value for phthalic acid diethyl ester?
I tried McLafferty but it didn't solve my problem.

First of all, you should indicate if you consider the fragmentation in electron ionization or in other ionization conditions. Considering it is the electron ionization conditions and that you refer to this spectrum, one can postulate for a reasonable mechanism, which can be derived from what is known on the fragmentation of benzoic acid and esters. I am not aware of any complete study of phtalate ester fragmentation, but I did not carry a complete litterature search.
In the figure below is a proposal for the mechanism leading to the formation of two possible structures for the $m/z$ 149 ion.
If one consider that is fragments as do benzoate esters, through radical induced $\alpha$-bond cleavage leading to the loss of $\ce{CH3CH2O^.}$ followed by standard decarbonylation of the acylium ion. For phtalate esters, the question at this point is whether the structure will remain as I or might undergo isomerisation to more stable structures.

I have provided a few alternative structures: simple addition of the neighbouring carbonyl would lead to structure II which although it has some constraints, allows a good stabilization of the positive charge both in the ester and by the aromatic ring. Other structures can be postulated, such as III after addition of the ethanoyl oxygen to the aromatic cycle. This four membered ring can reopen to lead to the acylium structure IV which can in turn undergo decarbonylation to form the ion $m/z$ 121, also present in the spectrum. So there is some indirect evidence that these structures might be involved in the fragmentation process.
Finally, in order to explain the formation of ions at $m/z$ 105 and 77, one can also propose a last isomer for ion $m/z$ 149, structure V, formed after 1,5 hydride transfer. This one can quite easily lead to loss of acetaldehyde, eventually followed by decarbonylation to lead to ion $m/z$ 77.
If phtalates present specific reactivity, the other option is that ion A at $m/z$ 177 can cyclize to structure B which is a very good candidate for ethene loss leading also to an ion at $m/z$ 149.
The question then becomes which of these two $m/z$ 149 is the actual ion. These two isobaric ions could be resolved by high resolution experiments, but after looking in publicly available databases, I did not find any high resolution spectra. Deuterium labelling of the ethyl groups could also prove handy, but once again I did not find any spectra of diethylphtalate labelled on the ethyl groups. (The usual diethylphtalate used in quantitation experiments seems to be the $d_{4}$ isotopomer, with deuterium on the aromatic ring.) Finally one can also have a look at isomers and related ions to see how they fragment. And indeed if one takes the dipropylphtalate ester, the ion at $m/z$ 149 is present (produced after propanoyl loss followed by propene loss) and there is no strong ion at $m/z 163$ which would tend to be a good indication of ethene loss for diethylphtalate. But, on the other hand, both the meta and para isomers of diethylphtalate have a much weaker ion at $m/z$ 149, which cannot be accounted so easily by an equivalent to intermediate B. So I would probably favor the concurrent occurence of both pathways, with a ratio that still needs to be measured.
In addition to the answer above, it should have been quite immediate that a McLafferty rearrangement was not an option: since there are no nitrogen, the nitrogen rule allows to conclude directly that the odd mass of the ion at m/z 149 indicates that it is an even electron cation, as opposed to radical cations produced by McLafferty rearrangements.

Since the two ester groups are present ortho to each other, fragmentation via loss of ethoxide ion(by ortho effect) and ethene in different steps can also give a compound with peak at m/z 149.

@user5255: You are right to point this out. If one considers the meta and para isomers, the peak at m/z 177 becomes the base peak, altough the peak at m/z 149 remains present, which might indicate either coexistence of two fragmentation pathways in competition (loss of CO vs loss of ethene from m/z 177) or that formation of m/z 149 is favored by the possible formation of more stable cyclic delocalized structures. High res EI mass spectra could give an answer to this question, but I did not find any available. I will edit my answer accordingly. BTW which ortho effect are you refering to?

@user5255: following your remark, I have completed the answer. After looking at a bunch of EI spectra of phtalate esters and their isomers, I feel like there are still a lot of open questions.

The MW of diethyl phthalate is 222.  The peak at 149 corresponds to loss of a MW=73 fragment.  This would correspond to loss of one of the ester groups ($\ce{CO_2Et}$).

Addition of a proton, then loss of $\ce{CO2Et}$.

the fact that dimethyl phthalate does not produce significant amounts of the 149 m/z fragment, but all other phthalates with a longer alkane chain do, would put decarbonylation in favour, leading to fragments I - V from above scheme.