Are the orders of reactants with respect to a reaction different for different starting concentrations? If the reaction is repeated with $\pu{2 M}$ ethyl iodide the pyridine concentration decreases as shown below. Give the rate law of the reaction in terms of pyridine and ethyl iodide. Explain your reasoning.
$$
\begin{array}{l|rrrrr}
\hline
t/\pu{min} & 0 & 20 & 40 & 80 & 120\\
\hline
[\ce{C5H5N}]/\pu{M} & 0.1 & 0.0746 & 0.0559 & 0.0312 & 0.0175\\
\hline
\end{array}
$$

I have found the rate constant and found that the orders of ethyl iodide and pyridine are both $1$ with respect to the reaction when their initial concentrations were both $\pu{0.1 M}.$ I would have thought this would be the case for any set of initial concentrations.
I would like to confirm this. In that case, I am thinking the only thing different would be the rate constant. But how would I go about finding that?
Could I plot the data, interpret the half life and go from there? Or is my approach totally wrong?

A reaction's rate does depend upon the consumption of its reactants, and the manner in which the reactants interact. Let's consider the following reaction: $$\ce{A + B -> final products}$$ The rate of consumption of  reactants $$\dfrac{-d[A]}{dt}= k[A][B]$$ and $$\dfrac{-d[B]}{dt}= k[A][B]$$
It seems that this reaction is a second order reaction since the rate order appears to be:$$\text{Rate} = k[A][B] $$
As @Andrew astutely pointed out in the comments, a reaction can be of a lower order if one of the reactant's concentration is much higher than the other's.
Let's suppose, that one reactant is in great excess (say $10 \text{M of B vs } 0.01\text{M A}$)
So what we're observing now is that the reactant B is barely consumed, yet the reaction terminates since reactant A, the limiting reagent is exhausted.
In other words, the rate equation looks something like this: $$\text{Rate}=k'[A]$$, where $k^{'}=k[B]_o$ .Now, the reaction appears to be a first order reaction. This is known as a pseudo first-order reaction, since the rate of the reaction ultimately depends only on one reactant.
An example that I found here:
$$ \rm{CH_3Br + OH^- \rightarrow CH_3OH + Br^-}$$

[...]"Imagine we had an initial concentration of CH3Br of 100 μM and and an initial concentration of OH- of 10 mM.   Now even if all of the CH3Br has reacted the concentration of OH- will be essentially unchanged.  Therefore during the course of the reaction, the concentration of OH- will be essentially constant.  This makes the reaction "like a first order reaction", thus the name pseudo-first order."

Effectively, the rate now is: $$\ce{rate= k'[CH_3Br]}$$
This can also be mathematically represented in the integrated rate law:
$$\dfrac{1}{\ce{[OH-]_0 - [CH3Br]_0}}\ln \dfrac{\ce{[OH-][CH3Br]_0}}{\ce{[CH3Br][OH-]_0}} = kt \quad \ce{where [OH-]\neq[CH3Br]}$$
Now, since $\ce{[OH-]_0 \gg [CH3Br]_0}$, we can write the equation as:
$$\dfrac{1}{\ce{[OH-]_0 - [CH3Br]_0}}\ln \dfrac{\ce{[OH-][CH3Br]_0}}{\ce{[CH3Br][OH-]_0}} \approx \dfrac{1}{\ce{[OH-]}}\ln \dfrac{\ce{[CH3Br]_0}}{\ce{[CH3Br]}}=kt $$
Remember, however that the rate and order of a reaction usually depends upon its mechanism.
For further reading, check here and here as well.


When writing out synthetic reaction schemes, why are some reactions combined in one arrow? An example of what I mean is here in the synthesis of compound 13. The reaction of the first arrow describes a single reaction, whereas the second arrow to form the final compound seems to have five stages combined into one. I understand that the compound is carried on as is without characterization when the parts are combined into a single step.

If one were designing a new synthesis, what factors would lead one to write (and characterize) each step of the synthesis versus combining many individual reactions in one apparent step?

P. S. This is not an organic chemistry question per se, but meta-ochem (sorry if my tags are limited in description).

I want to start off by saying that your example is a particularly bad one. Usually, the arrows in a scheme should at least state a list of reagents and steps. Instead, I would love to discuss this using scheme 7 of Zhang et al.’s Beilstein J. Org. Chem. article.[1] (Beilstein J. Org. Chem. is a open access journal, the full text should be accessible from the DOI link provided below.)

The main reason for combining reactions like this is brevity. For example, every practicing chemist will immediately understand that the homoallyl Grignard under 1. will attack one of the carbonyl functions, that triethylsilane will reduce an oxygen and that TBAF will remove the silyl protecting group on the hydroxy function. None of these steps really warrant any discussion and most people would be able to draw out the missing structures in a few minutes. Likewise, I am not entirely sure why they did not include the preceeding protection into that same reaction scheme because there too there is noting interesting going on.

In more advanced synthetic schemes, the steps that are shortened and written together on a single arrow are often typical name reactions or very easily performed ones:

protections and deprotections
oxidations of alcohols to aldehydes with standard reagents
reductions with DIBAL, LAH or $\ce{NaBH4}$
ozonolyses
double bond isomerisations
simple Wittig/HWE reactions
etc.

They are typically chosen in such a way that the path is obvious and that the structure on the end of the arrow can be drawn up quickly by going through the steps. They will practically never include key steps or steps whose stereochemical outcome is uncommon. Cross-couplings of more elaborate structures or lesser known name reactions are also less common.

However, there is no rule as to what can be combined and what cannot. If anything, the reviewers might mark a scheme as too elaborate (i.e. can be shortened) or too brief (needs elaboration).

Reference:

[1]: J. Zhang, H.-K. Zhang, P.-Q. Huang, Beilstein J. Org. Chem. 2013, 9, 2358–2366. DOI: 10.3762/bjoc.9.271.


The difference between heat of reaction at constant pressure and that at constant volume The $K_\mathrm{p}$ value of a reaction is $\pu{10 atm}$ at $\pu{300K}$ and $\pu{4atm}$ at $\pu{400K}$. The incorrect statement about the reaction is-

  a. The reaction is exothermic

  b. The rate of forward direction is more than that of backward
      reaction.

  c. The rate of backward reaction increases more than that of forward
      reaction with increase of temperature

  d. The difference between heat of reaction at constant pressure and that at constant volume is $RT$.

The correct answer given is option b.

But then how can we say that the difference between the heat of reaction at constant pressure and that at constant volume is $RT$ for the above reaction?

How do we go about approaching this problem? Do we need to use the van't Hoff  equation? I tried using the ideal gas equation also but didn't get this $RT$. Also I think since change in enthalpy is negative so the reaction is exothermic.

So the question is: Why is the difference between the heat of reaction at constant pressure and that at constant volume equal to $RT$?

First Part: heat of reaction

From the first law of thermodynamics we know

\begin{equation}
  \mathrm{d}U = \delta Q + \delta W = \delta Q - p \mathrm{d} V
\end{equation}

where $U$ is the inner energy, $Q$ is the heat, $W$ is the work, $p$ is the pressure and $V$ is the volume.
So, for a reaction at constant volume you get

\begin{equation}
  \mathrm{d}U = \mathrm{d} Q \ .
\end{equation}

Now, for a reaction at constant pressure you can invoke the relationship

\begin{align}
  \mathrm{d}H &= \mathrm{d}U + p \mathrm{d}V + V \mathrm{d}p \\
              &= \delta Q - p \mathrm{d} V + p \mathrm{d}V + V \mathrm{d}p \\
              &= \delta Q + V \mathrm{d}p \\
              &\overset{\mathrm{d}p = 0}{=} \mathrm{d} Q
\end{align}

where $H$  is the enthalpy.
This gives you the relationships that describe the change in heats released during reactions at constant volume and constant pressure.
The so called heat of reaction is defined as

\begin{equation}
  \Delta Q = \biggl( \frac{\partial Q}{\partial \xi} \biggr)
\end{equation}

where $\xi$ is the extent of reaction.

Second Part: temperature dependence

Now using the equality of mixed partials and the definitions of the heat capacities at constant pressure $C_{p} = \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p}$ and constant volume $C_{V} = \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V}$ you get for constant volume ($\mathrm{d}U = \mathrm{d} Q$):

\begin{align}
  \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta U}{\partial T} \Biggr)_{V} &= \left( \frac{ \Bigl( \frac{\partial U}{\partial \xi} \Bigr)_{T, V}}{\partial T} \right)_{V} = \Biggl( \frac{\partial^{2} U}{\partial \xi \partial T} \Biggr)_{V} \\
  &= \Biggl( \frac{\partial^{2} U}{\partial T \partial \xi} \Biggr)_{V} = \left( \frac{ \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial  C_{V}}{\partial \xi} \Biggr)_{T, V} = \Delta C_{V}
\end{align}

and for constant pressure ($\mathrm{d}H = \mathrm{d} Q$):

\begin{align}
  \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta H}{\partial T} \Biggr)_{p} &= \left( \frac{ \Bigl( \frac{\partial H}{\partial \xi} \Bigr)_{T, p}}{\partial T} \right)_{p} = \Biggl( \frac{\partial^{2} H}{\partial \xi \partial T} \Biggr)_{p} \\
  &= \Biggl( \frac{\partial^{2} H}{\partial T \partial \xi} \Biggr)_{p} = \left( \frac{ \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial  C_{p}}{\partial \xi} \Biggr)_{T, p} = \Delta C_{p}
\end{align}

where $T$ is the temperature and $\Delta C_{i}$ is change in the molar heat capacities of the reactants and products during the reaction at constant $i$ (where $i$ is either $p$ or $V$):

\begin{align}
  \Delta C_{i} = \sum_{k} \nu_{k} c_{i_{k}}
\end{align}

where  $\nu_{k}$ and $c_{i_{k}}$ are the stochiometric coefficient and the molar heat capacity at constant $i$ of the $k^{\text{th}}$ component in the reaction, respectively.

When you integrate the equations for the temperature dependences of the reaction heats you get:

\begin{align}
  \text{constant volume:  } &\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} \, \mathrm{d} T = \left( \Delta Q \right)_{V} = \int \Delta C_{V} \, \mathrm{d} T \\
  \text{constant pressure:  } &\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{p} \, \mathrm{d} T = \left( \Delta Q \right)_{p} = \int \Delta C_{p} \, \mathrm{d} T \ .
\end{align}

So, the difference of the reaction heats at constant volume and constant pressure would be:

\begin{align}
  \left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T \ .
\end{align}

Third Part: ideal gases

Now, why would this boil down to $RT$? This works when you consider reactions that involve only ideal gases. For ideal gases you have (see here):

\begin{align}
  c_{p} - c_{V} = R \ .
\end{align}

So, the integral simplifies to

\begin{align}
  \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T &= \int \sum_{j} \nu_{j} \bigl( \underbrace{ c_{p_{j}} - c_{V_{j}} }_{= \, R} \bigr) \, \mathrm{d} T \\
  &= R \int \sum_{j} \nu_{j} \, \mathrm{d} T \\
  &= R \sum_{j} \nu_{j} \int \mathrm{d} T \\
  &= R T \sum_{j} \nu_{j}
\end{align}

where $\sum_{j} \nu_{j}$ will be some rational number but not necessarily equal to 1 - it depends on the reaction.
This is basically the result you were looking for although the reaction heat difference will only be proportional to $RT$ with the proportionality constant being $\sum_{j} \nu_{j}$:

\begin{align}
  \left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = RT \sum_{j} \nu_{j} \ .
\end{align}


H2Te dipole and electronegativities of H and Te (high school level chemistry) I saw somewhere that $\ce{H2Te}$ is a polar molecule, but as far as I know the only time dipole-dipole forces arise (giving molecules the possibility of becoming polar) are when there is a difference in electronegativity. I know $\ce{H2Te}$ is not a linear molecule but if there is almost no (less than 0.4) difference in electronegativity values the shape shouldn't matter as electrons wont be more drawn to either atom (or 'side' of the molecule). In other words, it seems to me the intermolecular forces should be induced-dipole induced-dipole (making the molecule non-polar), not dipole dipole.

I should clarify, I first saw this statement in the answers to a high school chemistry exam. I then googled it, and got the same answer (i.e that the molecule is polar).

Information in this question is all over the place and often out of context. Let me get a few things straight:

Van der Waals forces, also known as London interactions, occur between every pair of molecules regardless of polarity. They are the only intermolecular force observed for non-polar molecules such as $\ce{I2}$. They are explained by induced and spontaneous dipoles at a molecular level and femtosecond timescale.
Dipole interactions are present between macroscopic dipoles and account for the boiling point difference between e.g. tetrahydrofurane and cyclopentane ($\Delta \vartheta_\mathrm{b} \approx 15~\mathrm{^\circ C}$). For those, you need a notable difference in electronegativity resulting in an overall dipole (so $\ce{CO2}$ is not a dipole because the dipole moments of both $\ce{C=O}$ bonds cancel each other out). For strong dipoles, these interactions are orders of magnitude larger than the underlying van der Waals forces.
Electronegativities are, of course, atomic properties (and also depend on things like oxidation state if one is truly accurate). For no two different elements is the electronegativity difference ever zero. Every $\ce{X-Z}$ bond is always polarised ever so slightly in one of the two directions. Sometimes, as in the case of $\ce{NCl3}$, this is best determined experimentally. ($\ce{N-Cl}$ bonds in $\ce{NCl3}$ are polarised towards nitrogen.)
However, only bonds which are polarised to a certain extent (usually a cutoff value of $\approx 0.4$) are called polar on a high school level as per convention.
$\ce{H2Te}$ has non-polar bonds as per convention of the fourth bullet point. Therefore, you should consider it a non-dipole and hence only consider van der Waals forces/London dispersion forces. (DavePhD’s source is likely correct, but the dipole moment is neglegible at high school level.)
I would love to see a link to the pages that say $\ce{H2Te}$ be notably polar/display dipole dipole interactions. (Compare the $0.2~\mathrm{D}$ to the $1.78~\mathrm{D}$ dipole moment of ethyl acetate.) Calling van der Waals forces dipole interactions is usually considered wrong (although they build up upon the same principles: electrostatic interaction).

I tried to stick to a high-school level as well as possible. Hence a number of simplifications in this answer. At research levels, every molecule without improper rotation (of which inversion and symmetry planes are a special case) should be considered to have a non-zero dipole moment as DavePhD’s answer states. Thus, every molecule of that type will display dipole-dipole interactions however weak they may be.


Why does the pH before the equivalence point of a titration depend on the initial concentration only when the acid is strong? SO, in this picture, two different titration curves are shown: one for a weak acid, one for a strong acid. I'm confused as to why for a strong acid, if you vary the concentration of the acid, (and the concentration of the acid and titrant are the same) the pH lines are spread apart below the equivalence point, but for a weak acid, where the pHs are initially slightly different, the lines converge to a common path

To understand what is happening with the two sets of titration curves, start by considering the titration of acetic acid ("HOAc" for short) with sodium hydroxide solution. The titration curves are shown below:

In the figure, the mutual concentrations are simply the concentrations of the reactants. Thus, 0.1 M HOAc is titrated with 0.1 M NaOH, 0.01 M HOAc is titrated with 0.01 M NaOH, and so on.

Acetic acid is a monoprotic weak acid, with $\mathrm{p}K_\mathrm{a}=4.756$. The titration curves for the three highest mutual concentrations almost coincide at the half-equivalence points, i.e, at the center of the buffer region. This is as expected from the Henderson-Hasselbalch equation, which ignores the autoionization of water. However, the three lower concentrations demonstrate that the titration curves do not coincide at the half-equivalence points: as the concentrations decrease, the pH values at the half-equivalence pints move toward the assumed pH of 7 for pure water.

As the concentrations are lowered, the effect of water's auto-ionization can no longer be neglected and the Henderson-Hasselbalch approximation is no longer viable. The calculated pH values at the half-equivalence points are shown in the second column of the table below:

Note that for 0.1 M, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $, as expected. Even at 0.001 M, the approximation is not bad. However, as concentrations continue to decline, the pH at the half-equivalence points simply move even closer to that assumed for pure water.

Now consider the titration of hydrochloric acid with sodium hydroxide solution. The titration curves are shown below:

As shown in the figure and in the third column of the table, the pH values at the half-equivalence points are one pH unit apart until the concentration is down around $\mathrm{1 \mu M}$.  Then, as concentrations continue to decline, the pH at the half-equivalence points must invariably move toward that of pure water.

For hydrogen ion concentrations high enough to neglect the auto-ionization of water, the pH at the half-equivalence point is easy to calculate. First, divide the initial hydrogen ion concentration by 2, to account for the fact that half of the number of hydrogen ions, from the 50 mL of starting HCl solution, were neutralized by addition of 25 mL of equimolar NaOH solution. Then multiply the hydrogen ion concentration by 2/3, because the solution volume increased from 50 mL to 75 mL. The net result is that the hydrogen ion concentration, at the half-equivalence point, is 1/3 of the initial value. Then take the negative logarithm base 10, as usual:

$$\mathrm{pH} = -\log{({\ce{[H+] \times (1/2) \times (2/3)) = - \log{(\ce{[H+]/3)}}}} }$$

If $\ce{[H^+]}$ = 0.1 M, then pH = 1.477. N.B. Significant figures were ignored in this answer, so 0.1 M is treated as exact, etc.


Is water stream deflecting in an electric field really based on water's polarity? A common highschool chemistry example for the polarity of water is the experiment where stream of water is deflected under the influence of a charge (electric field).

Now I have come across some condradicting claims, namely this Youtube video from Veritasium (from around 1:50) and more importantly a publication by Ziaei-Moayyed et. al..[1]

These sources suggest that this phenomenon is not based on the polarity of water but charged impurity ions in the water stream. They also claim that the electric field could also only re-orient the molecules, but not deflect them as the electric field also has an opposite effect on the other side of the dipole. This has been bothering me and it is hard to find concluding evidence either way. What are your thoughts?

References:

Ziaei-Moayyed, M.; Goodman, E.; Williams, P. Electrical Deflection of Polar Liquid Streams: A Misunderstood Demonstration. J. Chem. Educ. 2000, 77 (11), 1520. DOI: 10.1021/ed077p1520.

An electrostatic field will attract both conducting and nonconducting materials even though those materials have no initial charge through electrostatic induction. That is why you can pick up lint or aluminum foil with a charged comb or balloon. See this demo with a neutral aluminum can.


What is the molecularity of the RDS of a zero order complex reaction? If we have a zero order complex reaction (although uncommon in chemical practices), say the decomposition of ammonia:

$$\ce{2NH3 -> N2 + 3H2}$$

This is a zero order complex reaction. In my textbook, it says:

  For complex reaction, the order is given by the slowest step and molecularity of the slowest step is the same as the order of the overall reaction. (sic)

At the same time, molecularity of an elementary reaction cannot be zero.

How is this possible and what am I missing?

Terminology
There are two term which you must understand and use carefully.

Molecularity - The [integer] number of reactant molecular entities that are involved in the 'microscopic chemical event' constituting an elementary reaction. (From IUPAC Gold Book)
Order of reaction - If the macroscopic (observed, empirical or phenomenological) rate of reaction (v) for any reaction can be expressed by an empirical differential rate equation (or rate law) which contains a factor of the form $\ce{k = [A]^\alpha[B]^\beta\text{ ...}}$ (expressing in full the dependence of the rate of reaction on the concentrations A, B, ...) where $\alpha$, $\beta$ are constant exponents (independent of concentration and time) and k is independent of A and B etc. (rate constant, rate coefficient), then the reaction is said to be of order $\alpha$ with respect to A, of order $\beta$ with respect to B, ... , and of (total or overall) order $n = \alpha +\beta + ...$ The exponents $\alpha$, $\beta$, ... can be positive or negative integral or rational nonintegral numbers.

Equilibrium reaction

For a reaction like the decomposition of ammonia you could write a bimolecular reaction like:

$$\ce{2NH3 -> N2 + 3H2}\tag{1}$$

Often chemists consider such a reaction reaching equilibrium such as:

$$\ce{2NH3 <=> N2 + 3H2}\tag{2}$$

where there is a forward reaction which (naively) could be consider as the forward reaction

$$\ce{2NH3 ->[$k_\mathrm{f}$] N2 + 3H2}\tag{2.a}$$

with the forward rate given by $r_\mathrm{f} = k_\mathrm{f}[\ce{NH3}]^2$ and the reverse reaction

$$\ce{N2 + 3H2 ->[$k_\mathrm{r}$] 2NH3}\tag{2.b}$$

with the reverse rate given by $r_\mathrm{r} = k_\mathrm{r}\ce{[N2][H2]}^3$.

However, if you consider the reverse reaction, you can see that there are four molecules involved. It is highly unlikely that four molecules would ever collide in just the right way to cause a reaction. So the reverse reaction occurs in multiple steps with intermediate species involved. The exponents $x$ and $y$ would be determined experimentally and would almost certainly not be integer values. Thus more realistically:

$$r_\mathrm{r} = k_\mathrm{r}[\ce{N2}]^x[\ce{H2}]^y\tag{3}$$

Catalytic reaction

Now, back to the more salient point of the question. The decomposition of ammonia is typically not done just by heating ammonia to some high temperature to form an equilibrium, but rather it is done selectively with a catalyst and a lower temperature.

$$\ce{2NH3 ->[catalyst] N2 + 3H2}\tag{4}$$

The catalytic surface absorbs ammonia preferentially (and very quickly compared to the reaction time of the ammonia decomposition reaction) and can only absorb a small amount of ammonia. Hence the rate of decomposition doesn't depend on the concentration of ammonia so the rate of decomposition is essentially constant until almost all of the ammonia is consumed. So with a catalyst there is a zero order kinetic reaction with the equation:

$$r_\mathrm{f} = k_\mathrm{f}\tag{5}$$


Determining perovskite crystal structure I've been reading that perovskite presents a sort of pseudocubic crystalline system, however it can be stretched and suffer modifications to become orthorhombic etc.

The thing is, if I consider it as cubic, it's sort of a combination between FCC and BCC, because it's an FCC with an atom in the center. I can't find a name for this sort of thing.

Can anyone help me out?

At Geoff's request I'll make comments into an answer:

Since it is a multi-component system you need to think of a crystal structure with a basis. Not surprisingly, the basis here is $\ce{ABO3}$. In your picture from Wikipedia, start with just the blue atoms - they form the simple cubic. Then each blue atom shares 6 red atoms with other blues, so pick three for a given blue. Then place the green atom at the cube center to go with the blue atom at the lower right front blue atom. This is now your basis, and placing that basis unit at each corner of the cube (repeating to infinity) gives you the crystal structure

To explain further: each red has two blue neighbors, so each red is 'half-owned' by a blue. For a basis, you want only whole atoms to keep your mind a bit saner, so pick three to be 'wholly-owned' by a blue, and leave the others to be owned by another blue. Taking the left lower center blue atom, I'd pick the red atoms in the +x, +y, and +z directions, then the green atom in the (x,y,z) direction to make a nice basis.

I'll also add some on deviations from cubic symmetry.  If the green atom is at (1/2,1/2,1/2) than it all remains simple cubic.  If the greens are big fat atoms compared to the blues, they may want a little more room, which leads to distortions.  Simply stretching along one dimension is pretty easy, since it leaves 2/3's of the blue-red pairs at the same interatomic distance, stretching the other 1/3.  This creates an orthorhombic crystal, still with the same $\ce{ABO3}$ unit as a basis, but now slightly elongated along one direction (commonly taken as z).


If concentration of K+ ions inside the cell is almost equal to concentration of Na+ ions outside of the cell why these ions cross the membrane? I'm a bit confused about electrochemical gradients of sodium and potassium ions. They have similar concentrations and are kind of similar ions but their distribution is different across the membrane. If that is true why they cross membrane. Why don't these gradients somehow cancel out? Why is there even need for sodium to enter the cytosol when potassium ions which are also positive are already inside?

In order to understand why this happens, you have to recall the permeability characteristics of the plasma membrane. It seems that the movement of ions to and fro the cystol is counterproductive however this is not so. Actually there is unequal distribution of ions in the extracellular fluid and cytosol

  Extracellular fluid is rich in Na+ and chloride ions (Cl–). In
  cytosol, however, the main cation is K+, and the two dominant anions
  are phosphates attached to molecules, such as the three phosphates in
  ATP, and amino acids in proteins.

Differences in intracellular and extracellular concentrations of ions result primarily from:

the Na +/K + pump and
the permeability characteristics of the plasma membrane.

Ions pass through the plasma membrane through ion channels. The two major types of ion channels are leak channels and gated ion channels.

i. Leak Channels

Leak channels, or non-gated ion channels, are always open and are responsible for the permeability of the plasma membrane to ions when the plasma membrane is unstimulated, or at rest. The plasma membrane is more permeable to K +and Cl- and much less permeable to Na+ because there are many more K + and Cl- leak channels than Na+ leak channels in the plasma membrane.
Because the plasma membrane typically has more K+ leak channels than Na+ leak channels, the number of potassium ions that diffuse down their concentration gradient out of the cell into the ECF extracellular fluid ECF is greater than the number of sodium ions that diffuse down their concentration gradient from the ECF into the cell. See below

The plasma membrane is relatively permeable to K + and much less permeable to Na + and to negatively charged molecules found inside of the cell. Consequently, positively charged K + tend to diffuse out of the cell through K + leak channels, leaving the negatively charged molecules behind, thus polarizing the membrane. The membrane is at equilibrium when the tendency for K + to diffuse out of the cell is resisted by the negative charge of the molecules inside the cell.

ii. Gated Ion Channels

These channels open and close in response to small voltage changes across the plasma membrane. The movement of ions into or out of the cell changes the charge difference across the plasma. membrane, which causes voltage-gated ion channels to open or close. Voltage-gated channels specific for Na+ and K + are most numerous in electrically excitable tissues

The Na+/K+ Pump

The differences in K + and Na + concentrations across the plasma membrane are maintained primarily by the action of the Na + –K + pump. Through active transport, the Na+/K + pump moves K+ and Na+ through the plasma membrane against their concentration gradients. Potassium ions are transported into the cell, increasing the concentration of K + inside the cell, and Na + are transported out of the cell, increasing the concentration of Na + outside the cell. Approximately three Na+ are transported out of the cell and two K + are transported into the cell for each ATP molecule used. These pumps help maintain the resting membrane potential by pumping out Na+ as fast as it leaks in. At the same time, the Na+/K+ ATPases bring in K+.

In summary it is important to note, there is uneven distribution of ions in and outside the cytosol.
Because the plasma membrane has more K+ leak channels than Na+ leak channels, the number of K+ ions that leave the cell is greater than the number of Na+ ions that enter the cell. As more and more K+ ions leave the cell, the inside of the membrane becomes increasingly negative and the outside of the membrane becomes increasingly positive. The inability of most anions to leave the cell also contributes to the negativity of the resting membrane potential. These trapped anions cannot follow K+ out of the cell because they are attached to non-diffusible molecules such as ATP and large proteins.

Nevertheless, sodium ions do slowly diffuse inward, down their concentration gradient. If not controlled, such inward leakage of Na+ would eventually destroy the resting membrane potential. The small inward Na+ leak and outward K+ leak are offset by the Na+/K+ ATPases (sodium pump)

This diffusion accross the plasma membrane is necessary because:

There is uneven distribution of ions which needs to be corrected.
The characteristics of the plasma membrane has lot to do with how transport of substances occur ( It is important to note that not only ions use this path, other substances vital for cell growth have to use this path as well, therefore strict control is necessary)

Hope this helps

References

Anatomy and Physiology (Seely)
Principles of Anatomy and Physiology (G. Tortora, B. Derrickson)


Why did the salt on the skin of my turkey react with the aluminum foil it was touching? I sprinkled salt on my turkey skin and left it to dry out over night. It was covered with aluminum foil. The next morning the aluminum foil had holes ‘burnt’ in it and the turkey skin turned silver at the spots where the salt crystals had touched the aluminum foil. Was my turkey poisoned with aluminum at this point? I did not eat it to find out.

If you don't mind, I'll start with a more colloquial and hopfully not to lengthy explanation:

Precious metals, such gold, do not corrode under "household conditions", others, such as iron or aluminium do. Rusting of iron is a typical example. Here, the corroded surface often is very porous and the corrosion carries on until the whole piece of metal is turned to rust (oxidized).

In the case of aluminium, the initial corrosive layer is very tight and serves as a protection for the rest of the aluminium foil. Such a very thin and transparent protection layer often is artificially applied. The item are sold as Eloxal (= electrolytically oxidized aluminium).

Now comes the table salt, which is sodium chloride ($\ce{NaCl}$). Sodium chloride is hygroscopic, it "attracts" the water and that's why you sprinkled it on the turkey. It is also very soluble in water. When it dissolves, sodium $\ce{Na+}$ and chloride $\ce{Cl-}$ ions are formed.

The chloride ions attach and move into the protective oxide layer of the aluminium foil and finally induce the corrosion of the aluminium metal. Aluminium ions are released to the water (= the wet surface of the turkey).

For a more scientific, in-depth explanation of this corrosive processes known as pitting, you might want to have a look at this article or this review.

  Was my turkey poisoned with aluminum at this point?

There seem to be some crazy myths concerning the "toxicity" of aluminium (ions), often spread by opponents of immunization and other quacks.

Fact is that

a healthy human body is capable to excrete aluminium ions though the kidneys
the British Alzheimer Society does not see any convincing relation between the uptake of aluminium and Alzheimer's Disease
the cancer information service of the dkfz (Deutsches Krebsforschungszentrum, German Cancer Research Centre) refutes the aluminium induced breast cancer myth
many antacids do contain aluminium hydroxide as one ingredient. The effect of "poisoning" is not known.

In summary:

Yes, aluminium ions were released to the wet surface of the turkey.
There's no reason to assume that the turkey is "poisoned".


What causes the really bad smell of rotten potatoes? Rotten potatoes have a distinctive unpleasant odour, and I can't find any definitive answer as to what the chemical involved is. I was thinking it might be Butyric acid, from the fermentation of the starch, but according to Wikipedia:

  Butyric acid is present in, and is the main distinctive smell of, human vomit.

And those puppies don't smell like vomit. They're more like rotting flesh – the reason I'm asking is that I've spent half the day looking for a dead animal that the cat brought in or something, only to discover that the horrible smell was coming from the potato box.

My first guess was that it was putrescine and other polyamines---but the smell of putrefying potatoes actually comes from methyl mercaptan,  dimethyl sulfide, and dimethyl trisulfide. Check it:

Source: A. Kamiya, Y. Ose, Study on offensive odor (Report IV): A Consideration of Putrefaction and Offensive Odor of Solid Waste, Journal of Japan Society of Air Pollution, 18(5), 1983, pp  453-463.

Notice that the headspace composition changes quite a bit with time. Most of the headspace chromatography studies I found dealt with early detection of disease organisms in infected potatoes, rather than potatoes in full putrefaction mode. You'll also find many references to solanine poisoning from potatoes; solanine is a toxic glycoalkaloid, is nonvolatile, and has nothing at all to do with the foul smell and toxic gas produced by putrid potatoes.

Methyl mercaptan ($\rm CH_3SH$) has an odor described as "rotting cabbage" by ATDSR; it's one of the major contributors to the smell of farts (oh, sorry, "flatus" if we're being polite). It has an odor detection threshold as low as 1 ppb.

Dimethyl sulfide ($\rm (CH_3)_2S$) is responsible for the smell of the sea (in low concentrations); it too has a cabbagy smell.

Dimethyl trisulfide ($\rm CH_3SSSCH_3$) is present in relatively smaller amounts but it has an even stronger odor. The detection threshold is around 1 part per trillion, and it is apparently a strong insect attractant.

The gases produced by rotting potatoes are quite toxic (see Rotting potatoes in basement kill four members of Russian family).

(To keep the notes below in context, in my original answer I said that a plot point in "The Walking Dead" was that zombies couldn't smell delicious humans if they were wearing coats smeared with rotting flesh. I suggested that when the zombie apocalypse arrives, packing your pockets with putrid potatoes might work, too. Now I'm not so sure. Can zombies distinguish between polyamines and sulfur compounds? Perhaps they're stench connoisseurs.)


Why are palladium and platinum carbonyls unstable at room temperature? By palladium and platinum carbonyls, I mean the mononuclear, homoleptic, charge-neutral, binary carbonyl complexes of the form $\ce{M(CO)_n}$ and not any heteroleptic complexes like $\ce{Pd(CO)(PPh3)3}$ and similar compounds of platinum which are abundant in chemical industry and used as catalysts.

So, I was reviewing carbonyls of transition metals when I encountered this statement:

  Group 10 elements with 10 valence electrons form metal carbonyls
  $\ce{Ni(CO)4}$ ($10 + 4 \times 2 = 18$ electrons). Curiously $\mathbf{Pd(CO)_4}$ and $\mathbf{Pt(CO)_4}$ are not stable.

I am proposing that heavy metal carbonyls tend to be unstable(?) and decarbonylate to form polynuclear carbonyls. For example: $\ce{Ru(CO)5}$ and $\ce{Os(CO)5}$ being unstable form $\ce{Ru3(CO)12}$, and $\ce{Os3(CO)12}$. But in the case of palladium and platinum, no carbonyl complex seems to exist in stable condition at room temperature. I reviewed two papers which described the Group 10 carbonyl complexes (links below):

  Complexes where the carbonyl group is the only ligand present have
  been reported for all the platinum group metals except
  palladium.[...] No substantiated reports have been published on
  platinum or palladium compounds similar to the nickel carbonyl
  $\ce{Ni(CO)4}$. Thus the polymeric platinum dicarbonyl is the only
  known carbonyl of these two metals and even this compound is unstable
  in air.[...]

  The unstable polymeric platinum dicarbonyl $\ce{Pt_n(CO)_{2n}}$ has
  been obtained as a cherry red precipitate by the action of carbon
  monoxide on aqueous or ethanolic solutions of Pt(II) compounds (e.g.
  $\ce{K2PtCl4}$) at temperatures up to 80 °C; or as a purple colloidal
  precipitate by the action of water on a benzene solution of
  $\ce{Pt(CO)2Cl2}$ under an atmosphere of carbon monoxide.

  It was not until 1972 that a definite existence of the congeners
  $\ce{Pd(CO)4}$ and $\ce{Pt(CO)4}$ was reported. Using matrix isolation
  and Raman techniques, data for the products of Pd atoms CO and CO/Ar
  cocondensation reaction were able to characterize the series of binary
  complexes $\ce{Pd(CO)_n}$ analogous to $\ce{Ni(CO)4}$, the most stable
  being $\ce{Pd(CO)4}$ but still unstable than $\ce{Ni(CO)4}$ which
  explains the failed attempts in its synthesis. Similar technique were
  used in making $\ce{Pt(CO)4}$ at 4.2-10 K which is stable up to 30
  K.  (slightly abridged)

The question is: why they are unstable/don't exist at room temperature?

Platinum Metals Rev., 1972, 16, (2), 50 "The Carbonyls of the Platinum Group Metals" By C. W. Bradford, Research Laboratories, Johnson Matthey & Co Limited (link)
J. Am. Chem. Soc., 1973, 95 (22), pp 7234–7241 "Binary carbonyls of platinum, Pt(CO)n (where n = 1-4). Comparative study of the chemical and physical properties of M(CO)n (where M = nickel, palladium, or platinum; n = 1-4) DOI: 10.1021/ja00803a009

Tl, dr:  It's not just a matter of size.  A combination of coordination geometry, node distribution in the $d$-orbital wavefunctions, and atomic size all conspire to weaken $\pi$ overlap in palladium and platinum carbonyls, not only relative to the nickel compound but also relative to earlier transition metal carbonyls in their respective periods.
Ziegler et al. [1] studied the relative stabilitie of metal-carbonyl bonds for $\text{G6(CO)}_6$, $\text{G8(CO)}_5$ and $\text{G10(CO)}_4$ carbonyl complexes, where $\text{G6}$ in the first formula is a Group 6 metal and similarly for the second and third formulae.  The following general trends were observed:

Within a period, the G6 hexacarbonyl is most stable.

Within each group, the 4th period carbonyl is most stable, the 5th period carbonyl is least stable and the 6tg period carbonyl is slightly more stable than the 5th.

From these trends it is seen that palladium and platinum carbonyls are literally in a bad place.
The authors attribute the stability order primarily to $\pi$ backbonding.  We may expect such backbonding to be strongest for the G6 hexacarbonyls where the octahedral geometry about the metal atom causes the metal-atom donor orbitals to have pure $\pi$ symmetry.  The five- and four-coordinate geometries (trigonal bipyramidal and tetrahedral, respectively) are progressively less favorable with regard to this symmetry.  This accounts for effect (1).
Effect (2) is related to relative atomic sizes within each respective group, but also to a more subtle effect.  In the $3d$ subshell, which is used as the valence subshell in the 4th period, the lobes have no radial nodes, a condition that allows maximum overlap with the molecular orbitals coming from the carbon monoxide.  The $4d$ and $5d$ orbitals used by the heavier metals have radial nodes so poorer overlap.  With the 6th period metals, relativistic effects intervene, but this effect (which I cannot tell from the abstract, and the full paper is behind a paywall) is less potent than having more compact orbitals with no radial nodes at all in the $3d$ series.
Reference:
1.
Tom Ziegler, Vincenzo Tschinke, and Charles Ursenbach, "Thermal stability and kinetic lability of the metal carbonyl bond. A theoretical study on M(CO)6 (M = chromium, molybdenum, tungsten), M(CO)5 (M = iron, ruthenium, osmium), and M(CO)4 (M = nickel, palladium, platinum)", J. Am. Chem. Soc. 1987, 109, 16, 4825–4837. Link


Can calcium carbonate scorch in a pan? Today a friend showed me something we could not explain:

She had a pan which she used for years to heat water. On the bottom it had a thick layer of calcium carbonate. She forgot the pan on the hot plate. So all the water evaporated.

Then the calcium carbonate layer at the bottom got really black. Is this possible if it was pure $\ce{CaCO3}$ or must there have been other organic compounds to explain the black colour?

May I have an explanation what happened chemically as well?

Like the comments have said, black residue is usually from organic material that has scorched.

If the pan somehow got to $840 ^{\circ}$C, then the calcium carbonate would decompose into calcium oxide and carbon dioxide, neither of which would lead to a black residue.

$$\ce{CaCO3 -> CaO + CO2}$$

Organic materials are mostly made of hydrocarbons. When you burn the organic matter, you're left with pretty much all carbon. This occurs when there is not enough oxygen to completely burn all of the carbon - when there is enough oxygen, some of the carbon combines with it to form $\ce{CO2}$ and $\ce{CO}$ and the hydrogen combines with oxygen to form $\ce{H2O}$. When there is a lack of oxygen, some carbon is left over, and this is what we see as the black substance.


Nomenclature of multiple double bonds? I know that in a compound, when there are two double/triple bonds its like :

1,3-diene ; 1,3-diyne and so on.

For three :

1,3,5-triene ; 1,3,5-triyne and so on.

But what about 4?

While doing a question I encountered this :

2,4,6,8-(What?)ene

Is it quad,quart,quater(like quaternary carbon) or something else?

Also, it would be helpful if you could extend it to like 5 or as much as you can double/triple bonds.

The multiplying affixes are given by Table 11 of the Blue Book, referenced in R-0.1.4 and R-4.1:
Number  Numerical term | Number  Numerical term
1       mono-          | 90      nonaconta-
2       di-            | 100     hecta-
3       tri-           | 200     dicta-
4       tetra-         | 300     tricta-
5       penta-         | 400     tetracta-
6       hexa-          | 500     pentacta-
7       hepta-         | 600     hexacta-
8       octa-          | 700     heptacta-
9       nona-          | 800     octacta-
10      deca-          | 900     nonacta-
11      undeca-        | 1000    kilia-
12      dodeca-        | 2000    dilia-
20      icosa-         | 3000    trilia-
30      triaconta-     | 4000    tetralia-
40      tetraconta-    | 5000    pentalia-
50      pentaconta-    | 6000    hexalia-
60      hexaconta-     | 7000    heptalia-
70      heptaconta-    | 8000    octalia-
80      octaconta-     | 9000    nonalia-

Therefore, the answer to your question is 2,4,6,8-tetraene.

Extra information
Note that there is no elision of vowels, specified by R-0.1.7:

R-0.1.7.1 Vowels are systematically elided as follows:
(a) the terminal "e" in names of parent hydrides when followed by a suffix beginning with "a", "i", "o", "u", or "y";
(b) in the Hantzsch-Widman system, the final "a" of an element prefix when followed by a vowel;
(c) the terminal "a" in the names of numerical multiplicative affixes when followed by a suffix beginning with "a" or "o", or a Hantzsch-Widman prefix or stem beginning with a vowel;
(d) the terminal "a" of an element prefix in "ababa" repeating unit names (see R-2.3.3.2) and the terminal "o" of a replacement infix when followed by a vowel;

The relevant part is (c), which states that the final "a" in the above affixes are elided when the next letter is "a" or "o", but in this case the next letter is "e" so it is not elided.
The following are given examples of elision:

benzenehexaol > benzenehexol
[1,1'-binaphthalene]-3,3',4,4'-tetraamine > [1,1'-binaphthalene]-3,3',4,4'-tetramine
1,3,5,7-tetraoxaocane > 1,3,5,7-tetraoxocane


How expensive are Hartree Fock calculations? Hartree Fock is a very simple mean field approach for finding the ground state energies of - for example - molecules. But how expensive is it from a computational view? So after calculating all one - and two-particle integrals, which is $O(n^4)$, how expensive is it to do HF then? How large molecules could one in general do with that?

I wouldn't call the Hartree-Fock methods 'very simple', I would consider it fairly complex both in terms of its derivation, interpretation, computational cost and implementation details. If Hartree-Fock calculations are now commonplace it is also because so many man-hours of study have been devoted to it.

In any case, as you say, its formal cost is $n^4$ ($n=$ number of basis functions): the number of two-electron integrals is $n^4$, and all these integral are necessary to build the Fock matrix, so the overall cost is also $n^4$. However, this is not the end of the story; as a system grows larger and larger in spatial extension many of the required integrals become vanishingly small and can be neglected, and other approximations become possible, at the cost of introducing a small and adjustable error. As long as the error introduced is (much) smaller than the intrinsic error of the Hartree-Fock method, it doesn't really matter. Some keyword often used in this context are 'linear-scaling' and 'density-fitted' methods.

The difficult, computationally expensive part of the Hartree-Fock method is dealing with the electron-exchange part of the Fock matrix, and most studies try to speed up this step. For example, a recent study of this type is:
Christoph Köppl and Hans-Joachim Werner, Parallel and Low-Order Scaling Implementation of Hartree−Fock Exchange Using Local Density Fitting, J. Chem. Theory Comput. 12, 3122−3134 (2016).

In this study the largest calculation they report on is for alpha-cyclodextrin, a polysaccharide with formula $\ce{C36H60O30}$ (126 atoms, 516 electrons); using the aug-cc-pVDZ basis set (2058 basis functions) the authors report that a density-fitted Hartree-Fock calculation took about 4 minutes on a beefy workstation with 20 cores (two Intel Xeon E5-2690 @ 2.8 GHz).


Balancing by oxidation numbers method multiple atoms In my Chemistry textbook, the rules for balancing a chemical equation using the oxidation-Number method are as follows:

Assign oxidation numbers to all the atoms in the equation.
Identify the atoms that are oxidized and the atoms that are reduced.
Determine the change in oxidation number for the atoms that are oxidized and for the atoms that are reduced.
Make the change in oxidation numbers equal in magnitude by adjusting coefficients in the equation.
If necessary, use the conventional method to balance the remainder of the equation.

I understand how to do this for simple equations, such as $$\ce{SnCl4 + Fe -> SnCl2 + FeCl3}$$

Following the steps:

$\ce{Sn^{4+} + 4Cl- + Fe^0 -> Sn^{2+} + 2Cl^- + Fe^{3+} + 3Cl^-}$
Oxidation: $\ce{Fe^0 -> Fe^{3+} (Fe)}$ reduction: $\ce{Sn^{4+} -> Sn^{+2} (Sn)}$
oxidation: +3; reduction -2
$\ce{3SnCl4 + 2Fe -> 3SnCl2 + 2FeCl3}$
(no change necessary, already balanced)

The Problem: Notice how Sn and Fe (the reduced and oxidized atoms) all have subscripts of 1. However, when their subscripts differ, I'm not sure how to balance it.

For example, I can easily balance the following equations by inspection, but not so easily by this method of oxidation numbers:

$$\ce{KClO3 -> KCl + O2}$$ (O is oxidized, subscripts of 3 and 2)

$$\ce{NH3 + NO2 -> N2 + H2O}$$ (N is oxidized and reduced, oxidation numbers of -3, +2, and 0)

And this gets even more complicated when the subscripts differ and there are more than two elements whose oxidation numbers changed (are reduced or oxidized), e.g.,

$$\ce{NH4ClO4 + Al -> Al2O3 + HCl + N2 + H2O}$$
(Al and N are oxidized, Cl is reduced)

For all of these, I get stuck on the fourth step. Determining oxidation numbers and calculating the change is not the problem, but figuring out how to do it with larger or different coefficients as in the examples above is the problem.

The five general steps in my textbook don't help me understand this method much with more complicated methods such as this. Could you give me a more thorough explanation - a better explanation or a general rule - of how to solve all of these?

In some (many?) cases you get a more parsimonious description by considering an entire reactant as an oxidant or reductant.  Separating out individual atoms makes you miss the forest because you get confused by the trees.

Take the ammonium perchlorate-aluminum reaction. Since aluminum is being oxidized let us render the ammonium perchlorate as the oxidizing agent, ignoring the fact that only some of its atoms are oxidizing.  You then have a reaction:

$\ce{NH_4ClO_4}\rightarrow \ce{N^0}+4\ce{H^I}+\ce{Cl^{-I}}+4\ce{O^{-II}}$

where the Roman numerals indicate oxidation states in the products.  These add up to $-5$ for all the product whereas the ammonium perchlorate, as a neutral compound, started with the states adding up to zero.  So five electrons must be added to the reactants:

$\ce{NH_4ClO_4}+5 e^-\rightarrow \ce{N^0}+4\ce{H^I}+\ce{Cl^{-I}}+4\ce{O^{-II}}$

Thus although some atoms might be oxidized, others reduced, and still others are just along for the ride, we identify a net oxidizing effect of the ammonium perchlorate as a whole, numerically five electrons per formula unit.

Now we know that the redox stoichiometry will be $3\ce{NH_4ClO_4}+5\ce{Al}$ and we can balance the reaction accordingly.  (We will need to double the coefficients to eliminate a fraction, that's just algebra.)


What is the term for the amount of volume change at a phase boundary? Apologies for the extremely basic question, but I would like to look up the amount by which the volume (or equivalently, density) of various substances changes when they freeze. However, I don't know the correct term for this, and Google isn't helping much.

To clarify, I'm asking about the vertical line segment on the plot below (from Wikipedia). If the $y$ axis was internal energy then the height of this line segment would be called "latent heat of fusion", and I could easily look it up for any given substance. I would like to know the equivalent term for the discontinuous volume (or density) change, in whichever units are most standard.

I realise that unlike latent heat, this will depend quite a bit on the pressure. However, I would hope that it would be tabulated for atmospheric pressure at least.

Phase transition can be distinguished by analysing their chemical potential, such classification has been introduced by Ehrenfest. For a given transition, lets call it $\phi$ transition:

$$ \ce{X_{\alpha} \rightleftharpoons X_{\beta}} $$

You can assess transition molar specific volume in the following way:

$$ \Delta_{\phi}V = v_{\beta} - v_{\alpha} = \left(\frac{\partial \mu_{\beta}}{\partial p}\right)_{T} - \left(\frac{\partial \mu_{\alpha}}{\partial p}\right)_{T}$$

Thus, by modelling your potential in term of pressure and temperature, you can assess molar volume change of a phase transition. This is the key to solve your question. In real life application, we use diagrams and tables to assess it. But building a model is always valuable for comprehension.

In a similar way, molar entropy is given by:

$$ \Delta_{\phi}S = s_{\beta} -s_{\alpha} = \left(\frac{\partial \mu_{\beta}}{\partial T}\right)_{p} - \left(\frac{\partial \mu_{\alpha}}{\partial T}\right)_{p} = \frac{\Delta_{\phi}H}{T_\phi} $$

Because $\Delta_{\phi}V$ and $\Delta_{\phi}H$ are not null for first order (in term of first derivative) transition, slope difference is not null and therefore potential slopes are different both side of phase transition when Gibbs parameters vary ($p$ and $T$). Thus there are discontinuities in specific volume, enthalpy and entropy when a first order transition occurs.

Ehrenfest did a classification of phase transition based on this criterion which led to: first order, second order (smooth, no slope discontinuity) and $\lambda$ (sharper) transitions regarding how such chemical potential discontinuity happens.

This discontinuity also impact heat capacity $c_p$. In case of water, whose phase transitions can be assumed as a first order, when a phase transition occurs, discontinuity leads to a virtually infinite heat capacity, explaining why fusion or vaporisation happens at constant temperature (when pressure is held constant). This is why we need an extra term called latent heat (temperature independent) which contrast with sensible heat (temperature dependent).

You will find detailed explanation (models, diagram and tables) of those phenomenons in Atkins, Physical Chemistry, Chapters 2-7.


Iodometric titration of copper – standard electrode potentials Upon inspection of the standard electrode potentials for the half reactions for both $\ce{Cu^2+}$/$\ce{Cu+}$ and $\ce{I2}$/$\ce{2I-}$, it becomes apparent it should be the iodide ions that act as the oxidising agent, as the standard electrode potential for the iodine half-cell is much higher.
$$\begin{alignat}{2}
\ce{Cu^2+ + e- \;&<=> Cu+}\quad &&E_\text{cell} = +0.159\ \mathrm{V}\\
\ce{I2 + 2e- \;&<=> 2I-}\quad &&E_\text{cell} = +0.54\ \mathrm{V}\\
\end{alignat}$$

To explain this, many websites that I’ve looked at suggest it’s the weak solubility of copper(I) iodide, increasing the $\ce{Cu^2+/Cu+}$ potential to around $+0.88\ \mathrm{V}$.

Such websites include:

http://www.titrations.info/iodometric-titration-copper
http://chemistry.tutorvista.com/analytical-chemistry/iodometric-titration.html

What I’d like to know, is how this value of $+0.88\ \mathrm{V}$ has been determined, and a little of the theory behind as to why it has increased from $+0.159\ \mathrm{V}$.


Simple tests for presence of CO₂ Am aware that one method for testing for the presence of $\ce{CO2}$ is passing the gas through limewater, with it going cloudy if positive.

Are there any other convenient methods?

There is not many simple tests. One of possible tests would be passing gas through neutral solution of suitable acidobasic colour indicator, e.g bromthymol blue, turning green to yellow while getting slightly acidic.

But it would not be specific, any acidic gas would do that.

You have not mentioned context, why do you need to know ? Or is it just homework-like question ?


Ethane and ethene: which is easier to burn? Which one burns hotter? Why? Anything that burns "easy", has a low activation energy ($E_\mathrm a$) for the burning process. Anything that burns hotter, will have a lesser enthalpy and thus, will have a more aggressive exothermic reaction.

Ethane is a molecule with two carbon atoms and 6 hydrogen atoms; all bonded with covalent single bonds. Ethene has the same number of carbons, but 4 hydrogen atoms; the bond between carbon atoms, in this one, is a double bond.

If you had to choose a fuel, and heat was your prior factor, which one would you have chosen? I imagined that for this, the activation energy and the enthalpy comparison could lead to an answer. I could easily have looked both $E_\mathrm a$ and enthalpies up, but I didn't.

Why and how do enthalpy and $E_\mathrm a$ change when two single bonds and two H atoms are replaced with a double bond?

Why and how do enthalpy and $E_\mathrm a$ change when two single bonds and two H atoms are replaced with a double bond?

Olefins are less stable than their corresponding alkane analogues.  For example, when ethene is hydrogenated to ethane roughly 33 kcal/mol of heat is given off because ethene has a higher energy content than ethane.

Why is this, why is a molecule with a double bond higher in energy than a molecule without a double bond?

There are two equivalent ways to describe the orbitals in ethene.  One is the "double bond" description where a pi and sigma bond exist together side by side.  We know that a pi bond is not as strong as a sigma bond due to poorer overlap with the pi bond compared to the the sigma bond.  It only takes something on the order of 60 kcal/mol to break a pi bond and produce cis-trans isomerization about the double bond, whereas it takes something like 90 kcal/mol to break a carbon-carbon single bond. This poorer overlap with the pi bond causes ethene to be higher in energy than ethane.  Alternatively one can view ethene as a two-membered ring, no pi bond, just 2 sigma bonds between the 2 carbons forming a two-membered ring.  It is easy to see that such a ring system would contain a significant amount of strain.

Whether you use the pi bond or two-membered ring approach to describe olefinic double bonds, both lead to the conclusion that olefins are destabilized (higher in energy content) due to poor overlap or ring strain (which in itself is really a reflection of poor overlap).  Because of this destabilization, alkenes will generally have lower activation energies and more heat will be given off when they react compared to alkanes.

Example:
Let's consider the bromination of ethene and ethane
$$\begin{align}
\ce{C2H4 + Br2 &-> BrH2C-CH2Br}\\
\ce{C2H6 + Br2 &-> 2 CH3Br}
\end{align}$$

The heats of formation of ethene, bromine and 1,2-dibromoethane are 52.5, 0 and -37.8 kJ/mol respectively; the reaction is exothermic by 90.3 kJ/mol.  On the other hand, in the case of ethane, the heats of formation are -84.7, 0 and (2x) -37.4 kJ/mol; this reaction is endothermic by 10.3 kJ/mol.  The olefinic reaction is significantly favored for the reasons discussed above.


How to think of solvated electrons? This excellent answer explains at length what's happening in this fascinating video entitled Liquid Electrons - Periodic Table of Videos. In the screenshot below, the metallic-looking solvated electron layer (middle) has a bronze color.

I am trying to understand what "solvated electron" even means. Is the way that these electrons move around in solution more like a solvated ion, or proton in an acidic aqueous solution, or more like the conduction electrons in a typical liquid metal we might see in a laboratory, such as mercury or gallium? Or are both of those completely inadequate analogies?

I chose proton as the lightest thing I could think of that could be roughly thought of as a "bare charge". It is, of course, the opposite sign, so maybe a lithium ion?

The analogy with a proton is actually a good one if you are careful to remember that an electron is nearly 2000 times lighter than a proton. What does that mean? It means that despite the fact that an electron is very "small", the electron is actually going to be very large because lighter particles will tend to spread out and have a much more diffuse wavefunction. For instance, if we just take a case where the environment sets up a more or less constant potential for the electron to sit in, then it becomes very clear that a proton will spend much more time towards the bottom of the well than the electron will. Of course the environment wouldn't set up the same potential for both of them because this potential is determined by interactions with the electron or proton itself.

Ok so that's a basic picture. Essentially, you have to think of the solvated electron as being a very quantum mechanical ion if you wish to think of it as an ion.

Before we can get to full solvation, however, it is useful to understand the behavior of an excess electron in simpler systems. One good example, because they're studied all the time and are usually fairly representative of what might be going in the liquid phase, is a water cluster. In Sommerfeld and Jordan's paper[1] studying the behavior of excess electrons on water clusters, they provide the following wonderful image:

This gives a great picture of what a solvated electron could look like. The point is that it can look like a lot of things. If you wish, you can think of these isosurfaces as being an "orbital" that the excess electron is free to move around in. The difference is that we draw orbitals as 90% probability densities, only these are drawn at a constant volume, and hence don't all contain the same "percentage of the electron". In fact many of these surfaces contain less than 50% of the charge density, which demonstrates these electrons are very diffuse!

Now, the arrangements where the electron is primarily inside the cluster is going to resemble a solvated electron the most. It is worth noting structures such as 20c and 20d. In these structures, the electron interacts electrostatically with the free $\ce{O-H}$ bonds. In structures such as 13a and 24b, the electron is actually bound by the dipole field of the cluster (these clusters all have large dipole moments). Because these electrons are bound by dipole interactions, they will be quite diffuse.

One way in which electrons like these are definitely not like a proton in solution is that correlation effects become very important for the behavior of these electrons. Indeed, some of the structures shown above only give a negative binding energy when correlation is included at the CI level. (See 1 for details on all this.)

Some more interesting papers on electrons bound to water clusters are 2 and [3].

So, this sets us up for thinking about fully solvated electrons (sticking with water because it's where the most work is done since water is very important). As it happens, our orital analogy applies quite nicely in liquid water. Below is the little cover art from a paper hot off the presses (2017) by Ambrosio et al.. [4]

They actually show that the solvated electron in water is bound strongly enough to have multiple electronic states (they match with experiment quite well), and they can be seen to be s-like and p-like. That is, one of them is roughly spherical while the other has a node with a smaller lobe. Again, keeping our analogy with a proton in mind, these electrons are much larger than how we think of a proton in water. Both, however, induce a local structure in the water molecules to stabilize the excess charge.

The fact this aqueous electron has at least two electronic states tells us already that it is probably bound more strongly than the electrons in the sodium/liquid ammonia solution (at least for the concentrated solution). The difference being that excess electrons in water are going to be very dilute (there's not many of them), while the sodium/ammonia ones are quite concentrated when we see them.

These solvated electrons in water are actually quite mobile, as we might expect. See for instance[5], in which it is shown that a surface-bound electron actually has almost the same energy as a fully solvated electron. That doesn't really make sense to me besides that the stuff from the water clusters might be relevant. That is, for this to be true, you would expect that the dominant interactions at the surface would be different than in the bulk. I'd guess dipole-bound at the surface and primarily dispersion and other correlation effects in the bulk.

As one last way to beat the proton analogy to death, we can maybe understand the mobility of the solvated electron by noting that protons are also very mobile. That is, if I drop $\ce{DCl}$ into a solution of $\ce{H2O}$, I will eventually end up with a bunch of $\ce{HDO}$ molecules and some $\ce{D+}$ and mostly $\ce{H+}$. This is to say that the proton which I drop in and say "that's my proton" is going to get attached to a water molecule and some other hydrogen is going to become the proton, and so on.

The big difference for the solvated electron is that electrons are indistinguishable from each other, so I can't say that this is the solvated electron and those are the valence electrons of the surrounding solvent. Rather, we have to picture the solvated electron as just being part of the wavefunction describing the electrons of all nearby waters (well really the whole solution but let's not get carried away). Maybe this is where we start talking about solvated muons...

In any case, if you use google scholar there is a ton of reading on solvated electrons, so feel free to check it out, and hopefully this helps you have a picture for whatever people might be working on.

Sorry I could only speak about water. In general an aqueous electron might behave quite differently in other solvents, but it's definitely always going to be quite quantum mechanical. The fact it can be bound by purely correlation effects tells us what kind of beast we're dealing with.

[1]: Sommerfeld, T., & Jordan, K. D. (2006). Electron binding motifs of (H2O) n-clusters. Journal of the American Chemical Society, 128(17), 5828-5833.

2: Sommerfeld, T., DeFusco, A., & Jordan, K. D. (2008). Model potential approaches for describing the interaction of excess electrons with water clusters: Incorporation of long-range correlation effects. The Journal of Physical Chemistry A, 112(44), 11021-11035.

[3]: Jordan, K. D., & Wang, F. (2003). Theory of dipole-bound anions. Annual review of physical chemistry, 54(1), 367-396.

[4]: Ambrosio, F., Miceli, G., & Pasquarello, A. (2017). Electronic Levels of Excess Electrons in Liquid Water. The Journal of Physical Chemistry Letters, 8(9), 2055-2059.

[5]: Coons, M. P., You, Z. Q., & Herbert, J. M. (2016). The hydrated electron at the surface of neat liquid water appears to be indistinguishable from the bulk species. Journal of the American Chemical Society, 138(34), 10879-10886.


Molar heat of fusion of ice? The molar heat of fusion of ice is 6.01 kJ/mol.
Can you explain what this means? What I think it means is that it requires 6.01 kJ to melt a mole of ice. I'm not positive though..

Your interpretation is correct.

The molar heat of fusion is the amount of heat necessary to melt a mole of ice at its melting (or freezing) point. That means, to get 1 mole of water from 1 mole of ice at its melting point (0 degrees Celsius) it would require an input of $6.01\ \mathrm{kJ}$.

The conceptual reasoning is that it requires so much energy to break the bonds and allow more motion (but not completely).

Conversely, $6.01\ \mathrm{kJ}$ would be released upon the freezing of water at that point (1 mole of water anyway).


How do scientists study liquid tungsten? Inspired by this What If article: https://what-if.xkcd.com/50/

In the above article, Randall Munroe mentions that liquid tungsten is difficult to study because of its extremely high melting point. Because of this property, containers for the tungsten tend to melt before the metal itself. This might be a fairly elementary question, but given this difficulty, what methods do modern chemists use to study it, or other materials with extreme melting points?

One way is to use electristatic or electromagnetic forces to hold the liquud in place.  This abstract refers to a "non-contact method" and an "electrostatic levitate", which us enough to reveal the basic strategy.  Unfortunately the article is behind a paywall, so you have to put up to get details on the good stuff.

Industrially, metals like tungsten are not processed as liquids.  They are ... see here.


What factors are influencing a woosh rocket? I asked this over at Physics, but no one seemed to know, so I'll try here:

If I fill a plastic bottle with a small amount of rubbing alcohol, shake it around and hold a match to the bottle neck, I'll get a flame wooshing out of the opening. This can easily be used to propel the bottle upwards, like a rocket. I tried it already, there are lots of youtube videos about this, and NASA has instructions for the experiment as well.

When trying to go for maximum speed or height of the rocket, there are a few factors influencing this experiment. Here are the ones that I can think of:

Some related to the bottle:

volume, shape, aerodynamics, weight, size of opening, possibly shape of the bottle (long and thin or short and fat) When the bottle neck is smaller it creates a stronger thrust.

Others related to the propellant:

How fast does it burn, how dense is it (especially compared to air), how heavily does it vaporize. If the vapor is denser then air it all floats out of the opening of my bottle before I can light it and I don't have as much fuel.

As you might be able to see, I don't know a lot about the propellants.

So here's my question:

What are the factors related to the propellant, that influence the launch height and speed of the woosh rocket.

What's the best propellant to choose for a woosh rocket, and why?

For home projects like this I believe one of the most important things is to ensure you have a good fuel to air (oxygen) ratio. It is surprisingly easy to use too much propellant which can take up air space within the combustion chamber. More fuel doesn't necessarily mean bigger bang.


How to transform a tertiary alcohol to a carboxyl? I am trying to transform a lipid with a tertiary alcohol into a carboxylic acid.

More specifically, I want to convert to diplopterol (see below) into something homologous to cholesteryl hemisuccinate (see below).

Any suggestions?

Diplopterol:

Cholesteryl hemisuccinate:

I am not sure that what you claim to want to do is what you actually want to do.

Cholesterol hemisuccinate is prepared from cholesterol, but not by converting the alcohol functional group into a carboxyl group. The alcohol oxygen becomes one of the oxygen atoms in the ester, but not the carboxyl oxygen. Cholesterol hemisuccinate can be prepared in the lab probably from cholesterol and succinic anhydride in the presence of a non-nucleophilic base (like pyridine or triethyl amine), which can act as a proton shuttle. An acidic workup will be needed to remove the base and protonate the carboxylic acid. The scheme below shows this reaction (sans acidic workup). The alcohol oxygen is highlighted in red.

An identical reaction could be done with your molecule, diplopterol. Dissolve both (1 eq) diplopterol and succinic anhydride (1.1 eq.) in dichloromethane or diethyl ether (or any low polarity aprotric solvent you like). Cool on an ice bath. With stirring, add pyridine (1.1 eq.) dropwise. Let the reaction warm to room temperature, probably about 2 hours. The product will likely precipitate as the pyridinium salt. Check the progress by TLC or some other means. The product will probably be more polar than the reactant. Work up by adding dilute HCl (probably 0.1 M). Extract the aqueous layer with your organic solvent of choice. Combine the organic layers. Wash with brine. Dry over magnesium or sodium sulfate. Remove the solvent by evaporation (rotary being preferred), and purify by chromatography if necessary.

Alternatively, if the product does precipitate at the pyridium salt, collect the salt by filtration, wash the precipitate with ether and then with cold ethanol (will wash away the reactants). Then, add the precipitate to the 0.1 M HCl and extract with ether if necessary. Further purification is probably not needed.


what happens to hydrogen released in addition polymerization? In formation of polyethylene(free radical method), the free radicals when attached to another ethene molecule release a hydrogen(I think).But addition polymerization does not release and by product, so what happens to that hydrogen atoms?

Consider the reaction stoichiometry.

You begin with ethylene (that is, CH2CH2).
The polyethylene chain is a chain of single-bonded carbons. Hence, it is a  -[...]-CH2-CH2-CH2-CH2-[...]- chain).

As you can se, no hydrogens are removed by the monomer.
So, no hydrogens are released during polymerization.

What is happening in that reaction is an electron transfer, not an atomic transfer: the double bond of ethylene, full of electrons, reacts with the radical and forms a more stable structure with a radical carbon (a radical carbon has three bonds, in this case with C, H and H, and an unpaired electron).
The carbon radical reacts with another ethylene monomer: the "old" radical carbon now forms four bonds (C, H, H, plus the new C from the ethylene monomer), the new ethylene monomer loses its double bond, and forms a carbon radical (again, the radical carbon has three bonds and an unpaired electron).

This is the logic of that tipe of polymerization: no hydrogen is involved.


Did Mendeleev predict the existence of neutrinos? According to this Wikipedia page, Mendeleev originally believed that the inert gases belonged in Group 0 (to the left of the alkali metals). Thus, helium would be placed in the second period to the left of lithium and argon would be placed in the third period to the left of sodium. Apparently this made sense to him at the time, since atomic theory was in its infancy (and Mendeleev didn't like it) and quantum mechanics was only a twinkle in the eyes of Planck, Bohr, and others.

Based on this arrangement, there is an open spot above helium and to the left of hydrogen for an ultralight inert gas he called "proto-helium". We now know that hydrogen must be the lightest known element, meaning Mendeleev was wrong about the placement of the inert gases. However, host of the other holes in Mendeleev's table were filled with real elements having properties nearly identical to what Mendeleev predicted.

Mendeleev predicted that proto-helium would have a miniscule mass (ca. $ 5.3\times10^{-11}$ amu) and high (near relativistic) velocity. Mendeleev believed that this near massless particle would be able to permeate all matter, rarely interacting chemically. He thought that they might be responsible for radioactive decay.

A neutrino is massless (or nearly so), and travels at the speed of light (if massless) or just under (if it has mass). Otherwise, it ignores the electromagnetic force, and so has no chemical interaction. Neutrinos are often produced during radioactive decay.

The relevant references point back to Mendeleev's original writings (in Russian), which I cannot access, so I cannot verify if he made a prediction about "proto-helium". Did he predict this particle? Has anyone previously made the connection between Mendeleev's proto-helium and neutrinos?

While it's an interesting thought, I don't think it's very useful to suggest Mendeleev predicted neutrinos. It's far more sensible for it to be a curious coincidence. His suggestions of ghost-like properties similar to neutrinos (very hard to detect, tiny mass, little interaction with matter, etc) are more easily explained as an attempt to justify (properly) why it was so hard to detect: if it had a much larger mass, someone should've weighed it by then; if it interacted more, someone should've managed to capture it physically or chemically, and so on. It's not uncommon for theoretical predictions to temporarily "take refuge" just beyond the limit of observability (for example, proton decay), with varying degrees of justification. In this case, Mendeleev's prediction was wrong in its original terms, and happened to almost match something else entirely because neutrinos really are hard to detect! Also, the connection with radioactivity could very likely have been made merely due to good timing with its discovery. Since no one had a good explanation, why not forward your own!

And as a sidenote, neutrinos definitely have mass, otherwise they could not suffer the observed flavour oscillation.


Why does screening effect decrease due to d-orbital? In 13th group, atomic radius increases from boron to aluminium. From aluminium to gallium, atomic radii decreases. From gallium to indium, atomic radii increases. And from indium to thallium, atomic radii decreases.

The reason for this irregular trend given is screening effect. While explaining my teacher told that due to d-orbital screening effect decreases.

I can't get why?

The reason d-orbitals make a difference is that electrons in d-orbitals do not screen nuclear charge as effectively as those in s and p orbitals.  This is because of something called penetration .

The mathematical shapes of d-orbitals prevent them from allowing electrons to penetrate very closely to the nucleus, compared with electrons in s or p-orbitals.  In gallium, you have $10$ electrons in the filled $3$d-subshell, and each of these electrons is doing a slightly worse job (relatively speaking) of screening the nuclear charge than the electrons in the s and p orbitals. Therefore, the effective nuclear charge in gallium is slightly higher than that in aluminum, so the increase in the radius is a quite a bit smaller than would be expected based on the difference between boron and aluminum, or gallium and indium.
The trend goes:

$\pu{82 pm}$ ($\ce{B}$)
$\pu{118 pm}$ $\ce{(Al)}$
$\pu{126 pm}$ $\ce{(Ga)}$
$\pu{144 pm \ce{(In)}}$

[covalent radii from www.webelements.com].

This effect is generally known as the d-block contraction.  (It can be more or less pronounced depending on how you define the atomic radii.)

A similar thing happens (in principle) when you go from indium to thallium; except in this case you are now dealing with adding a filled f-subshell to the valence shell.

Electrons in f-orbitals are even worse at screening nuclear charge than those in d-orbitals, therefore again, the effective nuclear charge in thallium is a bit larger than it is in indium, so again the jump in radius is fairly small (from $144$ to $148$ pm).  This effect (of the filled f-subshell) is generally known as the lanthanide contraction.


What is the efficiency of UV lamp in ozone generation (Ozone g vs. energy consumed ) re.:

What is the efficiency of UV lamp in ozone generation (Ozone g vs. energy consumed )

I read a lot of papers from the Internet on using UV lamp for generating ozone but none discussed the efficiency of such process.

How much energy (kWh) gets consumed by UV lamp to generate 1g of ozone ?

I read  Ozone Formation from UV Light
https://www.oxidationtech.com/ozone/ozone-production.html#answer2

and related web links
and all I get as pros and cons is:

"
Ozone production from UV light for commercial applications
Advantages of UV Ozone Generator

Simple Construction – only a UV lamp with ballast and a fan

Low cost – due to simple construction

Lower nitric oxide production

Disadvantages of UV Ozone Generator

Low ozone output – g/hr

Low ozone concentration – less than 1% by weight

Bulbs and ballasts will need to be replaced

"

Let me know your opinion

I read about UV light source efficiency from one boomarked site but just visit UV Light Ozone Generators boie.us/uv-ultraviolet.php claiming ozone output: 10g/h and still no data on power consumed in the specification "Lamps: Long-life VH** 9,000-hour high-output ozone lamps not rated "Electrical Rating: 120V~60Hz 2.6 Amps So in theory, power consumed is less than 120V x 2.6 Amps = 312W. Hard to believe, low efficiency UV lamps can generate 10g/h ozone comsuming less than 1,000W x hr – a a Apr 28 at 20:11

great image of the opened box BoiE UltraViolet 10K Shock Model boie.us/images/image-slider-uvozone.jpg not sure 10K stays for what " 10,000 mg/hr Eureka!!! 10K stays for 10,000 mg/hr to be verified – a a Apr 28 at 23:07

and one more 12V model of 12-volt corona discharge ozone generator boie.us/dc-sportsman-pro.php since ozone output is selectable, matching electrical rating, so "Electrical Rating: 12V DC 1.55A (Low) 4.45A (High) matches selectable ozone output (1,000 or 4,500 mg/hr) so 1,000 mg/hr consumes 18.6W x h 4,500 mg/hr .. 53.4W x h 4.5 x 18.6W = 83.7W so a single generator consumes less energy than 4.5 stacked in parallel, generating the same amount of ozone. – a a Apr 28 at 23:17

ok, ..20W rated power, 12-volt corona discharge ozone generates 1,000 mg/hr vs. 312W UV light ozone generator generates 10g/hr, so 30W UV ... generates 1g/hr 20W (corona) matches 30W (UV) rated 1g/h since corona generator is much more complicated HV I am going to build UV ozone generator – a a Apr 28 at 23:24


How does malachite react with acid? I performed an experiment where I did this:

  Add one spatula of malachite to a test tube, add dilute hydrochloric acid to the malachite until the test tube is one third full. Note observations and identify gas produced.

The green solid turned to a blue solution and there was an effervescence of hydrogen. However, I now realise that copper doesn't react with acid and malachite is a copper carbonate hydroxide mineral...
I researched online and found that

  It dissolves in sulfuric acid to give Copper sulfate, Carbon dioxide
  and Water.
  http://wiki.answers.com/Q/What_happens_when_malachite_is_dissolved_in_sulphuric_acid?classic=true

So how did I get that malachite reacted with acid and did produce hydrogen with a squeaky pop test?

Thanks in advance

No, pure malachite $\ce{Cu2CO3(OH)2}$ usually doesn't produce hydrogen because it is a copper carbonate. Copper is already in a higher oxidation state (+2) so it can't reduce hydronium to hydrogen. The right reaction is the following:
$$\ce{Cu2CO3(OH)2 +2H2SO4 -> 2CuSO4 + CO2 ^ + 3H2O}$$

However, if there are traces of metallic copper some hydrogen can be produced but not with sulfuric acid because:

$$\ce{Cu + 2H2SO4 -> CuSO4 + 2H2O + SO2 }$$

Nor with $HCl$ (copper doesn't react with $HCl$)  but only with nitric acid or aqua regia you can produce hydrogen from the metallic form. So, in fact, the gas you see is carbon dioxide.


Does the "extent of reaction" have a physical meaning? I am going to argue that it really doesn't have a useful or "comprehensible" physical meaning.

Lets say we have the following reaction:
$$3A\rightarrow2B$$

Chemists define the extent of reaction as:

$$\xi=\frac{n_{A_f}-n_{A_i}}{v_A}$$
where $n_{A_f}$ is the moles of $A$ once the reaction reaches equilibrium, $n_{A_i}$ is the moles of $A$ before the reaction began, and $v_A$ is stoichiometric coefficient of A for this reaction.

The numerator makes physical sense, it is simply how many moles of A are consumed over the course of reaction. The denominator makes sense because it states the amount of molecules needed to be consumed to generate the product.

But once you put the numerator over the denominator, it starts to lose any physical meaning to me and simply becomes a useful mathematical conversion factor. If there is an exact physical meaning here, what is it?

P.S. It has the units of moles which is weird because it doesn't equal the change in moles for this reaction or the amount of moles consumed and produced on either side.

The stoichiometric coefficients $\nu_i$ are arbitrary anyway; the only thing that matters is the ratio of them. $\ce{3A -> 2B}$ and $\ce{6A -> 4B}$ both describe the same physical process occurring (unless it is implying something about mechanisms, but in this context of thermodynamics, that's not the case). Clearly, even for a single unchanging system, its value of $\xi$ will vary depending on how you define the stoichiometric coefficients. So, I wouldn't expect $\xi$ to have any physical meaning. Personally (in my limited knowledge of thermodynamics) I just view it as a useful, but ultimately arbitrary, definition.

[In the same vein, I would say that the magnitude of the equilibrium constant $K$ doesn't have any physical meaning either, since it also depends on how you define the stoichiometric coefficients. But I digress.]


Atomic Orbital and Wave Function I was reading about the atomic orbital in my chemistry textbook. It says that the atomic orbital (psi) is a mathematical wave function that depends on the coordinates of the electron. It is also mentioned that psi does not carry any physical meaning. That part is understood, however I could not understand what a psi vs r graph represents.(r is the distance from the nucleus) Can someone please clarify ??

$\psi$ does carry a great amount of physical meaning. The annoying part is, that it doesn't represent anything directly observable. The way (non-relativistic) quantum mechanics is constructed, makes $\psi(x_i, s_i, t)$ ($x_i$ - physical coordinates of $i$-th electron, $s_i$ - spin projection on $z$ of said electron) -  a direct analogue to trajectory, or, in case $\psi$ doesn't depend on $t$ (time),  orbit. *

Orbitals themselves, however, do not have direct physical meaning in that sense that they are nothing more than building blocks we use to construct collective $\psi$ of the electrons in the chemical system after we apply Born-Oppenheimer approximation (that we can ignore nucleus movement when we talk about electron movement). It's a good approximation, so it is almost always in effect in quantum chemistry.

In some special cases (hydrogen-like atoms) the electron wave-function represents exactly one electron of one atom and, ignoring spin contribution, orbital equals wave-function of that one-electron system.

Orbitals of hydrogen-like atoms can be represented ** as $\psi(x)=\psi(r,\theta,\phi)=\psi_r(r)\psi_{angle}(\theta,\phi)$. It is a useful representation, because even if $\psi_r$ changes from level to level, $\psi_{angle}$ doesn't, so you can see parallels between, say, $2p$ and $3p$ orbitals, which differ*** only by $\psi_r$. Again, neither part of the orbital is observable. What IS observable is "electron density" i.e. a probability to observe electron in this particular finite (small) volume. Those can be extracted from various electron or X-ray diffraction experiments. What, however, it is good at is to point out, how good outer electrons shield inner electrons and that chemical properties are mostly guarded by outermost reaches of outer shells.

$\psi_r$ on its own cannot be observed and has no simple interpretation. It has no inherent meaning on its own, and I'm unaware of obvious interpretation. It IS useful when one has to compare different orbitals between each other or to compare various approximations of the same orbital though.

$\psi^2_r$ is fairly useless, imho. It is observable for $s$ orbitals as electron density at given radius, but it isn't very useful. What is useful IMO is density integrated over spherical surface, which for $s$ orbitals is $ |rR|^2 = 4\pi r^2 |\psi_r|^2 $, which shows how much time electron spends at given distance from nucleus. It shows, how diffuse the orbital is, and when plotted against observed total electron density, it shows how good electron is shielded against outer influence.

I hope this helps.

* Note: Here is one of the non-obvious differences between quantum and classical systems lie: you CAN have a quantum system with large kinetic energy and $\psi$ independent on time.)

** Note: it is not the case for all mathematical functions, we are lucky that it is the case for atomic orbitals.

*** Note: orbitals on the same level CAN differ by $\psi_r$. The general rule is that orbitals on the same level have the same number of surfaces where $\psi$ is zero that is equal number of the level minus one. That means that we have only $s$ orbitals on the first level with zero such surfaces, but $2s$ and $2p$ orbitals. For $2p$ orbitals the "zero" surface is one of coordinate planes and for $2s$ orbital$ it is a sphere around the nucleus.


Bond angle of nitrous acid Why are there two possible bond angles for the nitrous acid molecule?

After perusing Wikipedia I found that there are both cis and trans forms of the nitrous acid molecule. I also note that the cis form is sterically hindered. Could this be a reason for the two different bond angles - there are two different geometric isomers and in one case there are larger van der Waals repulsions, which in turn expands the bond angle?

This ism the microwave structure of nitrous acid, 1971.  Your mileage may vary condensed phase.

DOI:10.1039/TF9716702179 (1971)
The structure of cis-nitrous acid is found to be: O—H = 0.982, N—O(H)= 1.392, N—O = 1.185 Å, ∠ NOH = 104.0 and ∠ ONO = 113.6°, and for trans-nitrous acid: O—H = 0.958, N—O(H)= 1.432, N—O = 1.170 Å, ∠ NOH = 102.1 and ∠ ONO = 110.7°.

http://nzc.iap.ac.cn/uploadpdf/Wang_et_al._CSB_2007a.pdf
  2007 pape with more recent citations.


What is the correct way to transfer large volumes of chemicals? I am an engineer starting a film developing and processing lab in my house for personal use. I am currently using a bathroom, and am concerned about contaminating my chemical process and the bathroom using what seem to be accepted practices in developing labs.

Many developing chemicals (e.g. Kodak D-76) come in exact quantities to mix 1 gallon of stock solution. I will mostly be using quantities 500 mL or less, though certain chemicals will be larger arbitrary volumes.

What methods/equipment are used for transferring large volumes?

Given that the chemicals in question aren't air sensitive (i.e. theres no need for an inert atmosphere), a measuring cylinder would suit your needs.

They're available in glass and various plastics, in sizes upto around 3000 mL. The glass ones are generally better as theres less risk of the chemicals reacting with the glass compared to with plastic, but it really does depend on what you're using.

Once you're done with the measuring cylinder, just rinse it with water (into a suitable toxic waste container if the chemicals cannot go down the drain, which I suspect they can't), and leave it to dry somewhere (else put it in a warm oven for an hour or so).


Why is this organic compound trans, and not cis? I'm reading a high school chemistry course online, and I'm now reading about geometric isomers.

They provide this example:

And in terms of naming it, they say this:

  Begin by identifying the parent chain - it is the continuous chain that contains the carbon-carbon double bond. The two groups that belong to the parent chain are on opposite sides.

Then supply this illustration:

And with that explains that it's trans-3-ethyl-3-hexene for that reason. I don't quite get their explanation though, becuase to me it seems like this could also be cis.

Why can't it be this one:

As far as I understand, that's also a valid way of expanding it, because it has the same amount of carbons in the chain and the numbers are still as low as they get. The only difference is that this one would be cis-3-ethyl-3-hexene instead of trans-3-ethyl-3-hexene.

So, why exactly is it that only one of them is right?

You are right, there is no difference between the two molecules that result if you rotate 180 degrees about the $\ce{C=C}$ double bond - they are the same exact molecule.  So for this molecule, the terms "cis" and "trans" would not be used, they have no meaning here.  The name of the molecule is simply 3-ethyl-3-hexene.  If the on-line course says otherwise, it is wrong.


Relation between number of equivalents of reactants Consider the reaction $\ce{Sb2S3 + F2 -> SbF5 + SF6}$. If the milliequivalents of antimony, sulfur and fluorine are x, y and z respectively, then prove that $x+y=z$.

This is what I know: Number of equivalents of each reactant in the reaction is equal, and that number of equivalents is (mass of species) /(equivalent weight). The problem I'm facing is that $\ce{Sb}$ and $\ce{S}$ are in the same compound, so I think $x$ should be equal to $y$ (this is clearly incorrect). Also, the mass of reactants used is not mentioned, so I don't know how to proceed with the question.

Step 1: Balance the equation. Your equation as written $\ce{ Sb2S3 + F2 -> SbF5 + SF6}$ is not balanced. You do not have the same numbers of sulfur, antimony, or fluorine atoms on each side of the equation. You cannot say anything about equivalents or stoichiometry with an unbalanced equation. Hint: Start with $\ce{Sb}$ and $\ce{S}$ and balance $\ce{F}$ last.

Step 2: Determine the relationship between $x$ and $y$. Since $x$ is antimony and $y$ is sulfur and they are in the same compound initially, they have a relationship (but they are not equal). In $\ce{Sb2S3}$, there are 2 sulfur atoms for every 3 antimony atoms. What does that tell you that the ratio of $\frac{x}{y}$ must be?

Step 3: Use what you now know about $x$ and $y$ and the correct stoichiometric equivalents (which should give you $z$) to show that $x+y=z$.


Preparing a 0.030 M solution of potassium ferrocyanide trihydrate? I am a little bit confused on what I am supposed to do. The instructions say to prepare 250 mL of this solution by measuring out in grams the amount of PFT using a 250 mL volumetric flask. Then it says I have to add dI water until te meniscus hits the full mark on the flask, then it says to calculate the actual molarity of the solution. Am I supposed to multiply the molarity by the liters given to me to find the grams of PFT I am supposed to use? How would you go about this?

It is quite trivial. Just calculate how many moles of potassium ferrocyanide trihydrate you need to measure out (n = C x V), then multiply with molar mass to see how many grams you need to measure out. After measuring the sample on analytical balance, quantitavely transfer it to the volumetric flask. Add some water and dissolve the sample. After homogenizing the solution, add water carefully to the full mark. You can't actually calculate the exact molarity of the prepared solution: you need to determine it experimentally, by performing a standardization with silver-nitrate standard using chromate salt as an indicator.


Is HPLC the best way to detect residues from anti-coagulant rodenticides I am a beginner chemistry student studying anticoagulant rodenticides. The literature shows that high-performance liquid chromatography (HPLC) with fluorescence or UV light is the most commonly used method for the detection of residues in tissue samples. I want to understand on a basic, conceptual level why this particular analytical method is the preferred method.

I understand that HPLC can separate complex organic compounds and enable detection at extremely low concentrations, but I do not know whether there are other analytical methods that can also do this.

Is this particular analytical method more effective in detecting the presence of residues from anticoagulant rodenticide in tissue samples? If so, why? Is the answer related to the chemical properties of the compounds in rodenticides (hydroxycoumarins and indandiones)?

UV(/VIS) spectroscopy is the most common HPLC detection method, it's quite cheap and powerful but you can only detect substances which absorb light in the range you are using for detection. You also don't get any real structural information, most of the time identification is only possible using reference material. But most HPLC systems will be equipped with such a detector.

Fluorescence detectors can be used for fluorescent molecules and usually you got a much lower limit of detection. Like for UV/VIS you don't get structural information, but identification works quite well, since you can selectively only detect molecules with the used excitation and emission wavelengths.

Other detectors would be for example refractive index detector, which only really work well if HPLC is run isokratic or MS, which gives you structural information.


Chlorinating specifically one carbon on a symmetrical alcohol The final target molecule is 1-chloro-2-amino propane.

The amination step would be simple via reductive amination of 1-chloro-iso-propanone.

Obtaining 1-chloro-iso-propanone would be from nucleophillic addition-elimination of 1-chloro-iso-propanol by H2O under acidic conditions (or oxidation using potassium permanganate but no need for such a reagent when a simpler route available).

However, assuming you started with only iso-propanol and you wanted to chlorinate only one of the two terminal carbons, how would you do this? I understand $\ce{SOCl2}$ would of course be the chlorinating reagent, but with no blocking group it would simply attack both carbons. Attempting to add a blocking group would block both carbons also as they are symmetrical and identical.

It might be overkill for your purposes, but there is a nice reaction that coverts secondary alcohols to $\alpha$-chloroketones in a one-pot synthesis, using trichloroisocyanuric acid (for instance, they got 75% yield of 2-chlorocyclohexanone from cyclohexanol).  The paper is 'Direct conversion of alcohols to $\alpha$-chloro aldehydes and $\alpha$-chloro ketones' (Jing, Daniliuc, Studer 2014, Org. Lett. 16 (18) pp. 4932-4935).

  Direct conversion of primary and secondary alcohols into the
  corresponding α-chloro aldehydes and α-chloro ketones using
  trichloroisocyanuric acid, serving both as stoichiometric oxidant and
  α-halogenating reagent, is reported. For primary alcohols, TEMPO has
  to be added as an oxidation catalyst, and for the transformation of
  secondary alcohols (TEMPO-free protocol), MeOH as an additive is
  essential to promote chlorination of the intermediary ketones.

Trichloroisocyanuric acid has the advantages that it's inexpensive, widely available and easy to handle.


Why is delta-H negative here? Very basic question here, but I'm confused why Delta-H seems to be the reverse of what I would expect for bond enthalpy. For example, here's a problem in my textbook where the goal is to find delta-H:
$$\ce{H2(g) + Br2(l) -> 2HBr(g)}$$
Using the table I have, I see that an $\ce{H-H}$ bond has an average enthalpy of 436, $\ce{Br-Br}$ has 193, and $\ce{H-Br}$ has 366. 436+193 = 629, and 2*366 = 732, so the reactants have a total of 629, compared to the products, which have 732.
My intuition (and Google, as far as I can tell) would tell me that, because the H value increased when the reaction occured, the answer would be positive 103. But my textbook says the answer is -103. This makes no sense to me - what am I missing here? Why do all the Delta-H (AKA Change in H) values seem to be negative what you'd expect?

The bond enthalpy is conventionally the enthalpy change of bond breaking, not of bond forming, in the contrary to formation enthalpy of compounds.
This way they are all positive.
The consequence is, the order of products and reactants in computation the enthalpy difference has to be switched.
So if the summary bond enthalpy of products is higher than of reactants, the difference of enthalpy as the reaction enthalpy is negative, not positive, and the reaction is exothermic.
Formation enthalpies:
$$\Delta H_\mathrm{r}=\Sigma  \left( \Delta H_\mathrm{f, products} \right) - \Sigma \left( \Delta H_\mathrm{f, reactants} \right)   $$
Bond enthalpies:
$$\Delta H_\mathrm{r}=  \Sigma   \left(  \Delta H_\mathrm{b, reactants} \right)   - \Sigma  \left(  \Delta H_\mathrm{b, products}  \right) $$