Why must an oxidizer not touch rubber? Why is it that oxidizers (specifically $\ce{KClO3}$) are at risk of causing a fire when they come into contact with the rubber stopper used to contain them in their container, but can be slowly heated if they do not contact the rubber stopper?

Essentially, why is it that oxidizers are incompatible with rubber?

A lot of rubbers possess C=C double bonds. Alkenes are very easily oxidized by KMnO4 or KClO3, therefore in the case where you're not aware of the chemical nature of the rubber, it's best not to expose it to strong oxidizers.


Which metal ions precipitate in NH4OH? From Wikipedia-1, and from an answer here on Stack-Exchange, I can list at least the following elements which will dissolve in aqueous $\ce{NH4OH}$:

$\ce{Cr(III), Co(III), Ni(II), Cu(I, II), Zn(II), Ag(I),}$ along with  "several platinum group metals"

From reading Wikipedia-2, I also suspect the following elements to dissolve in liquid ammonia:

$\ce{Mg, Ca, Sr, Ba, Eu, Yb, Li, Na, K}$ and $\ce{Rb}$

As a matter of fact, according to Wikipedia "almost all metal ions bind ammonia as a ligand." I thought it would be interesting, and potentially very useful, to find out which metal ions are NOT soluble in $\ce{NH4OH}$.

I can list two for now:

$\ce{Fe(OH)3}$
$\ce{Mg(OH)2}$ with a solubility product of only $5.61×10^{−12}$ (credit to Oscar Lanzi)

For the case of magnesium ion, at least, it seems a stretch to say it dissolves in ammonia solution.  Assume that the ammonia solution has $\mathrm{pH}$ of $11$ and magnesium hydroxide has a solubility product of $5.61×10^{-12}$. We infer that the magnesium ion solubility at equilibrium with the hydroxide in ammonia solution is only $\pu{5.6μM}$ which is not really all that soluble.

Magnesium ion is appreciably soluble not in an ammonia solution but in a buffered solution of ammonia and an ammonium salt.


Calculating concentration and volume in acid metal reactions In the reaction:
$$\ce{3 Zn + 2 H3PO4 -> Zn3(PO4)2 + 3H2}$$

I know the initial mass of $\ce{Zn}$ to be 13.08 g and the volume of phosphoric acid to be 0.5 $\mathrm{dm^3}$. How do I work out the final concentration of $\ce{Zn(PO4)2}$ and the volume of $\ce{H2}$ gas evolved?

Once I work out the moles of Zn, are the moles of $\ce{Zn(PO4)2}$ just $\ce{Zn}$ moles/3?

Your equation uses molar ratios, rather than weights or volumes.
If you know the weight of zinc used, you know the number of moles. From this, you can calculate the number of moles of zinc phosphate generated.
Supposed that the phosphoric acid is sufficiently concentrated to react all zinc, the number of moles calculated above together with the volume of the phosphoric acid used allows to calculate the concentration of zinc phosphate. Note: Do not omit units in your calculations!
As far as the hydrogen is concerned: The number of moles produced can be calculated using the amount of zinc used. Additional knowledge of the molar volume of gases under standard conditions is helpful to figure out the volume.


Why are sodium and potassium displaced from soap by calcium and magnesium ions in hard water? Sodium and potassium are more reactive than calcium and magnesium still they get displaced. Why?

I have no external justification to point to on this, but I've always thought one aspect contributing to this behavior is the fact that the divalent $\ce{Ca^2+}$ and $\ce{Mg^2+}$ bind two soap fatty acid molecules, whereas $\ce{Na+}$ and $\ce{K+}$ as monovalent ions can only strongly bind one molecule.

As a result, the alkaline earth cations have two fatty acid "tails" that can 'glom' around them, shielding them better from the water molecules and making the resulting complex more stable, as well as less soluble.

On the other hand, since the alkali cations only have one fatty acid "tail" they're strongly associated with, that one tail can't wrap around nearly as completely, and so there is significantly greater accessibility of the solvent to the cation.


what surface would liquid wax adhere to best? I want to make a very thin sheet of wax by pour hot wax over a surface and letting it cool. What surface would wax spread out on best? I.e. which one would give the best wettability?
Many thanks everyone

THanks guys - will let you know how this goes!

I think that you need a surface with very low wettability, because as you said in the comments, you want to remove the solidified wax layer afterwards. It is possible to make a very thin layer of wax by spreading it between two hard plates pressed against each other. I would try something like this:

You spread the molten wax between the two sheets made from an appropriate material placed between two plates (glass or metal). To precisely regulate the thickness of the wax layer you use some separators placed near the edges. After setting you remove the plates, then carefully remove the sheets. The main problem is to choose a suitable material for the sheets. You can experiment with various materials. I think that a sheet made from, or covered with a silicone rubber would be perfect. You can try a parchment paper, aluminum foil, glassine and so on. If you need a really thin and fragile wax layer, you can try to use a tin or aluminum foil, and then remove it chemically using diluted hydrochloric acid.
EDIT: Now it hit me. There is another, simpler method you may try. If you want a wax plate with a uniform thickness and smooth surfaces use the procedure above. However, if you don't have those requirements, then simply pour the molten wax on the surface of hot water in a large pot. The wax should spread evenly on the whole surface. Leave it to cool down and collect the wax plate. It may be a bit thicker near the edges and its surface will be matte, but maybe it'll suit your needs.


Difference between electropositivity and electronegativity Sodium is more electropositive than hydrogen although they are in the same group.

Why is the electronegative value of hydrogen greater than sodium (metal)?

Background:
Electropositive  implies having tendency to lose electrons to form positive ions. Electronegative implies having tendency to gain electrons to form negative ions. Electronegativity is a concept used to describe the tendency of an atom to attract electrons in a covalent bond.

  Why is the electronegative value of hydrogen greater than that of sodium?

This is because sodium after losing an electron gets its octet configuration, but a hydrogen atom after losing an electron will not get the configuration of helium but it can get helium configuration if it accepts an extra electron or forms a covalent bond.

So this implies that sodium is more electropositive that hydrogen, even though they belong to same group.

Sometimes "more electronegative" is used to describe higher electronegativity and "more electropositive" is used to describe lower electronegativity. In this sense, elements are less electronegative (or more electropositive) as you go down any group in the periodic table. However, the difference in electronegativity between hydrogen and sodium is exceptionally large, and correlates with the very different chemistry these two elements show.


What is the entropy of mixing of two ideal gases starting out with different pressures? The entropy of mixing of ideal gas is given by this equation:

$\Delta S_{mix} =  -nR(x_1\ln x_1 + x_2\ln x_2)$

Does this equation works only when the initial conditions of both compartments are at the same pressure?
If ideal gas 1 at initial conditions $n_1,P_1,V_1,T$ and ideal gas 2 at $n_2,P_2,V_2,T$ are mixed, which equation should I use to obtain $\Delta S_{mix}$ given that $P_1 ≠P_2$? Assume $V_1 + V_2 = V_{tot}$, and $T_1 = T_2 = T$.

A strategy that often works is to look at the initial and the final states and then compare them. A good place to start is usually the Gibbs free energy of the system (with the intention to use $\Delta_{\mathrm{mix}} S = - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \big|_{p, \{ n_{i} \}}$ later): for isothermal and isobaric conditions it is

\begin{equation}
  G = \sum_{i} n_{i} \mu_{i} + \mathrm{constant} = n_{\mathrm{tot}} \sum_{i} \underbrace{x_{i}}_{= \, \frac{n_{i}}{n_{\mathrm{tot}}}} \mu_{i} +  \mathrm{constant} \hspace{4em} (1)
\end{equation}

with $n_{i}$, $\mu_{i}$, and $x_{i} = \frac{n_{i}}{n_{\mathrm{tot}}}$ being the amount, the chemical potential, and the mole fraction of the $i$th component in the system, respectively, and $n_{\mathrm{tot}} = \sum_{i} n_{i}$ being the total amount of gas in the system.
So, now we need an expression for the chemical potential of an ideal gas.
It can be shown that this is given by

\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\biggl(\frac{p_{i}}{P^{0}}\biggr) \hspace{18em} (2)
\end{equation}

where $\mu_{i}^{0}$ is the chemical potential of the $i$th component in a given standard state, $R$ is the universal gas constant, $T$ is the temperatur, $p_{i}$ is the partial pressure of the $i$th component, and $P^{0}$ is the standard pressure (i.e. the pressure connected to the standard state).
For ideal gases we can apply Raoult's law stating that

\begin{equation}
  p_{i} = x_{i} P_{\mathrm{tot}}
\end{equation}

where $P_{\mathrm{tot}}$ is the total pressure of the system.
Using this with equation (2) we get

\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\biggl(\frac{x_{i} P_{\mathrm{tot}}}{P^{0}}\biggr) = \mu_{i}^{0} + R T \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{tot}}}{P^{0}}\biggr) \hspace{6em} (3)
\end{equation}

Furthermore, it is known that the partial pressures in a system must add up to the total pressure, i.e.

\begin{equation}
  P_{\mathrm{tot}} = \sum_{i} p_{i}
\end{equation}

Ok, now that the stage is set we simply have to apply equations (1) and (3) to the initial and the final state of the mixing process.
At the initial state the chambers are seperated (whereby chamber $i$ contains only one sort of gas labeled by $i$) and so these subsystems are completely independent from each other.
Thus, the Gibbs free energy of the total system of chambers is simply the sum of the Gibbs free energies of each chamber, whereby the gas in each of the chambers feels the pressure $P_{\mathrm{tot}} = P_{i}$ in the system (whereby $P_{i}$ is the ambient pressure in chamber $i$), since $x_{i} = 1$ (each chamber/independent subsystem holds just one kind of gas).
According to equation (1) the Gibbs free energy of the $i$th chamber is then given by

\begin{align}
  G_{i} &= n_{i} \mu_{i} = n_{i} \biggl( \mu_{i}^{0} + R T \underbrace{\ln 1}_{= \, 0} + R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \\
        &= n_{i} \biggl( \mu_{i}^{0} + R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr)
\end{align}

and the Gibbs free energy of the total system at the initial state is

\begin{align}
  G_{\mathrm{ini}} &= \sum_{i} G_{i} \\
                 &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln\biggl(\frac{P_{\mathrm{tot}}}{P^{0}}\biggr) \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + \sum_{i} x_{i} R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + \sum_{i} x_{i} R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \ .  \hspace{7em} (4)
\end{align}

For a multicomponent system with more than two components I would simply use this equation. But for a system with exactly two components we can introduce some substitutions that will simplify the final equations a little. For that purpose we want to establish a relationship between the initial pressures $P_{1}$ and $P_{2}$ and the final pressure $P_{\mathrm{fin}}$. In order to get there we use the ideal gas law and the additivity relations for the volumes and amounts: From

\begin{equation}
  n_{\mathrm{tot}} = n_{1} + n_{2} \ , \qquad V_{\mathrm{tot}} = V_{1} + V_{2} \\
  P_{1} V_{1} = n_{1} R T \ , \qquad P_{2} V_{2} = n_{2} R T \ , \qquad P_{\mathrm{fin}} V_{\mathrm{tot}} = n_{\mathrm{tot}} R T
\end{equation}

it follows that

\begin{align}
  P_{1} V_{1} + P_{2} V_{2} = P_{\mathrm{fin}} V_{\mathrm{tot}} \ .
\end{align}

Now, we can express $P_{2}$ in terms of $P_{\mathrm{fin}}$ by introducing the deviation factor $\alpha$ which is defined by $P_{2} = (1 - \alpha) P_{\mathrm{fin}}$ and use this in the previous equation

\begin{align}
  P_{1} V_{1} + P_{2} V_{2} &= P_{\mathrm{fin}} V_{\mathrm{tot}} \\
  P_{1} V_{1} + (1 - \alpha) P_{\mathrm{fin}} V_{2} &= P_{\mathrm{fin}} V_{\mathrm{tot}} \\
  P_{1} V_{1} &= P_{\mathrm{fin}} \bigl( \underbrace{V_{\mathrm{tot}} - V_{2}}_{= \, V_{1}} + \alpha V_{2} \bigr) \\
  P_{1} &= \biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) P_{\mathrm{fin}}
\end{align}

which gives us the relation between $P_{1}$ and $P_{\mathrm{fin}}$. Introducing these equations into equation (4) yields

\begin{align}
  G_{\mathrm{ini}} &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl(\frac{P_{1}}{P^{0}}\biggr) + x_{2} R T \ln\biggl(\frac{P_{2}}{P^{0}}\biggr) \biggr) \\
  &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\frac{\bigl( 1 + \alpha \frac{V_{2}}{V_{1}} \bigr) P_{\mathrm{fin}}}{P^{0}} + x_{2} R T \ln\frac{(1 - \alpha) P_{\mathrm{fin}}}{P^{0}} \biggr) \\
  &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + \underbrace{(x_{1} + x_{2})}_{= \, 1} R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \\
 &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \ .
\end{align}

At the final state the chambers are connected and the gases mix, so the chambers are not independent subsystems anymore and the Gibbs free energy cannot be broken down into independent parts like before.
Also, the ambient pressure in the system changes to $P_{\mathrm{tot}} = P_{\mathrm{fin}}$.
Now, the complete system of connected chambers must be treated via equation (1)

\begin{align}
  G_{\mathrm{final}} &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} R T x_{i} \ln x_{i} + \sum_{i} x_{i} \underbrace{R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr)}_{= \, \mathrm{constant}} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \ .
\end{align}

Now, that we have the Gibbs free energies of the initial and final state we simply need to subtract the former from the latter one in order to obtain the change in Gibbs free energy during the mixing process:

\begin{align}
  \Delta_{\mathrm{mix}}G &= G_{\mathrm{final}} - G_{\mathrm{ini}} \\
                       &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\
                       &\quad - n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \\
                       &= n_{\mathrm{tot}} R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) - n_{\mathrm{tot}}  R T \biggl( x_{1} \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} \ln(1 - \alpha) \biggr) \qquad (5)
\end{align}

Now you see why we went through those substitutions for $P_{1}$ and $P_{2}$. This way the $n_{\mathrm{tot}} R T \ln\Bigl(\frac{P_{\mathrm{fin}}}{P^{0}}\Bigr)$ terms cancelled in equation (5).

Differentiating equation (5) with respect to the temperature we get the entropy of mixing since $\frac{\partial G}{\partial T} \big|_{p, \{ n_{i} \}} = - S$

\begin{align}
  \Delta_{\mathrm{mix}} S &= - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \bigg|_{p, \{ n_{i} \}} \\
   &= - n_{\mathrm{tot}} R \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + n_{\mathrm{tot}}  R \biggl( x_{1} \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} \ln(1 - \alpha) \biggr)
\end{align}

The first term $- n_{\mathrm{tot}} R \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right)$ is just the usual entropy of mixing. The second term $n_{\mathrm{tot}}  R \Bigl( x_{1} \ln\Bigl( 1 + \alpha \frac{V_{2}}{V_{1}} \Bigr) + x_{2} \ln(1 - \alpha) \Bigr)$ describes the contribution of the pressure equilibration to the entropy.


Calculating n factor of SO2 I'm having a hard time trying to figure out n factor of $\ce{SO2}$ in the reaction

$$\ce{FeS2 + O2 → Fe2O3 + SO2}\qquad \text{(unbalanced)}$$

To balance it, I saw that in 1 mole of $\ce{FeS2}$, it looses a net total of 11 moles of electrons, while $\ce{O2}$ gains 4 moles of electrons per mole of it. That means they will have to react in a $4:11$ ratio, from this, the equation balances out to

$$\ce{4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2}$$

The question was (wrt to this equation), what will be the equivalent weight of $\ce{SO2}$, if the molecular weight of $\ce{SO2}$ is $M$?

This means I will need the $n$ factor of it to divide by $M$,

I tried to reverse this reaction

$$\ce{2 Fe2O3 + 8 SO2 → 4 FeS2 + 11 O2}$$

But the problem is that now it's hard to tell from which $\ce{O}$ atoms (of the now-products), the oxidation state is getting changed to $\pm0$ (zero). It's now confusing me.

The answer is given as $M/5$. I don't know why, according to the solution they only concidered the oxidation number change of $\ce S$ while ignoring the others, I don't think it's right.

As this is clearly a homework question, I will only point you in the right direction.

Balance the oxidation reaction:

$$\ce{S2^2- -> 2 SO2 + $n$e^-}$$

Balancing the charges will give you $n$. (I'll leave the balancing to the OP, as this is homework).

The equivalent mass of $\ce{SO2}$ in this reaction is then:

$$N=\frac{2M}{n}$$

The factor $2$ is needed because the electronically balanced reaction has two molecules of sulphur dioxide in the equation.


Why is it possible to form bromoalkanes but not iodoalkanes from free radical substitution? For a free radical substitution (FRS) reaction between iodine $\ce{I2}$ with ethane and bromine $\ce{Br2}$ with ethane $\ce{CH3CH3}$:

$E_\mathrm d(\ce{H-I})=299\ \mathrm{kJ\ mol^{-1}}$

$E_\mathrm d(\ce{H-Br})=366\ \mathrm{kJ\ mol^{-1}}$

$E_\mathrm d(\ce{C-H})=410\ \mathrm{kJ\ mol^{-1}}$

I understand that the reaction for iodoethane to form from FRS is energetically unfavourable, since the energy released from the formation of a $\ce{H-I}$ bond (from reaction between an iodine radical and an ethane molecule) is only $299\ \mathrm{kJ\ mol^{-1}}$, which is insufficient to compensate the energy needed to break the $\ce{C-H}$ bond to even form the first ethyl radical in the first place for further propagation of the reaction.

However, it is to my knowledge that bromoethane can form from FRS, despite the formation of a $\ce{H-Br}$ bond releasing only $366\ \mathrm{kJ\ mol^{-1}}$ of energy, which is still lower than the $410\ \mathrm{kJ\ mol^{-1}}$ of energy needed to break the $\ce{C-H}$ bond. This doesn't really make sense if the justification for iodoethane is correct.

I would appreciate if anyone could explain to me the reason why bromoethane can form from FRS despite being apparently energetically unfavourable too, thanks!

Mechanistic Summary

Initiation: Involves homolytic cleavage of the $\ce{X-X}$ bond to generate $\ce{X^.}$ radicals. Initiation is endothermic, and the energy requirement is usually met by thermal treatment or photoirradiation.
Propagation I: Abstraction of a $\ce{H}$ atom by the $\ce{X^.}$ radical, to produce a $\ce{R-H2C^.}$ radical. This transformation usually takes place via a $[\ce{R-H2C...H...X}]^\ddagger$ transition state. Also, this process is endothermic
Propagation II: This stage involves the abstraction of $\ce{X}$ atom from $\ce{X-X}$ by the the $\ce{\ce{R-C^.}}$ to prduce $\ce{R-H2C-X}$. This process occusrs via a transition state that looks so something like this: $[\ce{R-H2C^....X...X}]^\ddagger$. This process, if it is sufficiently exothermic, compensates for the previous endothermic step and the reaction proceeds.

Termination: Involves radical-radical recombination.

Now let us look at the energetics of the reaction. The first propagation step is exothermic for fluorine and this explains its extreme reactivity. For the subsequent halogens, it becomes increasingly endothermic from chlorine to iodine. This trend can be explained on the basis of decreasing bond strengths of $\ce{H-X}$ bond. (Data given in the table above)

Now let me compare to extremes, fluorination of methane and iodination of methane. As one can observe, the first propagation step for fluorination is not only exothermic, but it also has a small activation barrier.

Thus in the transition state, $[\ce{H3C...H...F}]^\ddagger$, the $\ce{H-CH3}$ distance is only slightly greater than that in $\ce{CH4}$, and $\ce{H-CH3}$ bond is weaker than the $\ce{H-F}$ bond. Thus only a slight shift of the hydrogen atom towards the fluorine radical is sufficient to break the carbon-hydrogen bond and form the fluorine-hydrogen bond. This kind of a transition state is called an early transition state, and is typical of fast,exothermic processes.

In contrast, iodination displays a late transition state which is typical of slow,endothermic processes.The activation barrier for the first propagation step is relatively high, in fact it is almost the same as the endothermicity of the reaction. The transition state is not reached till the the carbon-hydrogen bond is almost fully broken, and the hydrogen-iodine bond is fully formed.

These rules about transition states are called Hammond's Postulates, and I encourage you to read about them.

Now, indeed the reason why iodination doesn't proceed is that even though final step is exothermic it doesn't release sufficient energy to compensate for the extremely endothermic first step. Thus, the net reaction is endothermic.

However, it is not the exothermicity of $\ce{H-X}$ bond formation that drives the halogenation reaction, rather it is the energy released upon $\ce{-C-X}$ bond formation (in the second propagation step) that ensures reaction spontaneity. (this can be seen in the data presented in the table)

This is why bromination, or for that matter chlorination can take place.

My apologies, the question asked about iodethane and bromoethane, however, the examples I took up was that of iodomethane (not the first time I have misread a question in my rush to post an answer). However, the ideas presented here still apply

References: Data table, and reaction profile taken from Organic Chemistry: Structure and Function by Vollhardt&Schore.


Oxidation state of fluorine in HOF What should be the oxidation state of $\ce{F}$ in $\ce{HOF}$.

As fluorine is the most electronegative element in the periodic table, it should be $-1$. But when I googled it, I found that many sources say it is $+1$ based on the fact that it is an oxidising agent.

Is there any theoretical way in which we can predict its oxidation no. or at least explain this anomaly.

There is no anomaly in here. Fluorine’s oxidation state in $\ce{HOF}$ is $\mathrm{-I}$ as the theory says. Hydrogen’s is $\mathrm{+I}$. This leaves oxygen with an oxidation state of $\mathrm{\pm 0}$.

That final fact is where the oxidative power comes from. Oxygen, the second-most electronegative element in the periodic table, has the same oxidation state as in molecular form. It is also bonded to an even more electronegative fluorine. Even in hydrogen peroxide oxygen would me more reduced. Thus, any reaction of $\ce{HOF}$ involves oxygen gaining electrons, not fluorine.

Remember that many sources on the internet can be wrong.


Reaction of cyclobutylmethanol with HX I am interested in the reaction of cyclobutylmethanol with HX where these are the only reagents.

My question is whether the reaction proceeds via SN1 or SN2. My reference text insists that primary alcohols are eliminated by SN2 even if the protonated alcohol is a strong leaving group. So if the reaction were SN1, there would be ring expansion.

What if I conducted this reaction in the presence of a Lewis acid like ZnCl2?

This is actually a trickier question than first inspection.

Your statement about primary alcohol substitution is more or less correct. Unless you have a very good reason to form the primary cation, you won't.

The problem in this example is that cyclobutylmethyl cation is not a primary cation even though if you were to draw it out, you might naively think that. Cyclobutylmethyl cation belongs to a special class of cations that are considered non-classical in the sense that the positive charge is not heavily localized to one atom, for example like methyl cation or tert-butyl cation. This essentially means that the positive charge is spread out over several atoms, and this increases the relative stability of the cation, making it more likely to form.

The more famous example is a 2-norbonyl cation, which is covered in many introductory classes.

For cyclobutylmethyl, the literature I found (hard to link) seems to suggest that the contribution from cyclopentyl cation is much larger than the cyclobutylmethyl cation, so we might consider this to be more neighboring group participation (from the C-C bond) than a raw non-classical ion. This is likely due to the fact that the ring expanded version is so much better in terms of strain.

Either way, a dissociated cation is stabilized, so I would think that this mechanism is more $S_{\mathrm{N}}1$-like.


Mechanism of Dehydrogenation Reaction In the hydrogenation of alkenes over a metal catalyst, the two hydrogen atoms are added with syn stereochemistry.

In the metal-catalysed dehydrogenation of alkanes, is the removal of hydrogen likewise concerted with syn stereochemistry? And if so, why?

The answer you seek can be found here uwyo.edu/roddick/site/alkane_dehydrogenation.html , the catalytic cycle of dehydrogenation of alkanes by Iridium complexes does not follow a simple syn addition such as in the hydrogenation of alkenes. In the hydrogenation of alkenes, the $\ce {H2}$ dissociates and is added syn to the substrate in a well understood mechanism.

In the dehydrogenation though, the coordination of the alkane to the Ir metal center occur first, then a migratory insertion forming a 5-coordinated trigonal bipyramidal complex, then a β-elimination occur forming an unstable 6-coordinated complex which readily loses the alkene (final product). The catalyst is then regenerated by a hydrogen acceptor which can be a bulky alkene (so its product can't participate in the cycle due to steric hindrance). If no regio inducing group is present in the catalyst, the formation of the Saytzeff product (most substituted alkene) is favored.


I was told that the more exothermic the solution, the more soluble the salt. Is this statement wrong? I was also told that $$\Delta H(\text{solution}) = \Delta H(\text{hydration}) - \Delta H(\text{lattice E})$$

However, for $\Delta H(\text{solution})$ to be exothermic, it has to have a negative value <0. Which means lattice energy is going to have to be greater than $\Delta H(\text{solution})$ to produce a negative value, based on the equation above.

Yet I am also told that hydration energy released needs to be enough to compensate for the lattice energy required to break down the solid. So it can't be lesser than $\Delta H(\text{lattice E})$.

Isn't this contradictory? I am confused.

I have read a similar question at   Enthalpy of solution
but the answer given is that dissolving is dependent on Gibb's free energy instead.
Does that mean that I should disregard the statement "the more exothermic the solution, the more soluble the salt."?

Does that mean that I should disregard the statement "the more exothermic the solution, the more soluble the salt."?

No, based on the Gibbs free energy equation $$\Delta G = \Delta H -T \Delta S$$ and the condition that $\Delta G<0$ for spontaneous dissolution, if you assume $\Delta S$ is constant then the statement is true.

The point of that other post is that as a general rule you should not simply ignore the $\Delta S$ term since sometimes it is the decisive factor determining solubility (such as in the case of endothermic dissolution reaction).


Find the order of acidic strength Arrange the following in decreasing order of acidity

I think it's S>P>R>Q

(Q) forms an anti aromatic conjugate base. So its the least acidic compound here.

(R) forms forms a conjugate base which is only stabilized by electron withdrawing (inductive) effect of $\ce{S}$.

(P) forms an aromatic conjugate base and (S) forms am ion stabilised by resonance (two resonance structures). But negative charge in $\ce{O}$ is more stable than negative charge on $\ce{C}$. So (S)>(P).

Is this correct?

(Q) is the least acidic compound because the conjugate base formed is anti aromatic and hence unstable.

(R) is more acidic than (Q) because the conjugate base is stabilized by the electron withdrawing inductive effect of two sulphur atoms.

Conjugate bases of both (P) and (S) are stable due to resonance. Among the two, conjugate base of (S) is more stable because a negative charge on Oxygen atom is more stable than a negative charge on carbon atom. Here are the resonance structures:

This is verified by @Ivan Neretin's comment on the OP.

  Let's look at some data. pKa of cyclopentadiene is about 16, while that of acetylacetone is about 9.


carbon dioxide in liquid and gaseous phase with liquid water What are the differences between carbon dioxide solubility (miscibility, reaction) in liquid water when in gaseous ($\pu{<38 bar}$) and liquid ($>\pu{38 bar}$) phase? The temperature in my experiments is usually between $0$ and $\pu{5\!^\circ C}$. Is there more phase - interface for the reaction between water and $\ce{CO2}$ in the gaseous phase?

According to Henry's law for real gases (including a Poynting correction), the solubility should rise with the fugacity of the vapor. How is it however, when the carbon dioxide is liquid? When observing a pressure drop in a closed stirred vessel with liquid water and $\ce{CO2}$ on top of it, the pressure drop is much larger for gaseous $\ce{CO2}$ than for the liquid $\ce{CO2}$, this does not however imply a larger solubility? In my understanding it just means, that the liquid $\ce{CO2}$ is not that compressible and will mix in a different manner than gas and liquid water, resulting in a lower pressure drop.

At a given temperature a mixture of gaseous $\ce{CO2}$ and liquid $\ce{CO2}$ will have a particular pressure. Thus more liquid $\ce{CO2}$ doesn't mean that more $\ce{CO2}$ will dissolve in the water.

Obviously liquid $\ce{CO2}$ has a lot more matter per unit volume than gaseous $\ce{CO2}$. So a little liquid $\ce{CO2}$ makes a lot of $\ce{CO2}$ gas.


What is the amount of nitrate ions in 20 g of Fe(NO3)3? What is the amount of nitrate ions in $20.0\:\mathrm{g}$ of $\ce{Fe(NO3)3}$?

The chemical formula for the nitrate ion is $\ce{NO3}$, I think.

The molar mass of $\ce{Fe(NO3)3}$ is $242~\mathrm{g~mol^{-1}}$ of which $186\ \mathrm{g\ mol^{-1}}$ belong to $\ce{NO3}$. That means that around $76.85\:\%$ of the substance is nitrate ion.

$20.0\ \mathrm g$ of the substance are equivalent to $20.0\ \mathrm g/242\ \mathrm{g\ mol^{-1}} = 0.0826~\mathrm{mol}$ of $\ce{Fe(NO3)3}$.

Since roughly $76.85\:\%$ are nitrate ions, there are about $0.06~\mathrm{mol}$ of nitrate ions in those $20\ \mathrm g$ of substance.

One mole is $6.02\times 10^{23}$, so if I multiply $(0.06\ \mathrm{mol})(6.02\times 10^{23}\ \mathrm{mol^{-1}}) = 3.612\times10^{22}$.

My Answer: There are $3.612\times10^{22}$ nitrate ions in $20.0\:\mathrm{g}$ of $\ce{Fe(NO3)3}$.

However, that is wrong. The options in the website are:

$1.49\times 10^{23}$
$4.98\times 10^{21}$
$60.0$
$8.25\times10^{21}$

And my answer is not even close to any of them.

What did I do wrong?

The molar mass of iron (III) nitrate is $m(\ce{Fe(NO3)3}\approx 242~\mathrm{g\, mol^{-1}}$. Pay close attention to the unit.

You correctly calculated the amount of substance of iron (III) nitrate to be $n(\ce{Fe(NO3)3}= 0.0826~\mathrm{mol}$

Now you should ask yourself the question: How many nitrate ions are in one formula unit of iron (III) nitrate?

   There are three $\ce{NO3^-}$ per every $\ce{Fe(NO3)3}$.

What does that mean for the number of moles of nitrate ions?

   It means that $n(\ce{Fe(NO3)3} = \frac13\cdot n(\ce{NO3^-})$, so you have to multiply the number of moles by three. $n(\ce{NO3^-}) = 0.248~\mathrm{mol}$

Now you know the number of moles of nitrate ions and you simply have to multiply with Avogadro's constant.

   $N(\ce{NO3^-}) = n(\ce{NO3^-}) \cdot L = 1.49\cdot10^{23}$


Stability of cis vs trans isomers? Recently, while reading about isomers I learned that cis isomers are more stable than trans isomers.

I referred many books as well as searched around the internet but I didn't find a suitable explanation.

Could someone please explain this to me?

As far as I am aware, and from a quick browse though some textbooks, the trans isomer is generally more stable than the cis isomer. This is due to the reduced steric hindrance of the substituents in the trans configuration versus the cis configuration.
For example trans-but-2-ene is more stable the cis-but-2-ene because there is less steric interference between the two methyl groups either side of the double bond. However, there are exceptions to this rule which I couldn't find a quick explanation for. The rule applies to both cyclic and acyclic compounds although there are some obvious cyclic exceptions such as cyclic alkenes, where the cis isomer is more stable, especially for smaller rings, due to the large ring strain caused by the orientation of the trans isomer. This page has a good illustration of this:


Determine heat flux from temperature profile The heating
  element and the insulator are of equal thickness L. Heat transfer in
  the air film adjacent to the heater is assumed negligible.

I've noticed that I find these type of problem the hardest to understand, so before I ask my question, does anyone know a good video or website which explains in detail how you should interpret temperature/concentration profiles?

My answer to (a) would be that both of the points should have the same heat flux. I know that one of the relationships used for layered wall is that q=constant, meaning that it is the same through every layer, however is that the correct way of thinking for this problem? I don't think so and don't understand how I should interpret the profile.

Another way I thought about it was that the heat flux for the layers should be:

$$q_1=(k_1/L)\cdot(T_\text{rear} - T_{12})$$
$$q_3=(k_3/L)\cdot(T_{23} - T_\text{front})$$

and that:

$$k_3>k_1$$

$$(T_\text{rear} - T_{12})\gt(T_{23} - T_\text{front})$$

If I just for simplicity assume that $L=1\ \mathrm m$. Then I would get that

$$q_1=(\text{something small})\cdot(\text{something big})$$
$$q_3=(\text{something big})\cdot(\text{something small})$$

Hence another reason why I think that the heat fluxes are the same.

Your first line of thinking to part a) is correct. For conductive heat transfer through rectangular layers, the heat flux needs to be constant and equivalent through all layers for steady state, or you would have a 'build up' of heat at between one layer. This would locally increase the temperature, and we wouldn't be at steady state.

For b) your second approach to a) is useful. You know the relative sizes of the temperature gradients from the sketch, and know from a) that the heat flux must be the same for each slab. Combine these two and you will find a requirement on the relative magnitudes of the conductivities.


Equilibrium constant Kp for vaporization of water? For the reaction H2O(l) -->H2O(g) it is given that Kp = p(H₂O(g))/p° where p° is the atmospheric pressure. Why is it not just equal to p(H₂O(g))?

Recall the origin of the equilibrium constant.  It comes from the fact that the chemical potential of an ideal gas can be written in the form u(T,p0) + RTln(p/p0), with p0 some reference pressure.  What this lets you do is measure the equilibrium composition at a reference pressure, then calculate the equilibrium composition at a slightly different pressure with a simple formula, without having to know the general dependence of chemical potential on pressure and temperature.  Very handy!  But the existence of the reference pressure is essential.  It goes into the numerical value of the equilibrium constant.  For example,  for just two reactants with stoichiometric coefficients c1 and -c2, i.e. the equilibrium is c1 R1 = c2 R2, you have:

dG = 0 = c1 (u1 + RTln(p1/p0)) - c2 (u2 + RTln(p2/p0))

-(c1 u1 - c2 u2)/RT = ln((p1/p0)^c1 / (p2/p0)^c2)

exp(-(c1 u1 - c2 u2)/RT) = (p1/p0)^c1 / (p2/p0)^c2

The thing on the left is just a function of T and the reference pressure p0, and we set it equal to the measured equilibrium constant K(T,p0), which of course we get by measuring the equilibrium composition at T and p0.  The thing on the right you'll then recognize as the usual expression for Kp.

K(T,p0) = (p1/p0)^c1 / (p2/p0)^c2

With this equation in hand (and the value of K) we can calculate the equilibrium composition {p1,p2} at some temperature and pressure not too
far away from the reference T and p0, using the usual freshman chemistry tricks.

Now you can certainly choose a different reference pressure if you like, it doesn't have to be 1atm, that's just convention.  If you chose a different pressure, K would be different, but the expression for Kp would be the same, only you'd be dividing by the new reference pressure.  But you have to have some reference pressure for this method to work.


Acid promoted Enol ring formation Mechanism I know this has to be an enol mechanism but which carbon gets attacked by the $\pi$ electrons?

This reaction proceeds in a few steps. In acidic media, the upper carbonyl oxygen is proton. This results, in greater electron deficiency of the upper carbonyl carbon, which strongly attracts nucleophilic attack. Due to strong activation by the methoxy substituent, the arene's $\pi$ electrons can attack the electron-deficient carbon in the carbonyl group. This nucleophilic addition to the carbonyl group gives a tetrahedral intermediate with a positive charge delocalised in the neighbouring ring.

Note that two types of reactions are happening here: Electrophilic aromatic substitution at the position para to the methoxy substituent and also, a nucleophilic addition to the upper carbonyl group.

A chloride ion now comes in to remove the proton from the para position of the arene. We then get back the aromatic ring with the methoxy group. In the next phase of the reaction, we now want to remove the $\ce {-OH}$ that we have created on the previous carbonyl carbon. This would entail us first protonating it, to make it a good leaving group. How do we create the double bond then?

Well... It is now important to note that $\alpha$ hydrogens in carbonyl compounds are relatively easily lost due to the electron-withdrawing effect of the carbonyl oxygen. Furthermore, the negative charge generated from this loss of the proton can also be delocalised into the the neighbouring aromatic ring on the right of the figure, enhancing the stability of this conjugate base, increasing the acidity of this hydrogen. What happens now is simply elimination to form the double bond. The protonated $\ce {-OH}$ first leaves as water to generate a tertiary carbocation. Then, a chloride ion acts as a base to remove the acidic $\alpha$ hydrogen, with the electrons going into forming a $\pi$ bond between the two carbons. This is essentially an $\ce {E 1}$ process. The desired product is now obtained.


Work done in an isothermal irreversible process In calculating the work done in an isothermal irreversible process, why do we consider the external pressure only (using formula: $W_{\text{ext}} = P_{\text{ext}} \mathrm{d}V)$ but not the internal pressure? When the system expands, the internal pressure does the positive work, but even in this scenario, we use the equation as mentioned above. Why is this happening?

The key word here is irreversible. Irreversible processes are more like sudden jerks or a quick change in the system and therefore the ${P_{\text{ext}}}$ is not same as the ${P_{\text{int}}}$ throughout the process.

  Why do we NOT use the internal pressure (${P_{\text{int}}}$) to find the work done?
  It is very difficult to find the ${P_{\text{int}}}$ exerted by the gas in the container since ${P_{\text{int}}}$ is varying with time. For this, we would have to use integration. Something like this:$${W_{\text{int}} = P(V)_{\text{int}}\mathrm{d}V}$$
  (where $P(V)_{\text{int}}$ is a function of $V$)
  But we know the value of ${P_{\text{ext}}}$ and this is constant throughout the process. This is why we use ${P_{\text{ext}}}$ instead of troubling ourselves with integration. Something like this: $$W_{ext}= P_{\text{ext}} \mathrm{d}V$$

So if a system undergoes a decrement in volume suddenly, it is the external pressure that does the positive work throughout this decrement. The equation would be: $$W_{\text{ext}}= P_{\text{ext}}\mathrm{d}V$$ (However we can use ${P_{int}}$ as well, please read the NOTE at the bottom)

And if a system undergoes an increment in volume suddenly, ${P_{\text{int}}}$ is doing the positive work. However, to find the $W_{int}$ we would have to use integration since ${P_{\text{int}}}$ is varying with time. It is much easier if we just find the negative work done by ${P_{\text{ext}}}$ and since we know that $${W_{\text{int}} = - W_{\text{ext}}}$$ we can find the work done. This is why we consider the ${P_{\text{ext}}}$ and according to the situation in the question, we apply the signs.

NOTE: Magnitude of work done by the internal pressure and external pressure is always same. We only use ${P_{\text{ext}}}$ because it is constant whereas ${P_{\text{int}}}$ varies with time. If we use integration to find ${W_{\text{int}} = P(V)_{\text{int}}\mathrm{d}V}$, where $P(V)_{\text{int}}$ is a function of $V$, we will get the same magnitude as we would from $W_{\text{ext}}= P_{\text{ext}} \mathrm{d}V$. Only signs are opposite.


Is perchloride the same as perchlorate ion? I read about perchloride in a talk and after trying to search for its chemical formula, the search results imply that perchloride is actually named perchlorate. Is this correct?

Short answer: Yes, sometimes "perchlorate" can be called "perchloride" in older literature, but because of the classical definition of a "perchloride", it's so confusing and technically wrong that the $\ce{ClO_4^-}$ anion should be called either "perchlorate" or "chlorate(VII)".

In older naming systems, pre-IUPAC, a "per-ide" is quite simply "any substance containing an unusually high proportion of the named element". So, a "perchloride" is any substance with more chlorine than is "normal" for that mixture of elements. This name was typically given to metal halide salts where the metal has multiple possible oxidation states; the highest oxidation state of the metal forms the metal "per-X-ide". for instance, titanium perchloride aka titanium tetrachloride, $\ce{TiCl_4}$ (IUPAC name titanium(IV) chloride), as differentiated from the trichloride and dichloride salts of titanium, and iron perchloride aka ferric chloride (that name's rarely used as iron only has two stable chlorides, and so the traditional "ferric" and "ferrous" descriptors work fine).

However, the perchlorate anion isn't actually a "perchloride" by that definition; it's more a "peroxide", as there's only one chlorine (which is relatively "normal") that has been "superoxidized" by four oxygens (not normal).

This use of "perchloride" as a synonym for "perchlorate" is more the result of a bad simplification of the traditional naming system, where salts of "-ic" acids are "-ides". While the rule follows for the hydrohalic acids (HF, HCl, HBr, HI) and most oxoacids (which, when they end in "-ous", form "-ites", as in hypofluorous acid, chlorous acid and sulfurous acid), the highly oxygenated oxoacids like chloric, perchloric, sulfuric, nitric, phosphoric and boric acids form "-ates", not "-ides", to mirror the low-oxygen "-ous" acids and their "-ite" anions/salts.

When not talking about the perchlorate anion, you sometimes hear about organic molecules that have had all their hydrogens replaced by halogens. Perchlorocarbons and perfluorocarbons are the chlorine and fluorine derivatives of hydrocarbons (for instance, the dry-cleaning chemical perchloroethylene, "Perc", and the perfluoroethylene monomer of polytetrafluoroethylene, aka Teflon).

"Peroxide" has found a home in the IUPAC system as the official name of the $\ce{(O_2)^{2-}}$ anion, composed of two single-bonded oxygens (often as the related "hydroperoxide", which is an OOH functional group bound to a carbon or to hydrogen). However, because each chlorine anion in a "perchloride" is typically bound to whatever the chlorine is oxidizing, the multiple chlorides are not a single anion and so aren't named as such in IUPAC.


How to calculate mole ratio of two volatile compounds in distillation? I have two volatile compounds (such as ethyl acetate and ethyl propionate). I am going to separate these compounds by simple and fractional distillation.

How do I calculate mole ratios of the two compounds and determine the boiling point temperatures of the first drop based on the liquid and vapor phase diagram?

If you have the vapor-liquid phase diagram, then it is easy to determine the vapor composition above a boiling liquid mixture (which should approximate the composition of your first drop in a simple distillation).

I'll assume you diagram looks like the one below, which is from the Wikipedia article on vapor-liquid equilibrium(and is mostly at the physical chemistry level):

Let's say you have a mixture of components 1 and 2 that is 40 mole percent (mole fraction = 0.4) component one (approximately the blue line labeled liquid on the diagram). Draw a line vertically from that composition to the bubble point (or liquid) line. Then draw a horizontal isotherm at that point (red line on the diagram above). Where that isotherm crosses the dew point (or vapor) line, you can read down to get the composition of the vapor above that boiling liquid (blue line labeled vapor), about 15% Component 1 (and 85% Component 2). This represents a single vaporization-condensation cycle, which is what happens in a simple distillation.

In a fractional distillation, you can have multiple vaporization-condensation cycles (called theoretical plates). If you know how many theoretical plates your distillation column will have, then you repeat the process for the simple distillation several times. For example, if there are two theoretical plates, you have two vaporization-condensation cycles. After the first one, start over with a new liquid composition equal to the vapor composition after the first cycle. Draw a line up to the liquid line, draw the isotherm to the vapor line, and draw a line back down to composition. If you use the example I have been working with on the diagram I have shown, you should get about 5% Component 1. Repeat as necessary.

If you do not know how many theoretical plates you have in your fractionating column, then you can only make an educated guess.


How to get or understand a sample in chemistry without knowing its chemical name or compound or formula How to get or understand a sample in chemistry without knowing its chemical name or compound or formula? I was thinking about whether there is a way to understand what the chemical compound sample that one might have. He/She doesn't know what its formula or its chemical properties or anything .He/She just got that one and was thinking what was that chemical and was interested to know about its structure. Would it be possible?
I am totally new at Stack Exchange so I doesn't know is this the proper way to ask.
And I am 15 year old so I doesn't know if these are taken in my upcoming classes but when I googled it or searched in many websites,I was not able find answers,So I am asking here.I have seen that we can guess the compound using many experiments but so also if the compound has been not found yet in human history would we be able to understand it?
Edit: This question had been closed stating it need to be more focused. I just wanted to clarify that what I wanted to know was whether there is a specific procedure to identify the chemical structure of the compound from just a sample alone.

Learning the unknown sample is called analysis and is a subject of analytical chemistry.
But your question is probably not about this.
You mean "can i guess something without knowing chemistry at all?"
And answer is yes!
Look at the food salt.
Do you know it without any knowledge of chemistry? Yes, you do!
But how much?
You know what it looks like,
you know its taste,
you know that it's soluble in water...
and probably that's all.
Now suppose you have a sample of something not found yet in human history.
You can easily know it similar to knowing food salt. Just check:
How it looks like?
What it's taste?
Drop it in water - is it soluble?
Now you "know" this substance just like you "know" food salt, without any chemistry!
But I guess you already suspect that there is something wrong with this knowledge.
It's very incomplete.
Chemist can learn more about food salt or your magic unknown sample.
Most of that serious knowledge require understanding of sample structure
at microscopic levels, and using theories that connect the structure to properties.
That was all about "knowing the substance".
Another question is knowing what a substance is, which is rather philosophy than chemistry.
Does a substance exist? What is existance? Does a substance have hidden fundamental purpose? Was it "created" or "appeared by itself"?
These questions are probably more important than chemistry, and you may be the one who one day will guess some answers here.


Why are discharge tubes filled at reduced pressure? Why are the gas sample tubes of hydrogen filled at a reduce pressure when observing hydrogen's emission spectrum? In my HS chemistry lab, we looked at gas sample tubes of hydrogen through a spectrometer and I am wondering why they were filled at a reduced pressure?

Not only does it take a high starting voltage and a current-limiting device (e.g. inductor) to maintain a continuous discharge at higher pressures, but the quality of the light emitted is changed by atomic (and ionic) interactions, tending to broaden lines into a quasi-continuous spectrum.

A good example is the sodium-vapor lamp, often used for street lighting. In the example cited, compare the line spectrum of a low pressure lamp with that of a high pressure one (though there is also a contribution from other gases in the second image) .

The low-pressure hydrogen glow discharge corresponds well with calculated values and is a useful tool in showing how theoretical calculation of orbitals can be translated into an observable spectrum. Calculation of high-pressure phenomena is much more difficult.


How would you find the mass in grams, if you only know the number of particles of a portion of the formula? If you were given a question like below:

  What is the mass in grams of a sample of $\ce{Fe2(SO4)3}$ that
  contains $3.59 \times 10^{23}$ sulfate ions, $\ce{SO4^{2−}}$ ? The molar mass
  of $\ce{Fe2(SO4)3}$ is $399.91 \, \text{g}/\text{mol}$.

How would you go about solving a question like that?

I'm confused about how to use only the quantity of sulfate ions to find the mass in grams of the whole sample, given it's grams per mole.

The formula $\ce{Fe2(SO4)3}$ tells us that 3 molecules of $\ce{SO4^{2-}}$ exist for every $\ce{Fe2(SO4)3}$ molecule.

$\frac{\ce{1 Fe2(SO4)3 molecule}}{\ce{3 SO4^{2-} molecule}} * 3.59*10^{23} \ce{SO4^{2-} molecule}$ can be used to find the number of molecules of ferric sulfate.

You can then use $\frac{1 \ce{mole}}{6.022*10^{23} \ce{molecule}}$ to find the moles of ferric sulfate.

You can then use the given information to solve for grams of ferric sulfate.

Write a comment if there's any ambiguity.


Why does benzene bend in this reaction? From here

I can see why hapticity changes here, but not why it is necessary for the benzene ring to bend. The resonance energy of benzene is pretty large, and there's no balancing increase in stability elsewhere--unless the decrease of hapticity increases stability, which I doubt (more delocalization=more stability, not vice versa). Also, the concept of a "bent benzene" is a bit strange to me.

I'd like to know what exactly is going on in this reaction and why it happens.

@F'x mentioned that the bent benzene may just be a way of drawing the molecule to clearly show the change of overlap. If this is the case, I'd still like to know how the hapticity changed (because my initial logic behind change of hapticity is invalid if the structure doesn't actually bend).

That diagram is misleading because it’s technically in reference to bis(hexamethylbenzene)ruthenium(0). This is a famous example by the late and great E.O. Fischer. This Ru(0) complex is 18e, three methyl resonances were observed for the bent ring at −10 °C. Crystallography later confirmed this. This structure is surprising to the new student, perhaps, because the classic 18e bis(benzene)chromium is sandwiched between two planar arenes.

This is not an organic chemistry question. It is deep in the realm of organometallic chemistry, with a proper explanation requiring a molecular orbital energy diagram and then one has to invoke a Jahn Teller instability argument that requires lowering symmetry from $D_{6d}$ or $D_{6h}$. Suffice it to say, stability is increased by changing the hapticity.


What are the steps to integrate the common rate law to find the integrated rate law for any order I cannot find any sources that actually list the steps for integrating a rate law, and it’s been driving me crazy. My Ap chem 2 teacher doesn’t know how to do it and just wants me to memorize the common rate laws, however I don’t memorize equations unless I know how to write them out and where they come from. I have the equation

$$ \text{rate} = \dfrac{\mathrm d[A]}{\mathrm dt} = k \cdot [A_0]^{n} $$

where $n$ is the order of the equation
$[A_0]$ is the concentration at time $0$
$[A]$ is the concentration with respect to time
$k$ is the rate constant

I know that for a first order ($n=1$) reaction the equation becomes

$$ [A] = [A_0] \cdot \mathrm e^{kt} $$

and that a second order ($n=2$) reaction the equation is

$$ [A] = \dfrac{[A_0]}{1 + kt \cdot [A_0]} $$

But I don't know what the steps are to find these equations. If I find that the order of my equation needs to be 1.5 ($n=1.5$), I will have no Idea what the equation will be.

From these equations I will know most of the time $[A_0]$, $[A]$ and sometimes Rate.

Edit:

After following the links I learned a portion of the process to solve the equation for any $n$. However after testing it out I am either getting a incorrect answer of a incomplete answer

$$ [A] =\int_{[A_0]}^{[A]}{\frac{1}{[A_0]^2}\,\mathrm d[A]} = \frac{[A_0]^{1}}{1} = \int_{0}^{t}{k\,\mathrm dt}$$

the answer should be

$$ [A] = \dfrac{[A_0]}{1 + kt \cdot [A_0]} $$

what parts am I missing?

The integrated rate laws can be derived by equating two expressions for the rate of the reaction. This gives us a first order differential equation which can be manipulated to get the integrated rate law. The steps that you need to follow are actually quite simple and structured. Here is an example for deriving the first order integrated rate law:

Step 1: Equate the two expressions for the rate of the reaction

For first order reactions, the two expressions for the rate of the reaction are: $$ r = -\frac{\mathrm{d}[A]}{\mathrm{d}t}$$ and $$r = k[A]$$ Equating them we get: $$-\frac{\mathrm{d}[A]}{\mathrm{d}t} = k[A]$$

Step 2: Rearrange so that all the same variables are on the same side to form an first order differential equation

So we rearrange this expression to get all of the same variables on the same side (note that k is a constant, not a variable, so it doesn't matter which side it is on) to get: $$\frac{\mathrm{d}[A]}{[A]} = -k\mathrm{d}t$$

Step 3: Integrate both sides

Now we integrate both sides: $$\int_{[A_0]}^{[A]}{\frac{\mathrm{d}[A]}{[A]}} = -\int_{t_i}^{t_f}k\mathrm{d}t$$
$$\mathrm{ln}[A] - \mathrm{ln}[A_0] = -kt$$

Step 4: Using some algebraic manipulation, rearrange expression to get an equation for [A]
$$\mathrm{ln}\frac{[A]}{[A_0]} = -kt$$
$$\frac{[A]}{[A_0]} = \mathrm{e}^{-kt}$$ Therefore we get as required:
$$[A] = [A_0]\cdot e^{-kt}$$

Using this guide, you should be able to derive the other integrated rate laws. If you have any trouble, feel free to ask.

Edit:

Since you have also mentioned that you are struggling with deriving the second order integrated rate law, I will show you how to derive it, using the steps that I have outlined above.

Step 1:

$$r = - \frac{\mathrm{d}[A]}{\mathrm{d}t} = k[A]^2$$

Step 2:

$$\frac{\mathrm{d}[A]}{[A]^2} = -k\mathrm{d}t$$

Step 3:

$$\int_{[A]_0}^{[A]}\frac{\mathrm{d}[A]}{[A]^2} = \int_{t_i}^{t_f}k\mathrm{d}t$$

$$\frac{-1}{[A]} + \frac{1}{[A]_0} = -kt$$

Step 4:

Okay now everything up to here has been exactly the same thing we did for deriving the first order integrated rate law. We have gotten an equation relating $[A], [A]_0$ and $t$. To make $[A]$ the subject, we multiply by $[A]$ and put all the terms containing $[A]$ on one side:
$$\frac{[A]}{[A]_0} + kt[A] = 1$$ Now we factorise $[A]$ out on the left hand side and divide both sides so that the left hand side equals to $[A]$:
$$[A]\left(\frac{1}{[A]_0} + kt\right) = 1$$
$$[A] = \frac{1}{\frac{1}{[A]_0} + kt}$$
Now we multiply the right hand side by $[A]_0$/$[A]_0$ and we get the required equation:
$$[A] = \frac{[A]_0}{1 + kt\dot~ [A]_0}$$


endothermic dissolution process Can anybody give me an example of an endothermic dissolution process, preferably one in which name of the substances involved are easy to remember. I have searched the Web thoroughly but could not find an example, only definitions. Thanks in advance.

An example of an endothermic dissolution is dissolving potassium iodide in water.

Added: Here is a table of Enthalpy of Solution for various compounds. Since this is the enthalpy of solution, some compounds will release energy when dissolve when the table indicates it would require energy to dissolve, and vice versa. If the number is negative, it is likely to be exothermic, and positive number are likely to be endothermic.


Deciding E1/E2/SN1/SN2 for cyanide ion The $K_\mathrm{a}$ for $\ce{HCN}$ is $6.17 \cdot 10^{-10}$ with a $\mathrm{p}K_\mathrm{a}$ of $9.21$

Thus, the cyanide ion is a strong base.

Also, the cyanide ion is a good nucleophile.

So in the reaction of alkyl halides with $\ce{KCN}$, a mixture of products must be formed depending on the solvent and alkyl group.

However my text suggests that the reaction  proceeds only via $\mathrm{S_N2}$.

Shouldn't the product formed depend on the alkyl group, say for a tertiary alkyl group, the reaction can go via $\mathrm{S_N1}$ mechanism for polar protic solvents and via $\mathrm{E2}$ for polar aprotic solvents and as the cyanide ion is strong base, the reaction is unlikely to go via $\mathrm{E1}$ and due to steric factors $\mathrm{S_N2}$ is also ruled out.

Also is there any  difference if we use $\ce{KCN}$ or alcoholic $\ce{KCN}$?

A general method for estimating $\mathrm{p}K_\mathrm{b}$ values is to use the equation $\mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} = 14$ where $\mathrm{p}K_\mathrm{b}$ is that of the conjugate acid; in this case, applying it gives us $\mathrm{p}K_\mathrm{b}(\ce{CN-}) \approx 4.79$. We can generally assume that ‘strong’ and ‘weak’ in $\mathrm{p}K_\mathrm{b}$ terms are similarly to be defined as for $\mathrm{p}K_\mathrm{a}$ values. Therefore, cyanide is a weak base and the premise of this question is incorrect. This observation is consistent with the fact that neutral cyanide solutions are relatively safe as long as no acid is introduced to protonate $\ce{CN-}$.

Note that the general statement ‘the conjugate base of a weak acid is a strong base’ is incorrect. The opposite is often true, namely:

  The conjugate base of a strong acid is generally a weak base.

And vice-versa if you swap all occurences of acid and base. The word generally is required here, since compounds such as $\ce{H2Cl+}$ are very strong acids whose conjugate bases are still very strong — but these compounds are not typically easily accessable and certainly not in aquaeous solution. A weak acid needs to have $\mathrm{p}K_\mathrm{a} > 15$ for the conjugate base to be a strong one but those are seldomly still considered ‘acidic’. An example would be water ($\mathrm{p}K_\mathrm{a} = 15.7$ therefore $\mathrm{p}K_\mathrm{b}(\ce{OH-}) \approx -1.7$) which is a weak acid with a strong conjugate base.

Since it is such a weak base, it has no access to $\mathrm{E2}$ pathways and $\mathrm{E1}$ pathways are disfavoured against $\mathrm{S_N1}$ if either is applicable. We don’t even need to consider $\mathrm{E1_{cb}}$. Thus, cyanide will generally react via an $\mathrm{S_N2}$ pathway unless the conditions for $\mathrm{S_N1}$ are much better.


questions about the synthesis of iodobenzene from aniline I am conducting an experiment about synthesis of iodobenzene from aniline, referring to this site(and various sites) (http://www.prepchem.com/synthesis-of-iodobenzene/). Below is my experimental process.

Put $\pu{9.5mL}$ of conc. $\ce{HCl}$, $\pu{9.5mL}$ of distilled water, $\pu{2mL}$ ($\pu{0.0215mol}$) of aniline in a $\pu{100mL}$ round flask with ice bath, and wait until the temperature of mixture drops below $\pu{5℃}$.
Dissolve $\pu{1.56g}$ ($\pu{0.0226mol}$) of $\ce{NaNO2}$ in $\pu{50mL}$ of distilled water, and cool it with ice bath.
Once the temperature of mixture dropped below $\pu{5℃}$, put the solution of $\ce{NaNO2}$. the solution must be poured very slowly not to raise temperature above $\pu{10℃}$.
Stir the mixture for $\pu{10 min}$.
Dissolve $\pu{ 3.58g}$ ($\pu{0.0216mol}$) of $\ce{KI}$ in distilled water. I used the water in very small quantities because $\ce{KI}$ is very soluble in water.
Put solution of $\ce{KI}$ to the mixture, then color of the mixture turns to be black, and gas bubbles (may be nitrogen gas) rise from the mixture.
After $\pu{1-2 h}$, gently heat the mixture (I set the temperature about $\pu{50-70 ℃}$).
Again after $\pu{1-2 h}$, it seemed that there are no more gas bubbles, so I stopped stirring and heating.
Layer separation occurred, (I guess) because iodobenzene is insoluble in water. so I moved it to separatory funnel and took the black oily organic layer presumed to be iodobenzene.
Here is the stage I am stuck on. according to the reference site, it says "use $\ce{NaOH}$ to decolorize the iodobenzene" (iodobenzene is colorless liquid). I put $\ce{NaOH}$ and made the pH of mixture 7, but its color is still black.

Question 1. Do I have to make the mixture basic to decolorize it? (actually, I tried making it basic, but it was not decolorised. Should I wait a moment after making it basic?)

Question 2. According to the reference site, iodobenzene have to be distilled using steam-distillation, but should I distill it with steam-distillation, not just ordinary distillation?

Question 3. Are there any problems with my experimental process?

This procedure looks basically OK. The key thing here is keeping the reaction mixture below 5C. It is good practice to cool your KI soln to 5C, and add it slowly so that the temperature of the reaction mixture stays below 10C until addition is complete. Using very concentrated KI soln may be an issue as the reaction is exothermic, better to have it more dilute and cold.

I would use an organic solvent (EtOAc or Et2O) at stage 9 to extract the product simply because you are doing this on a relatively small scale. On this scale chromatography is a better option than distillation for final purification.


Difference in application between Standard Type Bulb Pipette Filler and Universal Type? I want to ask what is the difference in terms of application between the HHH (Heinz Herenz Hamburg) Bulb Pipette Filler Standard Type (1081801) and the HHH Bulb Pipette Filler Universal Type (1081802)?

They both basically have same functions with same operations. However, universal type is longer than standard type with an adapter for pipette head (see the picture below: standard type is to your left).

Note: The bulbs shown are Deutsch-Neumann, but they are similar to HHH.


Rate law for A + B → C → P I need some clarification for the following assignment:

Derive the rate law for
$$\ce{A + B -> C -> P}$$
when $\ce{A + B -> C}$ is the slowest step and very slow.

My understanding is $\ce{A + B -> C}$ is the rate-determining step, so the rate law would just be
$$\mathrm{rate} = k[\ce{A}][\ce{B}].$$
I'm not sure if the question is asking about steady state approximation (SSA) or just a simple rate law. I know SSA happens if you assume the second step $(\ce{C -> P})$ is faster than $\ce{A + B -> C},$ meaning  $\ce{A + B -> C}$ is the slow step. I'm not so sure if my understanding of the question is correct.

The rate is determined by the slowest step, as you correctly assumed. The transition state of C->P is assumed to occur significantly faster than the first step in Organic Chemistry at least.
Unless you have an energy diagram for the reaction or any Le Chat, I think you have it right as rate = k[A][B].


What happens when acetone reacts with hydrogen cyanide? Which of the following is the major organic product when acetone is allowed to react with hydrogen cyanide in the presence of a catalytic amount of sodium cyanide?

The correct answer is (B).

However, I got this answer without using the catalytic sodium cyanide.  In my proposed mechanism, the dissociated proton from $\ce{HCN}$ activates the carbonyl, which facilitates the nucleophilic attack of $\ce{CN-}$.

What role does the $\ce{NaCN}$ play?

The acetone carbonyl is attacked by the CN$^-$ that comes from the NaCN.  This tetrahedral intermediate has a oxygen anion, which then grabs a proton from the HCN, regenerating the CN$^-$.


Hydrates In Aqueous Solutions Say you had copper(II) sulfate pentahydrate as a solid.

In an aqueous solution, is it still referred to by its hydrated form? Is it still called copper(II) sulfate pentahydrate (aqueous) or simply copper (II) sulfate (aqueous)?

I guess what I'm asking here is does the "hydrate" label apply exclusively to solids?

Yes, the ‘hydrate’ label applies exclusively to solids. Consider the following experiment: Dissolve both anhydrous $\ce{CuSO4}$ and $\ce{CuSO4 . 5 H2O}$ in water. You will get the same, blue solution.

What happened on a microscopic level is that the crystal lattice which once included the water atoms is broken up and the ions dissolve separately. What used to be copper(II) sulphate pentahydrate is now turned into a hexaaquacopper(II) complex.

$$\ce{CuSO4.5 H2O + H2O -> [Cu(H2O)6]^2+ + SO4^{2-}_{~(aq)}}$$


Bond order for carbonate ion for resonance I saw in a textbook that for carbonate ion, there are 3 resonance forms and the bond order is 1 and 1/3. So in general, how should we calculate the bond order for resonance structures?

Is there any definite way to do this?

There definitely is an easy way to do this.

One of the easiest examples that comes to mind is benzene.  If we look at the structure:

we see that each of the carbon bonds are either single bonds to the adjacent carbon or double bonds to the carbon in the other direction.  If we draw the other resonance structure these bonds shift by 1 atom to look like the following:

Thus, the overall structure can be obtained by looking at those two structure and averaging the bond order between them.  For benzene one resonance form for each bond will be single and the other double $(2 + 1)/2$. The "$2$" on the bottom of the equation is the number of resonance structure being considered.  Thus, for benzene the bond order is $1.5$.

Applying this same method to the carbonate ion, we have 3 resonance structures with bond orders of 2, 1, 1 when considering the bond between carbon and a single oxygen.  Since there are 3 structures the answer is $(2+1+1)/3 = 1.33333$ or $4/3$.


What are the major factors that influence LogP The partition coefficient is a very useful metric in characterizing compounds particularly drugs. What chemical properties influence that partitioning? Are chemical moieties additive in their LogP? What if the compound has both hydrophobic and hydrophilic ends?

Partition Coefficient, represented as P and defined as: P = [Organic phase] / [Aqueous phase]

Where "[]" represents concentration in the  phase and $\mathrm{\log (P)= \log_{10}(Partition Coefficient)}$ if log P value is higher then its more lipophilic ;those drugs were more absorbed by cells and elimination of those drugs from body will be less .The partition coefficient is a ratio of concentrations of un-ionized compound between the two solutions so for a chemical which is having both hydrophobic and hydrophilic ends also have log P values.

Eg: Oleic acid: $\mathrm{\log (P)= 7.64}$

chemicals which are hydrophopic(lipophilic) tends to have more Log P values and hydrophilic(lipophobic) chemicals have less log P.

eg: pentane($\mathrm{\log (P)= 3.45}$) [lipohilic], 1-pentanol ($\mathrm{\log (P)= 1.51}$) [lipophobic]


Is reverse osmosis a fancy term for filtration? In Reverse Osmosis (RO) systems, the water to be purified is pushed against a semipermeable membrane and comes cleaner the other side, the path of water is the reverse of regular osmosis, hence the name.
However, it seems to me that this process is pretty similar to a regular filtration.
So, is RO just a fancy term for a fancy filtration?
What is it that justifies it not being called filtration, ultrafiltration or some variation of that?

Ultrafiltration is ongoing at any pressure, even if slowly and liquid always goes to the side with lower pressure.
Reverse osmosis is ongoing only at pressure higher then osmotic pressure. At pressure lower than osmotic pressure, normal osmosis occurs and liquid flows to the side of higher pressure.


What is the substituent on this benzaldehyde? (MS attached) I have no clue what the substituent may be.

The molar mass of the unsubstituted benzaldehyde is 105 g/mol (subtracting one H, where the substituent will be). So the substituent is around 30–33 g/mol. Also I'm having trouble identifying the molecular ion. Is it the really small 138 peak or the 137 peak?

The molecular ion $\ce{[M]^{\bullet +}}$ peak is likely at 136 m/z. The peak at 137 m/z is due to the 1.1% natural abundance of $\ce{^{13}C}$.

The peak at 135 m/z is the $\ce{[M-1]^{+}}$ peak typical of aldehydes:

$$\ce{ [R-C(=O)-H]^{\bullet +} -> [R-C+=O <-> R-C\equiv O+] +H ^{\bullet} }$$

Another typical fragmentation of aldehydes is the loss of the CHO for $\ce{[M-29]^{+}}$, which you see at 107 m/z.

$$\ce{ [R-C(=O)-H]^{\bullet +} -> R+ + [HC=O] ^{\bullet} }$$

Since the main contributor to the $\ce{[M+1]^{\bullet +}}$ peak is $^{13}\ce{C}$ we can estimate the number of carbon atoms ($C$) in the molecule from the ratio of intensities of the $\ce{[M]^{\bullet +}}$ to the $\ce{[M +1]^{\bullet +}}$ peak:

$$\frac{I_{M+1}}{I_{M}}=\frac{1.1}{98.9}C$$
$$C=\frac{98.9I_{M+1}}{1.1I_{M}}=89.9\frac{I_{M+1}}{I_{M}}$$

The intensities appear to be $I_M \approx75$ and $I_{M+1} \approx 7$.
$$\therefore C=89.9\frac{7}{75}=8.4$$

Since the number of carbon atoms needs to be an integer, round down to $C=8$. We can now work backward to get a formula. Eight carbon atoms have a mass of $8*12=96$ g/mol, leaving 40 g/mol to be the rest of the formula. We know one is an oxygen (aldehyde) (16 g/mol), leaving 24 g/mol. The most number of hydrogen atoms that can be associated with eight carbon atoms is 18 (2n+2), so there has to be some other atom, maybe an oxygen or nitrogen?

I will leave rest to you:

Maybe there is a second oxygen?
How many hydrogens do you have?
What is the final formula?
How is it different from benzaldehyde?
How could those extra atoms be placed on the ring? (there are many possibilities)
Do you have any other evidence that suggests which of the possibilties from #5 is correct (IR, NMR, etc.)?


Why do most character tables include an $(x^2 + y^2)$ term? Why do most character tables (e.g. $C_{3h}$ but not $C_{2h}$) include an $(x^2 + y^2)$ term?
Is it an abbreviated form for $d_{z^2}$, applying only where $3d$ (and higher) orbitals might be involved, or just a generic binomial term with no specific relevance?
The only clue I can find is on this page where it lists the full expression for the angular wave function for $3d_{z^2}$ as
$$Y_{3d_{z^2}} = √(5/4){2z^2-(x^2 + y^2)}/r^2 × (1/4\pi)^{1/2} $$
where it also comments that "the $3d_{2z^2-x^2-y^2}$ orbital is abbreviated to $3d_{z^2}$ for simplicity".
Thoughts appreciated.

The functions on the right hand side of the table are the 'basis functions' for the irreducible representation, which are the rows of character, say $A_g$ or $B_g$ in C$_{2h}$. The basis functions have the same symmetry properties as the atomic orbitals with the same name (but other functions in $x, x^2, x^3$ etc and similarly for  $y,z$, could be used).
To determine where  an $xy$ function goes, look at a $d_{xy}$ orbital shape and subject it to each of the symmetry operations in turn,$E, C_{2}, i$ etc depending on point group.  (Assume z is the principal axis). Count +1 if the orbital (or function) is unchanged and -1 if not. So in C$_{2h}$, a $C_2$ operation leaves $xy$ unchanged as does $\sigma_h$ and $i$ so $xy$ is a 'basis' for A$_g$. The $xz$ transforms as B$_g$; a $C_2$ operation gives -1, as does $\sigma_h$, but $i$ produces +1.
(note in cases such as $z^2 -(x^2+y^2)/r^2$ the $x^2+y^2$ part is totally symmetric in the $xy$ plane and does not change anything so can be ignored; think of the donut around d$_{z^2}$.)


Can I somehow 'reuse' the chemicals in glow sticks? I particulary enjoy performing chemical experiments without buying any chemicals for them.

I was always sad about the fact that one glowstick lights out, you just throw it into trash can (I like all means of recycling). So I was wondering if I could (pick one or more):

Recycle the chemicals in it to make it glow again
Use the chemicals for other interesting (and maybe even useful) experiment

Here is a generalized version of what's going on chemically.  The key steps are dioxetane formation (the highly strained 4-membered ring) and its subsequent decomposition and energy transfer to a fluorescent dye, which then decays back to its ground state by emitting light.  By using different fluorescers, you can change the wavelength (color) of the light emitted.

Each company that manufactures glowsticks has their own patent(s) and proprietary process, hence the variation from product to product.

  So I was wondering if I could (pick one or more):
  1.Recycle the chemicals in it to make it glow again
  2.Use the chemicals for other interesting (and maybe even useful) experiment

1) Since the starting oxalate ester and concentrated hydrogen peroxide are consumed in the reaction, to make it glow again you would need to put in a new batch of both the oxalate ester and concentrated hydrogen peroxide.  You probably wouldn't want to open the glowstick because:

there might be some concentrated hydrogen peroxide remaining (if I were the manufacturer, I would want this cheap reagent present in a slight excess to make sure all of the ester is consumed)
many of the commonly used fluorescers are irritants, both externally and internally
phthalate esters are commonly used as solvents and they pose health risks

2) Even after all of the starting materials are used up, the fluorescer still remains in solution.  Exposing the glowstick to a black light will still cause it to fluoresce.


Reversible Sulfonation of Aniline? If I have aniline in acid and fuming $\ce{SO3}$, will it reversibly sufonate? I am worried that the nitrogen will be protonated and reduce the nucleophilicity of benzene ring to the point where sulfonation just isn't possible and instead you just get a salt? Or is it possible that you only get mono-sulfonation?

Here is my reaction scheme:

It sulfonates on N first and then rearranges on heating to give sulfonation in the 4 position. JACS 68 (1946) 969


How to calculate the parts-per-million concentration from a percent concentration? I am asked to find the concentration of magnesium ions in ppm for a $0.025\ \%$ solution of $\ce{MgCl2}$. I can use this formula:
$$\text{concentration in ppm} = \frac{\text{mass solute}_\mathrm{(g)}}{\text{mass solution}_\mathrm{(g)}}\times10^6$$

Assuming $100\ \mathrm g$ of solution, $2.5 \mathrm g$ solute/$\ce{MgCl2}$, and $97.5\ \mathrm g$ solvent, I get $\pu{95.205u}$ for $\ce{MgCl2}$, and then $0.026259\ \mathrm{mol}$ of $\ce{MgCl2}$, and thus $0.026259\ \mathrm{mol}$ of $\ce{Mg}$. That times $24.305\ \mathrm{g\ mol}^{-1}$ for $ \ce{Mg}$ gives me $0.63823\ \mathrm g$ of $\ce{Mg}$. $\frac{0.63823\ \mathrm g}{100\ \mathrm g}$ solution gives me $0.00638$, so that multiplied by $10^6$ would be $6382.28$. But the only options for the questions don't go over $500$. Like $64\ \mathrm{ppm}$ or $500\ \mathrm{ppm}$. Nothing even close to as high as I'm getting.

You're using $2.5\ \mathrm g$ of solute out of $100\ \mathrm g$ of solution, which is $2.5\ \%$, but the initial question is asking about a $0.025\ \%$ solution, which is why your answer is two orders of magnitude too large. You should be using $0.025\ \mathrm g$ of solute and $99.975\ \mathrm g$ of solvent.

As a general helpful hint for answering multiple choice questions, when you notice that one of the answers has the same coefficient but is off by a couple orders of magnitude (in this case the $64\ \mathrm{ppm}$ answer) go and double check all your initial concentrations and volumes, you probably forgot to account for a "$\%$" or a "$\mathrm k$" somewhere.


Why is the cis isomer of 1,3-dimethylcyclobutane more stable than the trans isomer? I expect there to be more steric repulsions between the two methyl groups if they are on the same face as each other. Why is this not the case?

The four carbon atoms in cyclobutane are not coplanar, as that would lead to a large degree of unfavourable eclipsing interactions. So, it is not a flat square; it adopts a puckered conformation. One of the carbon atoms makes a $25^\circ$ angle with the plane formed by the other three carbons, and this mitigates some of the eclipsing interactions, at the cost of a small increase in angle strain. The conformation is commonly called the "butterfly", and two equivalent puckered conformations interconvert rapidly. Here's a visual reference to help you

Pictured above is an unsubstituted cyclobutane.

Take a closer look at the figure on the left, and notice how there would be some transannular interaction between the two $\ce{H'}$ hydrogens at $\ce{C-1}$ and $\ce{C-3}$, marked in $\color{red}{\text{red}}$. This is bad. In the figure on the right, the $\ce{H'}$ hydrogens are no longer axial, but equatorial, and this is better.

In short, placing substituents in an equatorial position is better than placing them in an axial position.

Now, imagine what the cis and trans isomers would look like for 1,3-dimethylcyclobutane.

In the cis isomer both $\ce{H'}$ on $\ce{C-1}$ and $\ce{C-3}$ would be replaced by $\ce{-CH3}$ groups (since they are on the same side). Of course there will be considerable repulsions if they are arranged in the conformation on the left, which is why ring flip takes place and we will predominantly get the conformation on the right with both the methyl groups in equatorial positions. This is ideal.

The trans isomer would be like replacing one $\color{red}{\text{red}}$ $\ce{H'}$ and one $\color{blue}{\text{blue}}$ $\ce{H}$ at $\ce{C-1}$ and $\ce{C-3}$ with methyl groups. No matter what you do, you can only get one of the methyl groups at the equatorial position, but never both.


Do the compounds NaₓIrO₂ and NaOsO₂ exist? Based on some physics background, I want to know whether have scientists synthesized these two materials: $\ce{Na_{$x$}IrO2}$ and $\ce{NaOsO2}$ in the lab, if so, what properties do they have?

I will be very appreciative if someone could recommend me some relevant articles.

  Would you be willing to share your interest in these two materials? For example, is there a specific application in mind? If so, we might be able to suggest alternate materials should these two not exist.

I want to seek a material that has triangular lattice structure, strong spin-orbital coupling (e.g. 5d element) and large (or moderate) coulomb interaction between electrons. For example, $\ce{ABO2}$ type, $\ce{NaOsO2}$ compounds, where $\ce{A}$ and $\ce{B}$ are alkali and transition-metal ions, respectively. So far, I know $\ce{IrTe2}$ may be a one.
Do you mean that both $\ce{NaxIrO2}$ and $\ce{NaOsO2}$ do not exist?

  Is there any information on iridium and osmium oxides in these low oxidation states?

For example, $\ce{Na2IrO3}$ and $\ce{Li2IrO3}$. But these two belong to the honeycomb lattice structure, and what about the triangular lattice structure?

For $\ce{Na_{x}IrO2}$ the current answer seems to be "no" it has not been made but it is discussed in the literature theoretically.

From Kitaev-Heisenberg models for iridates on the triangular, hyperkagome, kagome, fcc, and pyrochlore lattices:

  However, no triangular lattice iridate with the relevant edge sharing octahedra structure is currently available; a compound of the
  type NaxIrO2 may or may not turn out to be stabilized.

and in Chiral spin liquid and topological superconductivity in the undoped and doped
triangular Kitaev-Heisenberg model:

  In addition, a class of ABO2 (where A and B are alkali and transition metal ions, respectively) type transition metal compounds and the possible material NaxIrO2 may be potential candidates for realizing the triangular Kitaev-Heisenberg model.

(But that's your paper isn't it? I don't have any information beyond what you already know, so really this is a useless answer.)


Do polyfluorosulfuric acids like H2SO2F4 exist? I was thinking about a compound resembling $\ce{H2SO4}$ but more acidic. I thought to replace the 2 oxygens with 4 fluorine atoms getting $\ce{H2SO2F4}$, which might be more acidic. Does such a compound even exist?

I also tried searching another compound i.e. $\ce{HOSF5}$ thinking of it to be more acidic than the previous but no results.

As the comments imply, no dice.  You can, of course, make a six-coordinate sulfur compound with the formula $\ce{SF6}$.  But trying it with a mixture of oxide, hydroxide and fluoride ligands instead of just fluoride presents the opportunity to evolve $\ce{HF}$ leaving the sulfur with a lower coordination number.  There is a five-coordinate compound $\ce{SOF4}$, but otherwise you should expect the sulfur to get down to four-coordination.  So the only stable protic acids you can get with one sulfur atom and oxide, hydroxide and fluoride ligands are plain old $\ce{H2SO4}$ and the more strongly acidic $\ce{HSO3F}$.


Is the strength of hydrogen bonding greater in hydrogen peroxide or water? When we compare water and hydrogen peroxide's hydrogen bond strength, which one should have the greater one? I know that peroxide has more hydrogen bonds, but when we compare the strength of each hydrogen bond, which one would be greater?

Online values of the boiling points indicate a greater bond strength in $\ce{H2O2}$. But peroxide has a larger number of hydrogen bonds.

This was a question for comparing bond strength in a test paper.

Online values of the boiling points indicate a greater bond strength in $\ce{H2O2}$.

The boiling point of water is $\pu{100 ^\circ C}$ while that of hydrogen peroxide is about $\pu{150 ^\circ C}$. Here is a list of factors that might contribute to this difference (adapted from source)

Stronger dispersion forces in $\ce{H2O2}$ (larger molar mass)
Higher dipole moment in $\ce{H2O2}$
More hydrogen bond acceptors in $\ce{H2O2}$

As for the number of hydrogen bonds, if we look at the structure of ice, each hydrogen atom of water is involved in a hydrogen bond (each oxygen is involved in two). If we assume that each hydrogen can make one bond at maximum, we would not expect more hydrogen bonds per molecule for  $\ce{H2O2}$. The crystal structure of solid  $\ce{H2O2}$ confirms that.

  But peroxide has a larger number of hydrogen bonds.

We can't infer that from the molecular structure alone. As stated above, in the solid state, the number of hydrogen bonds per molecule is the same. However, the density of solid $\ce{H2O2}$ (1.7 g/mL) is much higher than that of ice. This is true of the liquid forms as well (1.45 g/mL vs about 1 g/mL). This might allow for more dispersion forces (better contacts).

In liquid water, there are 3.4 hydrogen bonds per molecule on average (source). The higher density of $\ce{H2O2}$ compared to water suggests that atoms are closer in $\ce{H2O2}$, but you have to consider that the composition of the two substances are different. On a per-atom basis, the available volume is comparable.

  Is the strength of hydrogen bonding greater in hydrogen peroxide or water?

The question is a bit vague. If you compare one $\ce{H2O2}$:$\ce{H2O2}$ hydrogen bond to one water:water hydrogen bond, the hydrogen bond length is very similar (at least in the solid phase - source). Because the bond partners are the same atoms and the geometry around the oxygen is not that different in the two molecules, this would suggest that the strength of the bond is similar as well.


Are the canonical orbitals of Hartree-Fock also the natural orbitals? My question stems from the comments on my answer to After a unitary transformation, is Koopmans' theorem still valid?. There was some confusion relating to differing terminology referring to different, mostly unrelated density matrices. As of now, I'm uncertain as to what density matrix one diagonalizes to obtain the natural orbitals. Is it the density matrix that appears in the HF calculation or some different density matrix?

There is a bit of a terminology problem in the field that makes things very confusing and I will try to clarify some of this here. Part of the problem arises from the fact that sometimes only one kind of density matrix is relevant and so people drop the adjectives when talking about specific approximations.
The Density Matrix
We're talking electronic structure here (density matrices are used in other areas of quantum and physical chemistry. Honestly, it's the same density matrix but people use different language and representations and normalizations to discuss them and everyone gets confused when non-EST and EST people talk to each other about density matrices) so I will restrict myself to that topic.
The ($n$-particle) density matrix is given by:
$$\rho=|\Psi(x_1,x_2,\cdots x_n)\rangle\langle\Psi(x_1,x_2,\cdots x_n)|$$
where $x_i$ is the coordinates (space and spin) of the $i$th electron. That is, it gives the probability that the system is in a state with a given set of coordinates (space and spin) for all electrons in the system. The density matrix has many great properties, including being Hermitian and having the property that $\text{tr}{\rho A}=\langle A\rangle$ for any operator $A$. In other words, the $n$-particle density specifies everything about the wavefunction of your system. If you have the full wavefunction, you have the full $n$-particle density and vice versa.
Now, most operators are not $n$-electron operators. For example, the kinetic energy operator is a 1-electron operator (only the coordinates of a single electron are needed to calculate a kinetic energy), the standard electronic Hamiltonian is a 2-electron operator (because the Coulomb potential is only dependent on the positions of two electrons at a time). Hence, we don't need the full $n$-particle density matrix to get energies and properties.
The 1(or 2 or 3...)-particle (reduced) density matrix
Since we don't need all of those other coordinates we can integrate them out. Indeed, the trace operation that we described above would do just that. Say we want the expectation value of a 1-particle operator (like kinetic energy). We can define a 1-particle reduced density matrix (sometimes the word "particle" or "reduced" is omitted), often abbreviated as 1PDM, 1RDM, $P^{(1)}$, or $\gamma^{(1)}$ (also others).
$$P(x_1)=N\int \rho(x_1,x_2,\cdots x_n)dx_2\cdots dx_n$$
where $N$ is some normalization (commonly the number of electrons in the system ($n$) because, typically, people want the trace of the 1PDM to be the number of electrons in the system).
This is the matrix which is diagonalized to form the natural orbitals.
Relatedly, we can also define a 2-particle reduced density matrix (aka 2PDM, 2RDM, $\Gamma$, $P^{(2)}$, others) as
$$\Gamma(x_1,x_2)=N\int \rho(x_1,x_2,\cdots x_n)dx_3\cdots dx_n$$
where $N$ is a different normalization (usually, $n(n-1)$). Most operators that we care about (like the energy) are 2-particle operators so the problem of electronic structure is determining the 1 and 2PDMs.
The Hartree-Fock density
There are many ways to think about the Hartree-Fock approximation, but one way that is related to density matrices is that the HF approximation is that the 2-particle, 3-particle, etc. density matrices can be expressed as direct products of the 1PDM. Intuitively, since HF ignores electron correlation, it makes sense that the probability of finding electron 1 at position $x_1$ and electron 2 at position $x_2$ is related to just the product of the those individual probabilities (with appropriate fermionic antisymmetrization). In this way, the HF 1PDM specifies all density matrices in the HF approximation (which is why in HF you can compute $J$ and $K$ which are two-electron operators by contracting twice with the HF density).
$$P(x,x') = \sum_i^\text{occ} \phi_i(x)\phi_i^*(x')$$
where $\phi_i$ are the spin-orbitals. Note that this means for HF, the canonical orbitals are the natural orbitals. For a correlated method though, the 1PDM will generally be different. For example, CI methods like CAS express the 1PDM as the weighted sum of several HF-like 1PDMs, with coefficients determined by the CAS-CI optimization. Also, note that for Kohn-Sham DFT, the noninteracting reference system also looks like a system without electron correlation and so the 1PDM specifies terms that look like higher order density matrices. However, one does not obtain the 2PDM with KS DFT methods. One obtains the energy from the 1-particle density and an exchange-correlation functional although $i$-particle operators value can be extracted from this non-interacting $i$PDM.

That was a really long answer, but I sense there is a lot of confusion about these different density matrices, so I figured I would take the opportunity to give a very clear answer. In short, the answer to your question is that the HF (1-particle) density specifies all of the HF $i$-particle densities. Generally though, the $i$-particle densities are determined somewhat separately, as each are impacted by electron correlation differently. The orbitals that diagonalize the 1-particle density give the natural orbitals.


Does rate of auto-ionization of water change after adding a strong acid? We know that a reaction could have different rates at equilibrium. For example, lets take the reaction:
$$\ce{A + B -> C }$$
At equilibrium, the rates of formation and dissociation will be equal:
$$k_\mathrm{r}\mathrm{[A][B]} = k_\mathrm{c}\mathrm{[C]}$$
But that means at various concentrations of $\mathrm{[A],[B]}$ and $\mathrm{[C]}$, the actual velocity (rate)  of the reaction at equilibrium will be different.
In other words, for the above example: If I were to put additional [A] into the reaction after its reached equilibrium once, the new equilibrium it would reach would have a faster rate. Also, this means that a fold change increase in $\mathrm{[A]}$ doesn't require an equivalent fold change decrease in $\mathrm{[B]}$ to reach the new equilibrium.
Now, the autoionization of water is unique because it is independent of the concentration of water: $$K_\mathrm{w} = \ce{[H+][OH-]}$$
This means that a fold change increase in $\ce{[H+]}$  requires an equivalent fold change decrease in $\ce{[OH-]}$
Thus, I am guessing the rate would be equivalent to some constant: $$k_\mathrm{r}\ce{[H+][OH-]} = constant$$
Does the rate of auto-ionization of water change when you add a strong acid or base? An explanation at the level of molecules would be appreciated.

Ideal solutions
If we pretend for a moment that all the solutions under consideration are ideal, and that we have already reach equilibrium, we can consider the following reaction:
$$\ce{H2O(l) + H2O(l) <=> HO-(aq) + H3O+(aq)}$$
As Zhe said in the comments, the concentration of the reactants don't change much (even if you have $\pu{1 M}$ sodium hydroxide or $\pu{1 M}$ hydrochloric acid), so the forward rate at equilibrium is the same at any pH (again, assuming an ideal solution). That means the reverse rate is also the same at any pH. Because the product of the concentrations of hydroxide and hydronium ions is constant, the reverse rate constant does not depend on the pH.
Considering ionic strength and activity coefficients
The above argument does not work in a real solution with high concentrations of dissolved ions. If the pH is far from neutral, the concentration of ions (and counter-ions is high), and equilibrium concentrations (and with that the rates) will depend on the solutes present in solution.
Molecular level
At the molecular level, the auto-ionization of water is a special case because hydronium and hydroxide ions seem to diffuse faster than other ions (Grotthuss mechanism). At neutral pH, both species are rare. At acidic pH, there are more hydronium ions, but fewer hydroxide ions, so collisions are also rare (and reverse for basic). A high ionic strength would make it easier for ion pairs to form and separate, while a low ionic strength would be advantageous for the combining of hydronium and hydroxide.


Do I have to consider the water in the hydrate of a salt in a mass percent calculation? When calculating the mass percent of an element in a compound, I understand that you divide the molar mass of the element by the molar mass of a compound.

In a case such as $\ce{NH4MgPO4.6H2O}$ is the "$\ce{6H2O}$" part of the compound?
If asked to find the mass percent of Hydrogen in this, what number would you use.

As well, I don't fully understand what the $\ce{6H2O}$ in the molecular formula means? Does it just mean that it is aqueous?

Any clarification would be helpful!

As well, what would be the name of that compound? Would it just be Ammonium Magnesium Orthophosphate? Or would it be something else because of the $\ce{6H2O}$?

As stated in one of the comments to your question, the waters of hydration constitute part of the crystal structure itself, just as much so as the rest of the molecule. You can somewhat think of the waters as being functional groups of the compound. For example, the dehydrated form of a compound can have very different chemical and physical properties than the hydrated form.  The presence of waters of hydration in no way means that the compound is aqueous or dissolved in water.

Thus, when calculating the mass percentage of hydrogen of the hydrated compound, you do include the hydrogens from both the waters of hydration and the rest of the molecule.

The common name of  $\ce{NH4MgPO4⋅6H2O}$ is magnesium ammonium phosphate or struvite. According to this Wikipedia article:

  Struvite (magnesium ammonium phosphate) is a phosphate mineral with
  formula: $\ce{NH4MgPO4·6H2O}$. Struvite crystallizes in the
  orthorhombic system as white to yellowish or brownish-white pyramidal
  crystals or in platey mica-like forms. It is a soft mineral with Mohs
  hardness of $1.5$ to $2$ and has a low specific gravity of $1.7$. It is
  sparingly soluble in neutral and alkaline conditions, but readily
  soluble in acid. Struvite urinary stones and crystals form readily in
  the urine of animals and humans that are infected with
  ammonia-producing organisms.

Here is a photograph of struvite crystals from the same article: