Mass of chromium in ammonium dichromate What mass of chromium is in $\pu{100 mL}$ of $\pu{0.0400 mol/L}$ $\ce{(NH4)2Cr2O7}$?

The specific concept is how to find the mass of a certain ion/element in a substance, given the concentration. I am at a total loss to this.

At first I tried to find the molar mass of the chromium itself, which resulted in roughly $\pu{104 g/mol}$. I then tried to find the amount of moles in the concentration, which is

$$\pu{0.1 L} \cdot \pu{0.04 mol/L} = \pu{0.004 mol}$$

Then I multiplied the molar mass of chromium from the amount of moles in the concentration, and I got $\pu{0.208 g}$.

This is not the answer. The answer was $\pu{0.800 g}$

I then tried a different approach. I tried to find the molar mass of my original substance, then divided the molar mass of chromium from the molar mass of ammonium chromate, which will give me the amount of chromium per ammonium chromate. Then I thought all I needed to do was to multiply it by the concentration. This just messed up everything and I was back to square one.

How do I approach this problem?

To find the mass of chromium, one needs to know the amount $n$ ($m = nM$). Each formula unit of ammonium dichromate $\ce{(NH4)2Cr2O7}$ contains $2$ $\ce{Cr}$, therefore

$$n(\ce{Cr}) = 2 \cdot n(\ce{(NH4)2Cr2O7})$$

The amount of ammonium dichromate can be found from its concentration (as you correctly did):

$$n(\ce{(NH4)2Cr2O7}) = C(\ce{(NH4)2Cr2O7}) \cdot V$$

Putting it all together, the final expression for the mass of chromium is

\begin{align}
m(\ce{Cr}) &= 2 \cdot C(\ce{(NH4)2Cr2O7}) \cdot V \cdot M(\ce{Cr}) \\
&= 2 \cdot \pu{0.04 mol L-1} \cdot \pu{0.1 L} \cdot \pu{51.996 g mol-1} \\
&= \pu{0.416 g}
\end{align}

The answer deviates with the one provided by the textbook; however, if you were asked to find the mass of chromium(VI) oxide $\ce{CrO3}$ instead, it fits perfectly:

\begin{align}
m(\ce{CrO3}) &= 2 \cdot C(\ce{(NH4)2Cr2O7}) \cdot V \cdot M(\ce{CrO3}) \\
&= 2 \cdot \pu{0.04 mol L-1} \cdot \pu{0.1 L} \cdot \pu{99.99 g mol-1} \\
&= \pu{0.80 g}
\end{align}

I deduced $\ce{CrO3}$ by backtracking to the molecular weight of the compound, which should be around $\pu{100 g mol-1}$, and among common chromium substances this is the only one that comes to mind.


Can negative (abnormally) lattice enthalpies during the formation of ionic compounds exist? Why do compounds form bonds? I've got three words for you: Less potential energy. I don't remember a case of positive lattice enthalpies $\ldots$ - My comment

This makes me wonder. I remember seeing $\Delta H_{\mathrm{lattice}}$ of "common" ionic compounds like $\mathrm{NaCl}$ and considered them fairly positivea:
\begin{array}{|c|c|} \hline
\mathbf{Ionic~compound} & \mathbf{\Delta H^\circ (kJ~mol^{-1})} \\\hline
\ce{LiF} & 1022^\mathrm{\,b} \\\hline
\ce{NaCl} & 771 \\\hline
\ce{KBr} & 670 \\\hline
\ce{CsI} & 585 \\\hline
\end{array}

Yeah$\ldots$ so judging by the usual trends $\ce{CsI}$ should have the least positive lattice enthalpy in ionic compounds derived only from monoatomic ions. But that's still some lattice enthalpy!

So, could there be negative lattice enthalpies?

Notes:
a: Note that there are two major definitions of lattice enthalpy. Here, we're using the definition relevant to $\Delta H_r$ of the following endothermic reaction: $$\ce{MX(s) → M+(g) + X^- (g)}$$
While the other definition uses the enthalpy of the reverse process as the reference, only the sign might differ. Note that I had the other defintion in my mind when posting that comment.
b: All data obtained at $298~\mathrm{K}$. source

Short answer: Not for "normal" compounds.

First off, let's consider what the lattice enthalpy means. As you've written it, these lattice enthalpy is part of the Born-Haber cycle of formation of an ionic solid from a metal and halogen.

As you see from your table, the energies decrease with:

Small charges (i.e., these are higher for $\ce{Al2O3}$ or $\ce{CaO}$ because of higher ionic formal charges)
Large inter-atomic distances (i.e., increased ion distance between $\ce{Cs}$ and $\ce{I}$ yield lower electrostatic attraction).

The Born-Lande and Kaputinskii equations give some numeric estimates, e.g. (Born-Landé)

  $E = \frac{N_AMz^+z^- e^2 }{4 \pi \epsilon_0 r_0}\left(1-\frac{1}{n}\right)$

  where:

  $N_A$ = Avogadro constant;
  M  = Madelung constant
  $z^+$ = charge number of cation
  $z^−$ = charge number of anion
  e = charge on an electron (elementary charge)
  $\epsilon_0$ = permittivity of free space
  $r_0$ = distance to closest ion
  n  = Born exponent, typically a number between 5 and 12, determined experimentally or derived theoretically.

(Note that I've inverted the formula from Wikipedia's version to match your reaction.)

None of these can give you a negative lattice enthalpy.

How could it happen? Well, you'd need something where the solid is less stable than the gas-phase ions. Consider that in an ionic solid, the electrostatic attraction between anion and cation is always favorable. Since gas-phase ions aren't stable by themselves, I don't see how this is possible.


What is the mechanism for this nucleophilic conjugate addition? What is the mechanism for this reaction-

My working so far:

But my final product doesn't match with the one given. Where did I go wrong (if at all)? How should I have proceeded?

To figure out what's going on in an organic reaction, there are a few things you really should look out for:

what functional group(s) are present on the molecule(s);
what conditions the reaction is being run under;
which of the functional group(s) are most reactive under the given reaction conditions.

If you apply these steps to every stage of the reaction, then deducing the mechanism will be easier, and you also save yourself a lot of time by not having to memorise a bunch of mechanisms that you have been previously taught. If those mechanisms were taught by a competent lecturer, you should be able to apply the three steps above to see exactly why those reactions happen.

The solution (courtesy of @Jan)

So, without further ado:

  What functional groups are present on the molecule?

That's easy.

$\hspace{30 mm}$

  What are the conditions?

That's easy too. $\ce{EtO-}$ is strongly basic, but it can also act as a decent nucleophile.

  Which of the functional groups are most reactive in base?

Well, to begin with, the thiol is acidic, so it will be deprotonated easily. No question about that.

Now you have a couple of electrophiles - the two esters - as well as a couple of nucleophiles: the thiolate anion, as well as the ethoxide anion. A hard base such as ethoxide is not likely to react in a conjugate addition. Maybe it might react directly with the esters at the carbonyl carbon? But wait! Both esters are ethyl esters $\ce{RCO2Et}$. So, you can use ethoxide to attack those esters all you like, but you won't get anywhere doing it - you'll just get back the same product. In fact, that's exactly why we use ethoxide instead of a different base like methoxide, $\ce{MeO-}$: to prevent transesterification.

So, let's talk about the thiolate anion instead. It's pretty nucleophilic, and you also have an electrophilic α,β-unsaturated carbonyl, so these can react:

$\hspace{15 mm}$

Now, you have a nucleophilic enolate, as well as an electrophilic ester on the other side. They can immediately react with each other to form the product that you are given:

$\hspace{25 mm}$

This reaction has a name - it is called a Claisen condensation, but that's not important! The important part is that you realise that it is just the reaction of a nucleophile with an electrophile.

An alternative (my original answer)

There might actually be another possible product, though. The enolate that you formed above could pick up a proton from somewhere (either ethanol, or perhaps another thiol molecule) to give the conjugate addition product, which is exactly the same as the one you drew. However, your initial addition product had a positive charge, which is never a good idea in a basic medium. Under basic conditions, you should always deprotonate first and reprotonate at the end of any reaction.

If this were to be the case, you should probably go through the three steps again:

  Functional groups?

That's still pretty easy. You have a sulfide and two esters.

$\hspace{45 mm}$

  Conditions?

Still basic.

  Most reactive functional group?

Well, the sulfide is indeed nucleophilic and you could use it to attack either one of the electrophilic carbonyl groups. But there are a few things that don't really favour this:

Sulfur is a pretty soft nucleophile. It might love to react with $\ce{CH3I}$, but probably wouldn't like a carbonyl so much. In fact, this is the reason why the initial step consisted of conjugate addition instead of direct addition.
There's no more hydrogen on sulfur that can be taken off, so if sulfur does attack the carbonyl, it will be stuck with a positive charge. This makes the sulfur a pretty good leaving group, which means that any addition to the carbonyl group will be highly reversible.
If sulfur attacks the carbonyl group further away from it, it will form a four-membered ring. This is not particularly favourable. If it attacks the carbonyl group nearer to it, it will form a three-membered ring, which is even worse!

So, the sulfur is stuck where it is. You are better off thinking about the reactivity of the carbonyl group in base. You should know that they can be deprotonated to form enolates. Does it matter which carbonyl group is deprotonated? Of course it does. This molecule is not symmetric, so the two carbonyl groups are not equivalent. This obviously leads to the question of which carbonyl group should be deprotonated. Let's look at the possible anions formed from the deprotonation:

$\hspace{10 mm}$

The first of these two is more stable than the other, because the electronegative sulfur atom is capable of withdrawing electron density from the negatively charged carbon. (Actually, there are several theories as to why sulfur stabilises the carbanion, but I won't discuss that here. See this paper, for example.) So, it is that first conjugate base that will be preferentially formed.

At this stage, you have a good nucleophile (the anion) as well as a good electrophile (the carbonyl on the other side). So, what's left is to react them in the same Claisen condensation as above:

$\hspace{25 mm}$

There we go, side product.


Nomenclature of Halogen substituted Alcohol and longest chain For the following compound, the correct IUPAC Name is "4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol" but according to what I know, Chlorine is a substitutent and alcohol is a functional group. So including alcohol is necessary but not $\ce{Cl}$. And if we see the longest chain including alcohol then it would be a pentanol rather than a butanol. Why are we not naming it like that? Is it because the substitute $\ce{Cl}$ is necessary to include? Or what?

Your rationale is correct.

The proposed name “4-chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol” is unambiguous and describes the correct compound; however, it is not in accordance with IUPAC nomenclature.

The important simplified criteria for the choice of a principal chain in such compounds are:

greater number of suffixes (here: “-ol”)
longest chain
lower locants for suffixes
greater number of prefixes (e.g. “chloro-”)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

Therefore, the preferred IUPAC name for the compound that is given in the question is 3-(chloromethyl)-2-(propan-2-yl)pentan-1-ol.

The position of the chloro group or other substituent groups that are cited as prefixes is not relevant for the choice of a principal chain in this case. Only if there is still a choice, such substituent groups are taken into account according to P-45.2.


Is there a case when an atom of some element (e.g. carbon) does not have its expected number of protons (e.g. 6)? Page 12 of the Study Guide for Campbell Biology, 11th Edition has the following question:

  A carbon atom has 6 neutrons. How many protons are present in the nucleus of a carbon atom?

  A. 12
  B. 8
  C. 6
  D. Can't tell; more information is required.

The answer given on page 431 is D. I think the answer should be C since carbon has 6 protons.

What type of additional information might be required?

I understand that elements are defined by their number of protons, but maybe there is some strange edge case like a moment during nuclear fusion? Is there a case when a carbon atom does not have 6 protons?

A carbon atom has 6 neutrons. How many protons are present in the nucleus of a carbon atom?

Obviously this wording would make any astute student say the answer is 6 since carbon by definition has 6 protons. You are correct, elements are defined by their number of protons. Key word being defined, changing the number of protons changes the element.

I believe the key here that your book is getting at is you only know that you have an atom from a carbon sample that has 6 neutrons but nothing else. I think the question is trying to evaluate if you are aware that proton numbers are independent of neutron numbers thus knowing there are 6 neutrons does not help you determine protons, especially given one of the options is:

  D. Can't tell; more information is required.

If this is what they wanted then it was a very poor choice of words to say a carbon atom in the question then ask for number of protons. They should have said:

  Given an atom (or nucleus) with 6 neutrons, how many protons are present in the nucleus of the atom?


Unusual Result in Kinetics Lab The question: In the lab described below, why did the decolorization reaction described, which should be first order in each of the two reactants, go most slowly when the reactants were most concentrated?

As part of the introduction to our kinetics unit, I asked my students for factors that they thought would influence the amount of time it takes a reaction to occur.  Not surprisingly, they came up with a variety of things, including temperature and the concentration of reactants.  I set up a simple lab to compare temperature and concentration conditions based on the reaction of crystal violet with sodium hydroxide.  Later in the unit we will come back to the reaction and analyze its kinetics more carefully, but for the moment, I just wanted my students to do a more basic analysis.

The students did trials for 4 conditions, and each involved the same quantity of reactants: 25 ml of 0.10 mM crystal violet and 25 ml of 1.0M sodium hydroxide.  The reaction was performed by adding the crystal violet to a flask, adding water to dilute, adding the sodium hydroxide to the flask, stirring, and timing the reaction until it no longer showed visible purple.  (Yes, that's a sloppy endpoint, but it was sufficient for this activity's purposes.)

Trial 1: 25 ml crystal violet, 150 ml room temp distilled water, 25 ml sodium hydroxide (a 1:8 dilution of stock reactants)

Trial 2: 25 ml crystal violet, 150 ml hot distilled water, 25 ml sodium hydroxide

Trial 3: 25 ml crystal violet, 50 ml room temp distilled water, 25 ml sodium hydroxide (a 1:4 dilution of stock reactants)

Trial 4: 25 ml crystal violet, 0 ml distilled water, 25 ml sodium hydroxide (a 1:2 dilution of stock reactants)

The results:

The hot reaction went much faster than the cold.
The 1:4 dilution went faster than the 1:8 dilution.  (~6 minutes vs 9 minutes, respectively)
The 1:2 dilution went much, much more slowly than any other. (~16 minutes)

Result #3 is the anomaly.  It involved the most concentrated solutions and so should have been the fastest of the 3 room temperature reactions, but it was far and away the slowest.

Things I've considered:

The reaction isn't significantly endo- or exothermic, so the solution
didn't have a different temperature due to its minimal mass.  (All of
the reactions except for the one with hot water added were within a
couple of degrees of each other.)
The students didn't make a mistake in execution - this was observed by multiple groups over multiple trials.
The difference was not due to the sloppy endpoint of the reaction.  Though the reduced volume meant that a given amount of remaining crystal violet would be most purple in this trial, the difference in color was enormous and obvious.
The reverse reaction never becomes significant with such a massive excess of sodium hydroxide.
New idea: The unusual result is the only trial without distilled water - maybe the water is contaminated or something like that, but in a way that is promoting the reaction?  I was excited when I thought of this (I actually popped up out of a dead sleep with it in my head,) but it seems unlikely, I used the exact same container of water an hour earlier to make each of the stock solutions...

I've wracked my brain, but I can't figure out a reasonable explanation for what we observed.  Any suggestions would be welcome.

Your analysis of the expectation is correct, and we can look at it more quantitatively in this way:
Since the hydroxide concentration is much higher than the CV concentration in all of the reactions, let's approximate it as unchanging. Assuming the reactions are first-order in each reactant, we have a rate law for the concentrated reaction of:

$$-\frac{d[\ce{CV}]}{dt}=k[\ce{HO-}][\ce{CV}]\approx k[\ce{HO-}]_{initial}[\ce{CV}]$$

For simplicity, we'll define a $k'$ as $k'=k[\ce{HO-}]_{initial}$ and treat the reaction as pseudo-first-order.

The half-life of a first-order reaction is given by $t_{1/2}\approx \frac{0.7}{k}$, so for the concentrated reaction we have $t_{1/2}=\frac{0.7}{k'}$.

For the 1:4 reaction, the initial hydroxide concentration is halved, so we have a pseudo-first-order rate constant of $0.5k'$ and $t_{1/2}=\frac{0.7}{0.5k'}$ or 2 times the half-life of the concentrated reaction.

Likewise, the half-life of the 1:8 reaction is twice that of the 1:4 reaction and four times that of the 1:2 reaction, so it's $t_{1/2}=\frac{0.7}{0.25k'}$.

Let's say that the concentrated reaction needs 6 half-lives for the concentration to get to the point where the color disappears. Since the 1:4 reaction starts at half the initial CV concentration, it only needs 5 half-lives. But the half-lives are each twice as long, so it should take a time equivalent to 10 half-lives of the concentrated reaction, so 10/6=1.7 times as long.

For the 1:8 reaction, the same analysis says that it will take 4 half-lives, but each is 4 times as long as that of the 1:2 reaction, so it takes 16 equivalent half lives, meaning 16/6=2.7 times as long.

The only instance in which the more dilute reaction will be faster is if the concentrated reaction requires less than two half-lives to be observed to be complete. That can't be the case here or there would be no color at the start of the 1:8 reaction.

So we must look for an alternative explanation for your observations. 0.5 M NaOH is a very concentrated solution, and it is possible that other reactions are occurring besides the intended one. To test that, I would follow Maurice's suggestion and run the reaction at a wider range of base concentrations. It would also be preferable to have the same concentration of CV in every reaction, so only one variable is involved. To do this, used a fixed volume of (NaOH + water) in each reaction.

UPDATE: This paper1 reports doing this reaction at much lower hydroxide ion concentrations and also notes that the reaction rate decreases with increasing ionic strength, which could be at least a partial explanation for why your observed rate at high concentrations is lower than expected.

1 Felix, L.D. Kinetic Study of the Discoloration of Crystal Violet Dye in Sodium Hydroxide Medium. J Chem Appl Chem Eng Vol: 2 Issue: 1


Baking Powder instead of soda error Bit of a random one but I had an oven fire yesterday due to grease.  I decided to clean my over as a result but stupidly used baking POWDER rather than soda and wiped it off with vinegar.

Now I am really worried as I have a gas oven, and some of the baking powder has seeped through the holes at the bottom where the naked flame.  Do you have any idea whether this is dangerous?  Will the baking powder ignite or set fire? Or will it just burn off? I'm very worried about what will happen to the oven now.

Baking powder is baking soda ($\ce{NaHCO3}$) plus a mild acid salt or dry acid, such as $\ce{Ca(H2PO4)2}$ or tartaric acid. These ingredients should pose no hazard in a stove unless there is remaining powder blocking a vent hole or gas nozzle. In that case, just use a paper clip or other tool to open the vent.

Tartaric acid is theoretically flammable, but the amount you might have left is so small as to be innocuous. You've probably cleaned the stove about as well as baking soda would, once you remove all obvious deposits of the white powder. Bon appétit!


How much ozone is produced by ionization of air and how turn ozone into oxygen I want to make a plasma speaker. I'm worried about the amount of ozone being produced by the ionization of the air and it possibly not being a safe amount. Even if it doesn't produce a dangerous amount, I don't like the smell of ozone. I'd like this all to be continuous. I am very unfamiliar with the chemistry here but I have read that UV light, around 250nm, can cause $\ce{O_3 \to O_2 + O}$ and also $\ce{O_2 \to 2O}$

But also without any help it seems that $\ce{2O \to O_2}$ and also $\ce{O + O_2 \to O_3}$. So it seems that if I shined UV light of the right frequency on the spark I'd make a lot of $\ce{O}$ and perhaps that would cause more $\ce{O_2}$ than $\ce{O_3}$ but I'm not really sure.

I was also thinking that it might oxidize some sort of metal filter really well and I could just keep replacing those but I wasn't sure. I could also heat the filter slightly to speed up the reaction if necessary.

Even if the UV light idea doesn't work out very well I still want a way to do this. Is the amount of ozone produced by a continuous electrical current though air harmful? Is there a way of getting rid of the ozone to make some odorless non-harmful chemical?

Even if the UV light idea doesn't work out very well I still want a way to do this

With some photochemical background, I suggest to forget about running such a UV lamp

Remember that your plasma will be formed in air. For every molecule of ozone formed, there are much more harm- and odourless $\ce{O2}$ molecules around. If you cleave $\ce{O3}$

\[\ce{O3 + O2->[\mathrm{h}\nu][\lambda\ \lt\ \mathrm{300\ nm}] [O2 + O* + O2] -> O2 + O3}\]

the resulting oxygen atom will be captured by the next $\ce{O2}$ to form ozone again.

The overall process will not reduce the number of ozone molecules, unless $\ce{O*}$ reacts with $\ce{N2}$, which might lead to some other unpleasant species ;)

Note that these processes are fortunately effective in the ozone to protect us from short wave UV radiation, which brings us to the next aspect of the setup.

In order to photolyze ozone, you'll neeed a light source that emits UV at $\lambda\lt300\,\mathrm{nm}$. It's not that these light sources would not be available, but this UV range is not that of a tanning booth lamp! Remember that the ozone layer protects our DNA from running havoc! The least you can expect from short-wave irradtion is the sun burn of your life! Consequently, you will have to shield the whole setup properly!

In summary, I absolutely like the idea of building a plasma speaker yourself and there even seem to be some DIY kits around, but I suggest to forget about the UV part, which can be much more dangerous than the ozone formed.

UPDATE

Burak Ulgut brought up a very interesting point. Back in the days, plasma speakers using other gases (or gas mixtures) were tried, and even went commerical.

This site in German describes a helium driven plasma speaker system :)


How to calculate probability of finding an electron at a point? In school i learned how to calculate probability of finding electrons in some volume but how can we calculate the probability of finding a electron at a particular point.

Point here has negligible volume.

Consider the 1D case for simplicity.

The modulus square of the wavefunction is interpreted as a probability density in the sense that
$$
|\psi(x)|^2\mathrm{d}x
$$
gives you the probability of finding the electron in within $x$ and $x+\mathrm{d}x$. Now if you want to consider the probability to find the electron exactly in $x$ you have to take the limit $\mathrm{d}x\to 0$ and this gives you zero probability. Formally, this corresponds to compute
$$
P(x=x_0) = \int_{x_0}^{x_0} |\psi(x)|^2\mathrm{d}x = 0,
$$
which is zero since the limits of the integral are the same. Therefore there is $0$ probability to find an electron at exactly one point.


How exactly is activation energy defined? In a common interpretation of the Arrhenius rate equation

$$k = A\exp\left(-\frac{E_\mathrm a}{RT}\right),$$

the activation energy $E_\mathrm a$ is understood to represent the difference in the energy of the reactants and the minimum energy required to form the activated complex or the transition state, as shown in the following diagram

But this must be wrong! As $E_\mathrm a$ is the minimum energy itself which the molecules must possess to reach the activated complex. That is, it is not the difference but the minimum energy level itself. This is the only way in which it will remain constant i.e. independent of temperature (otherwise with increase in $T$, $E_\mathrm a$ will go down) and can be used in the Arrhenius Equation where the exponential term denotes the fraction of molecules having energy higher than or equal to $E_\mathrm a$ according to Maxwell-Boltzmann distribution. Sometimes, threshold energy ($E_\mathrm t$) is defined as the minimum energy required for reaction to proceed and $E_\mathrm a = E_\mathrm t-\mathrm{KE_{average}}$. If this definition is true, then threshold energy and not activation energy should be in the Arrhenius equation. I therefore need to clarify the actual definition of activation energy $E_\mathrm a$.

You are right that the energy of the initial state is a function of temperature. But in practice for this equation the energy difference $E_\mathrm a$ usually is fixed for an arbitrary temperature (in the middle of the diapason of experimental measurements) and assumed temperature-independent.
From Wikipedia:

While this equation suggests that the activation energy is dependent
on temperature, in regimes in which the Arrhenius equation is valid
this is cancelled by the temperature dependence of $k$. Thus, $E_\mathrm a$ can be
evaluated from the reaction rate coefficient at any temperature
(within the validity of the Arrhenius equation).

As to the physical meaning of activation energy, citing Wikipedia:

Both the Arrhenius activation energy and the rate constant $k$ are
experimentally determined, and represent macroscopic reaction-specific
parameters that are not simply related to threshold energies and the
success of individual collisions at the molecular level.

In other words, activation energy $E_\mathrm a$ is a semi-empirical quantity and is not considered to have a strictly defined physical meaning (at least on molecular level). Note that the Arrhenius equation
$$k=A\exp\left(-\frac{E_\mathrm a}{RT}\right)$$
is itself considered to be a semi-empirical relationship.
In the IUPAC Gold Book (DOI: 10.1351/goldbook.A00102), activation energy is also defined to be an empirically determined quantity, in line with the above:

An empirical parameter characterizing the exponential temperature dependence of the rate coefficient $k$:
$$E_\mathrm a = RT^2 \frac{\mathrm d (\ln k)}{\mathrm dT}$$


Precipitation of water-insoluble materials from shower or laundry greywater I am currently working on a project to build a filter that aims to separate or salt out the 'dirt' in shower or laundry greywater. The ideal result from this chemical reaction will be such that the 'dirt' will be in a solid-state (precipitate). Meanwhile, the residual water will be 'clean'. The degree of cleanliness of 'clean' I mentioned is such that this residual water is clean enough to be used for non-potable purposes such as mopping the floor.

Therefore, I would like to ask your help in figuring out how can I turn the soap micelles into precipitate, leaving out only 'clean water'?

Disclaimer: I am ignoring soap scum as I can filter that out easily.

Your ambitious project is not that trivial. Soaps are of several types.
a) If you are talking about salts of fatty acids, the classical soaps, then one can precipitate them out with calcium salts. Many other cations form insoluble salts with soaps. Now if you are trying to purify laundry water, dishwasher water, shampoos, liquid soaps, nobody can remove those surfactants from water by simple means. They (manufacturers) have specifically designed them so that they never precipitate out in water, otherwise your clothes and dishes would always have a layer of scum on them. Potassium ions may form ppts with certain surfactants like sodium dodecylsulfate but forget about ppt'ing out these typical surfactants from water used in ordinary household.

b) Alum can quickly help you in cleaning all the particulate matters in waste water, like magic.

c) Filtration on beach sand beds can remove small amounts of certain surfactants from water.

d) Filtration on activated carbon would be the best to remove surfactants.

Well it turned out to be more complicated than you desired. There is no single cure and certainly this water is not good for mopping floors because it will spread nothing but germs from using laundry water and shower waste. You can think of solar distillation as an alternative. That distilled water would very clean.

Note that in certain countries, household waste water including the sewage water after doing natural bacterial treatments is used for watering gardens and small farms. In one large pit, all wastewater is allowed to stand in open air, far away from residential areas. In the first ou all solids settle down, next pit further purifies via same settling and bacterial oxidation process, finally the supernatant water is transferred to third pit, which is now clean enough that gardens and small farms can be watered with it.  Of course, this water is rich in nitrogen and phosphates (from human and laundry waters) which is good for plants.


Would it be possible to create a (non-toxic) alcoholic drink that, when zapped with laser, changes color? Would it be possible to create a (non-toxic) alcoholic drink that, when zapped with a 0.5mW laser pointer, changes color?

This would be mainly for the visual effects (bartender show) so it should not require aiming the laser at the drink for more than a few seconds. The reason for choosing a 0.5mW laser is to minimize the need for eye protection.

It would not matter what color laser is used, but would this be possible?

If it is possible, what kind of chemical(s) would be needed in the drink?

I see a couple of problems here:

Beam diameter vs volume of the beverage

The volume of the sample that is actually hit by a narrow laser beam is very small, as compared to the total volume of the sample (= the drink). As a consequence, only a very small amount of the dye in drink will undergo a transformation. A wide beam of a switchable light source definitely is the better option.

Is is possible in general?

Yes, a number of compound classes undergo transformations upon iradiation and change their colour (spiropyranes, fulgides, etc.). Often, the reverse reaction can be induce thermally.

So what?

Food dyes are strictly regulated, you can't add a photochromic dye to a drink just because the colour change looks nice in the lab. If the dye isn't approved as a food dye, there's no legal way! To my knowledge, food dyes typically do not show photochromism but are photostable. Consumers usually don't like that their food changes the colour in the light.

In summary, the idea sounds funny but I rather doubt that it can be realized.

The only option to add a dramatic effect would be the use of fluorescent and food-compatible ingedients. Think in quinine (tonic water).


How to determine which compound will get oxidised and which will be reduced in a cross cannizzaro reaction? How to determine which compound will get oxidised and which will  be reduced in a cross cannizzaro reaction?

As a general rule of thumb, I have always been told that Formaldehyde is always oxidised, and the other is reduced.

But no general principle was ever given to me. What factors (eg electronic, steric, etc) affect this ??

There is this question already here, but the both the answers seems to be contradictory. @Jan says , the one with more hydrate forming tendancy oxidised , whereas
@RE60K says less Nucleophilic aldehyde oxidises (which I believe will have a lesser tendancy to form hydrates)

It correlates very strongly with ability to form a hydrate as @Jan's answer says, which is why formaldehyde is usually the sacrificial oxidised component. In general electron withdrawn aldehydes are more easily oxidised, electron rich aldehydes reduced.

This paper from J. Org. Chem. here compared several pairs of aldehydes and the conclusions were:

  m-nitrobenzaldehyde is (relatively) extensively oxidised as contrasted to p-anisaldehyde, which is much less extensively oxidised - i.e., much more extensively reduced.  On the bases of this and prior investigations, the five aldehydes studied can be arranged (in order of diminishing susceptibility to oxidation): m-nitrobenzaldehyde, furfural, p-bromobenzaldehyde, benzaldehyde, and p-anisaldehyde. With reference to benzaldehyde and the substituted benzaldehydes, this order correlates with expected combined resonance and inductive influences of the substituent groups on the formyl group, as these should contribute to ease of a) hydroxide ion attack and b) hydranion transfer


How to correctly mix solutions? Imagine you have two ingredients - 45 grams of 60% ethanol, and as much water as you wish. The only accurate tool you have is a scale, which measuring error is up to $\pm \pu{1gram}$. How would you create five different solutions of 1, 2, 3, 4, and 5% ethanol?

It can be a solution of % ethanol by volume or by weight, depending on which can be created more accurately with the tools at hand. It might be super easy to you, but as an electrical engineer the logic and math eludes me.

You should use the concentration dilution equation

$$C_1V_1 = C_2\cdot V_2$$
If you re-arrange it so the volume of 60% ethanol needed is:

$$V_1 = \frac{C_2\cdot V_2}{C_1}$$

You can then insert all the values you do know and there you go.
For example, if you need a $\pu{100 m}$ of 5% solution: $$V_1=\frac{5\% × \pu{ 100 ml}}{60\%}$$

Which means you need $\pu{8.33 ml}$ of 60% ethanol in $\pu{100 ml}$ solution.

As max pointed out, since you only have a scale, use d=m/v. D = density, m = mass and v= volume You re-arrange so mass = d.v


What is the major hydrolysis product of 2,2,2-trifluoroethyl 2-acetamido-3-phenylprop-2-enoate? In the below reaction we have to find the major hydrolysis product of 2,2,2-trifluoroethyl 2-acetamido-3-phenylprop-2-enoate.

My doubt is why the major hydrolysis is not when $\ce{OH-}$ at the $\ce{C}$ of $\ce{CF3}$ as it is more electropostive. However the major product is given by the following reaction, why?

Plain hydrolysis would just replace trifluorexthoxide with hydroxide.

But you might want to consider that your staring material in equilibrium with another tautomer:

Deprotonation of the $\ce{OH}$ in the right structure, followed by an intramolecular reaction (transesterification/lactonization) in which $\ce{CF3CH2O-}$ is kicked out should give rise to an oxazolone.


Side chain hydroxylation of pyrrole What reagents are required to form 1-[1,5-dimethyl-3-phenyl-4-(propan-2-yl)-1H-pyrrol-2-yl]propan-1-ol with 1,2-dimethyl-4-phenyl-3-(propan-2-yl)-5-propyl-1H-pyrrole as starting material?

Is it possible for the starting material to react with a halogen gas e.g. bromine to form a bromine on the carbon and then react with e.g. NaOH to form the OH with Br as the leaving group?

I'd like to propose a dark, non-radical pathway :)

Pyrroles, supposed that they are not fully substituted, undergo electrophilic aromatic substitutions with halogens.

For the given starting compound, a complete bromination isn't possible due to the lack of a proton that would have to be eliminated in the final step.

However, the addition of an electrophile is conceivable. Addition of "$\ce{Br+}$" at C-2 of the starting material, usually the prefered position in the reaction of unsubstituted pyrroles, is a dead end here. Addition at C-3 yields an $\alpha$-enamine cation intermediate from which a proton might be eliminated. The resulting species is both

a double enamine and
an allylic halide

This intermediate might rearrange upon elimination and addition of a nucleophile $\ce{Br-}$ in a way that the aromaticity of the pyrrole is reestablished and bromine is introduced in the side chain.

Edit 1

Supposed that you are mainly interested in the hydroxyalkyl-substituted pyrrole as the target molecule, you might want to consider an alternative strategy. Let's have a second look:

The target compound (1) has an $\ce{OH}$ group in a side chain. Alcohols may be prepared by different routes, such as

reduction of the corresponding ketone
reaction an aldehyde with carbon nucleophile

Let's focus on the second idea. This would leave us with the 2-formyl pyrrole (2), that could be transformed into (1) by a Grignard reaction with $\ce{EtMgBr}$.
Again, several methods are conceivable for the synthesis of the aldehyde (2), such as

selective reduction of a $\ce{-CO2R}$ substituent (or reduction to $\ce{-CH2OH}$ and subsequent oxidation to $\ce{-CHO}$)
direct introduction of a $\ce{-CHO}$ substituent at carbon 2 of  pyrrole (3)

The latter seems promising and you will probably realize that this an application of the Vilsmeier-Haack reaction, where the reactive intermediate, the $\ce{CHO}$ synthon, is prepared in situ from N,N-dimethylformamide and $\ce{POCl3}$.

This would leave you with a strategy for the synthesis of (3), but that's another question.


Why are bromine oxyanions uncommon? Sodium hypochlorite is used in bleach, calcium hypochlorite is used in public swimming pools, ammonium perchlorate is used in solid rocket fuel.

Sodium periodate is used in the cleaving of syn-vicinal diols, and is generally useful as an oxidizing agent.

Bromates are water contaminants, and perbromates weren't even successfully synthesized until 1968, but only through radioactive decay. Though alternate methods for their production exist today, what makes bromine oxyanions so unstable when chlorine and iodine oxyanions are widely used in chemistry?

I can only really speak for bromine(VII). Bromine(V) is pretty common and I'm not entirely sure what's the deal with Br(I) readily disproportionating to Br(V) + Br(-1). (The tendency of Br(I) to undergo disproportionation explains why we don't see it very commonly, but why it has this tendency I'm not so sure.)

This is an example of what is sometimes called the "alternation effect", or "d-block contraction" in p-block chemistry.

Due to the intervention of the ten 3d electrons, which are poorly shielding, the 4s and 4p electrons of the elements Ga through Kr are more tightly bound than one might expect. These elements show a distinct reluctance to take on their group valency. In $\ce{BrO4-}$, the higher-than-expected energy cost of ionisation to form Br(VII) is not fully compensated for by covalent bond formation (the strengths of which decrease linearly down the group). A selection of other observations that can, to various extents, be attributed to this are:

the anomalously high electronegativity of germanium (2.01), compared to silicon (1.90) and tin (1.96)
the anomalous instability of $\ce{AsCl5}$ (decomposes above $-50~\mathrm{^\circ C}$) even though $\ce{PCl5}$ and $\ce{SbCl5}$ are well-known,
the fact that nitric acid oxidises sulfur to $\ce{H2SO4}$, but selenium only to $\ce{H2SeO3}$

In a sense, it is exactly the same as the inert pair effect, which arises partly due to the lanthanide contraction - i.e. the intervention of the fourteen 4f electrons, which shield 5d, 6s and 6p electrons poorly. The difference is that the inert pair effect is even more pronounced because of significant relativistic stabilisation of the 6s electrons. Here's an insightful discussion on the topic: Why does tin form tin (II) compounds?


Why do all atomic masses have decimals if there are some elements that don't have isotopes? I understand that the way to calculate the atomic mass is to obtain a weighted average of the masses of all the isotopes of an element. I also understand that there are about 26 elements that are monoisotopic. If the reason the atomic masses have decimals is because of this weighted average, then it should follow that the monoisotopic elements should have a whole-number mass. Therefore, why do all the atomic masses have decimals?

The answer to your question has to do with Einstein's most famous formula
$$
E=mc^2
$$
which relates energy to mass, but what does it mean? Let's consider a simple example first. The simplest atom is of course the hydrogen atom which consists of one electron and of one proton. If you add the mass of the proton $1.672621898\times 10^{-27}$kg, to the mass of the electron $9.10938356\times 10^{-31}$ kg, you find a value of $1.67353284\times 10^{-27}$ kg. If you compare this to the mass of the $^1$H atom of $1.6735326915926\times 10^{-27}$ kg, you see that the hydrogen atom is lighter than the combined mass of the proton and the electron by $2.44017148\times 10^{-35}$ kg. What happened to the mass when the electron and proton got bounded together? Well, according to Einstein the difference in mass is converted to the binding energy of the electron! If you use Einsteins formula you find a binding energy of $-2.19311675\times 10^{-18}$ J or $-13.7$ eV which is very close to the Rydberg constant!

Let's now get back to other atoms. In the same way that the formation of a hydrogen atom from an electron and a proton lowers the mass of the atom by the corresponding binding energy, the binding energy between the neutrons, protons and electrons that make up heavier atoms also results in a reduction of the mass of the atom. Because the binding of the neutrons and protons is much stronger than the binding energy of the electrons to the core, the effect of the mass is also much larger. See for instance here for further details.


Decomposition of AsH3 In German problems olympiad 2014, the solution says that decomposition of arsine is not a redox reaction. Why would that be? Shouldn't H turn from -1 to 0 and As from 3+ to 0?

$$\ce{2AsH3 -> 2As + 3H2}$$

Often in chemistry "facts" depend on perspective. You're right that essentially H goes from -1 to 0 and As from 3+ to 0. But in general redox reaction involve two different chemical species. In this case it is better to think of the reaction as a decomposition reaction rather than a redox reaction.


Vapor Pressure - Sealed Vessel I have read that vapor pressure is only affected by temperature, but all of the examples have been in a sealed vessel. I have also read that pressure does not affect vapor pressure. I am looking for clarification of if and how pressure affects vapor pressure in sealed and unsealed vessels.

Also I would be appreciative if someone could provide an in depth, intuitive explanation of vapor pressure and why it is unaffected by pressure.

I assume you are talking about a 2 phase system of vapor in equilibrium with liquid.  Strictly speaking, the term "vapor pressure" only applies to single-component systems.  If there is a second component present, like air and water in a closed sealed container, then the equilibrium of each species is described by the partial molar free energy (or fugacity) of the species in the vapor as being equal to the partial molar free energy (or fugacity) of the same species in the liquid.  Even in the case of water and air in a closed container at 1 bar (where air is regarded as insoluble), the partial molar free energy of the liquid water will be slightly different from its value in the single component species at the same temperature (Poynting correction).  Even if the gas phase were ideal, such that the partial molar free energy of the water in the gas phase were the same as its value for pure water vapor at the same partial pressure, the effect of the higher pressure on the liquid water would slightly disturb the equilibrium and the partial pressure of the water vapor would differ from its value for a single component system.

If total pressure in the two-component system were high enough (i.e., higher air pressure), the partial molar free energy of the water in the gas phase would also be affected (say by the non-ideal interaction between the water vapor and the air), but not in the same way as the partial molar free energy of the liquid water changes.  So here, even more strongly, the total pressure times the mole fraction of water vapor in the gas (which would typically be defined as the partial pressure of the water vapor for an ideal gas mixture) will not be equal to the equilibrium vapor pressure of water at the temperature of the system.  With all this said, in practice, particularly for air and water, the effect can be expected to be pretty small.


Ideal Gas Equation Contradiction You have a regular balloon, and you decide to heat it up. The gas particles inside the balloon will gain kinetic energy, and so, the gaseous particles will collide with the internal wall of the balloon with a higher frequency and more energy. This is what we call pressure, which will, as a result, increase. This is more commonly generalised to be Gay-Lussac's law, which formally states that the pressure of a gas is directly proportional to its absolute temperature.

Now, as pressure has increased, meaning that the gaseous particles are colliding with the internal wall of the balloon with a higher frequency and more energy, the balloon will begin to inflate, or in other words, the volume of the balloon will increase.

This seems to contradict Boyle's law, which states that the pressure of a gas is inversely proportional to its volume. In the scenario I described above, it seems clear to me that it was the increase in frequency of the gaseous particles colliding with the internal wall of the balloon, that too with more energy, which altogether is known as an increase in pressure, that caused the balloon to inflate.

I perfectly understand Boyle's law when the idea of a pump is used, where you can use a slider to change the volume of a gas and clearly see its inversely proportional relationship with the pressure of the gas, however, can't seem to understand what is incorrect in my logic, from which I seemingly have contradicted Boyle's law.

I am a high-school student who just learnt about these laws, and out of curiosity decided to ask this question in hopes of clearing my doubts and fundamentally gaining a better understanding. Any help will be greatly appreciated, thanks in advance.

Your arguments do not contradict each other. You are trying to separate a process into two processes, which are coupled of course, but in first approximation this is a good way at looking it.

Let's state our model system again: A balloon with the pressure $p_1$, the volume $V_1$ is heated from temperature $T_1$. As a result we see an inflated balloon. Let's call those variables with $2$ as an index. Of course we treat everything as ideal as possible.

Now we look at the coupled process as two separate processes:

Isochoric heating, i.e. $V=\mathrm{const}$:
\begin{align}
&&p_1V&=n\mathcal{R}T_1\\
&&p_2V&=n\mathcal{R}T_2\\
\therefore
&&\frac{p_1}{p_2}&=\frac{T_1}{T_2}\\
\therefore
&&p&\propto T\\
\end{align}
That is Gay-Lussac's law.
Isothermal expansion, ie. $T=\mathrm{const}$
\begin{align}
&&p_1V_1&=n\mathcal{R}T\\
&&p_2V_2&=n\mathcal{R}T\\
\therefore
&&\frac{p_1}{p_2}&=\frac{V_1}{V_2}\\
\therefore
&&p&\propto \frac{1}{V}\\
\end{align}
That is Boyle's law.

You can also look at the whole process as the pressure stays the same while the balloon is heated. That then is an isobaric process, i.e. $p=\mathrm{const}$:
\begin{align}
&&pV_1&=n\mathcal{R}T_1\\
&&pV_2&=n\mathcal{R}T_2\\
\therefore
&&\frac{V_1}{V_2}&=\frac{T_1}{T_2}\\
\therefore
&&V&\propto T\\
\end{align}
That is known as Charles' law

Of course in reality there will be a mixture happening as the rubber of the balloon will expand, but the pressure that is possible is dependent on the material of it. In combination you will arrive at:
\begin{align}
&&p_1V_1&=n\mathcal{R}T_1\\
&&p_2V_2&=n\mathcal{R}T_2\\
\therefore
&&\frac{p_1V_1}{T_1}&=\frac{p_2V_2}{T_2}\\
\therefore
&&\frac{pV}{T}&\propto \mathrm{const}\\
\end{align}


How are pencil marks "adhered" to the paper? Pencils create marks by physical abrasion, leaving behind a trail of solid core material that adheres to a sheet of paper or other surface. Wikipedia

Normal paper is mainly composed of cellulose fibers, which were "funded" by the plants. And this "solid core material" is supposedly composed of graphite and a type of binder clay.

Image source

"Adhering" is usually caused by intermolecular forces. (e.g.: when sugar adheres to wet surfaces because of Hydrogen bonding). The questions are:

Is the case of a pencil mark some kind of intermolecular force too?
If so, what is it?

The graphite itself can adhere to paper fibers with London forces; it's not a strong adhesion because the marks can be erased easily. Some of the graphite is just mechanically stuck between the paper fibers; rubbing a mark from a soft pencil can easily blur it. Enough graphite is deposited that a line drawn by a pencil can conduct electricity.

The clay gives the lead structural integrity; an "H" pencil has a harder lead, and deposits less graphite; a "B" pencil is softer, and an "F" pencil can be sharpened to a finer point.

In my time as an art major, I made a lot of sketches using vine charcoal on newsprint. Vine charcoal has no binding clay, and the marks were easily smeared. On paper without much "bite", it was easy to lift the marks off completely with a piece of kneaded rubber, and if you wanted to keep a charcoal drawing (or a soft pencil drawing) you really had to spray it with a fixative.

So I'd say that it's possible that the binding clay also plays some role in binding the graphite or charcoal to the paper.

  In the manufacture of graphite pencils, air-classified grades of Volclay sodium
  bentonite and micronized hectorite are regularly used to bind the graphite
  compound in pencil lead. --- American Colloid Company, Industrial and Household Applications

Bentonite and hectorite both have ionic surfaces that probably help bind it to OH groups on the surface of cellulose fibers.


Why does copper form bubbles in vinegar in this situation? I watched this video in which the host tries to show the principles behind the voltaic pile.

The host first immersed zinc in vinegar and bubbles are observed forming around zinc. I've learned that this is because hydrogen ions in vinegar get electrons from zinc atoms and become gaseous.

Then the host take a conductor and attach at one end the piece of zinc and at the other a piece of copper. Bubbles start to form around copper too! (fig1)

Why does this happen?
Why do some electrons travel through the conductor instead of being attracted immediately by the hydrogen ions in the vinegar?
What if the conductor instead was very long and the pieces of copper and zinc far apart? (fig2)

When you put a zinc rod in a acid solution containing $\ce{H^+}$ ions, the zinc piece loses $\ce{Zn^{2+}}$ ions to the solution leaving electrons behind. These electrons attract hydrogen atoms to the zinc rod, and an electron transfer occurs, resulting in bubbles of hydrogen forming on the surface of the zinc. However, the evolution of gaseous hydrogen from zinc surface is rather slow, and that is why pure zinc reacts slowly with dilute acids.

The situation changes when a copper,or any other metal less electropositive than hydrogen is attached to zinc, by a wire. The copper can now be considered as an extension of the zinc rod: a place form where hydrogen can take electrons quickly and become gas. As the rate of evolution of hydrogen from the copper piece is faster, hydrogen bubbles out from the copper piece. This does not mean, however, that the zinc stops reacting with acid; it does, but at a very slow rate.

A common mistake is to think that the copper rod participates in the electrochemical reaction taking place. The copper is merely a pathway to release electrons to H+ ions in the solution. The only reaction taking place is

$\ce{Zn + 2H^+ -> Zn^{2+} +H2\uparrow}$

Copper has no place in the reaction.

So, to answer your questions,

  Why does this happen?

This happens because copper can do away with the electrons quicker than zinc, and when electrons get a pathway, they will take the pathway.

  Why do some electrons travel through the conductor instead of being attracted immediately by the hydrogen ions in the vinegar?

Well, some electrons do attach to H+ ions from vinegar directly, and hydrogen bubbles should form on the zinc plates, although they are less in amount. And I believe my earlier explanation answers this question.

  What if the conductor instead was very long and the pieces of copper and zinc far apart?

It does not have anything to do with distance. If you take a large solution, the reaction will certainly occur. On the other hand, if you separate the poles into two solutions the copper rod will not give off electrons as the circuit is not complete. Think about it this way, if the reaction was to occur, the solution with zinc will become progressively positive and the solution with copper will become negative. This can never happen.

Footnotes:

(1) As per your comment below, I now attempt to answer the question "What makes copper better than zinc in releasing electrons?"

The answer lies in a factor called overpotential, sometimes referred to as overvoltage. There is one type of overpotential which is important in this case— activation overpotential. This is caused due to the fact that when hydrogen atoms are produced on the surfce of a metal ($\ce{H+ +e-> [H]}$), they take some time before they can combine to form diatomic hydrogen and some more time before the hydrogen molecules can be released as gas. This results in the formation of a layer of hydrogen on the surface of metal effectively reducing the rate of reaction and halting it; this effect is known as polarization.

As you can see from the data chart in the Wikipedia page, Zn electrode has -0.77 V overpotential for evolution of hydrogen while there is no overpotential for copper electrode. This means that copper is not polarized easily.

(2) I mentioned in the beginning of my answer that pure zinc reacts with dilute acid slowly. Commercial zinc, on the other hand, reacts quickly with acids, owing of the same effect of polarization. In commercial zinc, impurities of other metals form cathode and the pure zinc forms anode in the acid medium, essentially producing a galvanic cell, where the anode and cathode are connected. Thus the reaction occurs quickly, without the hindrance of polarization.

(3) You should read this paper : http://pubs.acs.org/doi/abs/10.1021/ed078p516. It has some facts which you may find interesting.


Are the stereoisomers of nucleosides actually anomers? An anomer is an epimer at an (hemi)-acetal carbon like $\ce{R2C(OR')2}$.

Why is the 1'-position of a nucleoside referred to as the anomeric center despite the lack of acetal functional groups?

The C1 position of carbohydrates and carbohydrate derivatives is generally referred to as the anomeric center and compounds differing only in their configuration at this position are called anomers. In nucleosides, the 1'-position may not be part of a hemiacetal or acetal functional group, but it is at the same oxidation state. The IUPAC Gold Book defines anomers as:

  Diastereoisomers of glycosides, hemiacetals or related cyclic forms of sugars, or related molecules differing in configuration only at C-1 of an aldose, C-2 of a 2-ketose, etc.

The fact that the definition includes "glycosides" and "related cyclic forms of sugars, or related molecules" points towards why it is acceptable to describe the 1'-position as the anomeric center. Furthermore, this position in nucleosides is at the same oxidation state as a hemiacetal or acetal. Therefore, there are structural and chemical reactivity similarities between these functional groups and the 1'-position of nucleosides.

The correct name for this R1C(OR2)(NR3R4)R5 functional group is a hemiaminal ether. Hemiaminals have a free hydroxyl group: R1C(OH)(NR2R3)R4.

Also of note is that the anomeric effect refers to the preference for polar groups to be axial on pyranose derivatives, where the polar group may be not only oxygen substituents (eg. hemiacetals, acetals), but also other derivatives such as halides (eg. glycosyl halides). In all these cases, C1 would be referred to as the anomeric center.


How to determine polarity in large molecules? In case of small molecules like water things are more or less clear, but when I see a large complex molecule like Thyroid Hormone - how should I find out that this molecule is non-polar?

Once you move up to "larger" molecules like this, the definition of a "polar" molecule becomes a bit fuzzy. Rather, you will find that molecules contain polar and non-polar groups which all contribute to the overall characteristics of the molecule. It really depends on the context that you are interested in.

If you are curious about more macroscale properties, such as boiling point or solubility, a variety of group contribution methods have been developed to provide ballpark estimates based purely on the contributions of individual components of the molecule.

While group contribution methods will primarily give you an idea of bulk properties, you may also be interested in lower level behavior such as micellization or specific protein binding which will depend on the location and arrangements of all of the polar and non-polar groups on a molecular level.


Liquid crystals for academic outreach? I am interested in making a Schlieren texture of liquid crystals to show kids at an outreach event (see image below). I have the polarizers and camera for the imaging part, but I don't know of a place to find liquid crystals.

I would prefer something cheap and chemically inert because I don't want to bring anything sensitive or hazardous. I would imagine that it should be really easy to get solutions or powders of liquid crystal materials, but I am having a hard time finding a source.

Any suggestions? I don't have access to a proper chemistry lab, but I can mix basic things in my kitchen.

Assuming you already have the optical setup built around, aiming for the least amount of expensive chemicals, you may try semicrystalline extruded polymers, e.g. polyethylene foils, or PET bottles.

The «trick» is that you do not melt them, but that you reach just the glass transition temperature ($T_g$) which is below melting. Here, crystalline parts in the polymer break up, rendering the material isotropic and in transmittent observation with crossed polarizer / analyzer black.  Depending on material and degree of polymerization, this solid-solid transition already may happen in a range of 60 to 80 Celsius.  Put the material on a glass slide which is heated gently from beneath with an low-powered heating plate.  With a tooth-pick you may test that the material becomes softer.

Upon cooling you will see how spherulites re-appear, e.g.

(source)

Size and shape of them will depend on the rate of cooling, for example if let cool at air (or on the table of your microscope), or if you cool the glass slide by putting it on an ice bag.
If not overheated this may be repeated over and over easily for a dozen of times.  Apart from the heating plate, there is little potential danger.

A low-cost set up including a heating stage, where you could continue the observation of the sample's recrystallization was described in the February issue of the Journal of Chemical Education (J. Chem. Educ., 2019, 96, 823-826, doi 10.1021/acs.jchemed.8b00879).


Why wouldn't you see blue or purple when you heat metal and only either red or yellow? Question

Why wouldn't you see blue or purple when you heat metal?

Background and thought process

When metal gets heated the electrons gets excited and thus it emits the color that we see. But my question is why don't we see the other colors? If we heat metal to an extreme temperature it might turn white but not blue or purple. Why is this?

Because hot bodies emit a continuous range of light not a single wavelength

All warm objects emit a range of wavelengths that depend on how warm they are. The source of the light is a variety of mechanisms not all of which involve electronic transitions (some will involve transitions between molecular or solid state vibrations or rotations.) The important point for the perceived colour of the light they emit, though, is that the emitted radiation is continuous.

The chart below show the mix of light wavelengths from bodies at different temperatures:

The peak of the emission curve changes with the temperature. So the mix of light you see does change colour somewhat with the heat of the object but will never be a pure colour like blue or purple. Tungsten light bulbs, for example, reach temperatures of 2.5-3K °C and have peak emissions in the infra-red; their light has more red than blue in the visible so they are perceived as somewhat orange when compared to sunlight (which follows the curve for a body of 5-6k °C, the sun and has peak emissions in the middle of the visible spectrum).

The differences between different sources of black body radiation are harder to perceive than you might think because the eye and brain are good at adjusting what you see to a common perceived colour mix and it is only when comparing different light sources that the differences in the mix of light become easy to see (something also obvious when comparing other types of light bulb to tungsten bulbs).

So a hot object at ~7k °C will emit more blue light than red light but it will still be emitting a lot of red light so you will tend to see blue-tinged white rather than pure blue. Astronomers can, when their eyes are well adjusted to darkness, tell the subtle colour differences of different stars (which range in temperature from maybe 2k °C to maybe 20k °C). But the effect is small.

Of course, if you heat something enough you will turn it into a vapour or plasma and there will be enough energy to excite the outer electrons. The light resulting from transitions of outer electrons is often very narrow in wavelength and can be used to identify the atoms in the vapour. Low pressure Sodium street lights (the bright orange ones) are a result of this effect (the excited outer electrons sodium atoms emit yellow-orange light very characteristic of sodium). Many other elements have characteristic lines that can be just about any specific colour under similar conditions).

But normally when you heat things up you just see the continuous black-body emissions and that can't be a single colour.


What are the limitations of the Hendersson-Hasselbalch equation? Our instructor told us that prior to the equivalence point, the $\mathrm{pH}$ of a solution is dependent on the HH equation. However, when I tried practicing for polyprotic acid titration, I came upon this problem that does not make any sense at all. When I'm using the HH equation, the $\mathrm{pH}$ I get is lower than the initial $\mathrm{pH},$ which is illogical.

The problem goes likes this:

  What is the $\mathrm{pH}$ of the solution of $\pu{25 mL}$ $\pu{0.100 M}$ of $\ce{H2SO3}$ $(K_\mathrm{a1} = \pu{1.23e-2},$ $K_\mathrm{a2} = \pu{6.6e-8})$ when $\pu{5 mL}$ of standardized $\pu{0.1000 M}$ of $\ce{NaOH}$ is added?

The initial $\mathrm{pH}$ that I've calculated is $1.53,$ but every time I try to use HH for the computation of the first $\pu{5 mL}$ increment of titration, the $\mathrm{pH}$ is lower than the $\mathrm{pH}$ of the initial conditions.

The key approximation made in deriving the Henderson-Hasselbalch equation is that the equilibrium constant can be written as $$K=\frac{c_{H^+}c_{A^-}}{c_{HA}}$$
that is, we assume activity coefficients are unity. If you take the base-10 logarithm of this equation and rearrange terms you arrive at the HH equation:$$pH=pK_a+\log_{10}\left(\frac{c_{A^-}}{c_{HA}}\right)$$ Problems usually arise with approximations made beyond the HH equation, in the context of a polyprotic compound particularly  by ignoring one of the equilibria, including perhaps water autodissociation, when this shouldn't be done.

In the case of the problem you outline, you might

ignore the second dissociation equilibrium and
assume that added base neutralizes an equimolar amount of remaining acid $\ce{H2SO4}$ to form additional $\ce{HSO4-}$.

Then

$$\begin{align}c_{HA} = c_{HA,\,initial}-c_{added\,OH^-}&=\pu{0.0705 M} - \pu{0.100 M} \frac{\pu{5 mL}}{\pu{30 mL}}\\&=\pu{0.0538 M}\end{align}$$
$$\begin{align}c_{A^-} = c_{A^-,\,initial}+c_{added\,OH^-}&=\pu{0.0295 M} + \pu{0.100 M} \frac{\pu{5 mL}}{\pu{30 mL}}\\&=\pu{0.0462 M}\end{align}$$

Applying the HH equation we then have, as the approximate pH
$$\begin{align}pH &=1.91 + \log_{10}\left(\frac{0.0462}{0.0538}\right)\\&=1.84\end{align}$$ which is above the starting pH of 1.53.

A more accurate value (although still approximate) can be obtained starting from the condition of charge balance for the system, and by assuming that the $\ce{OH^-}$ concentration is negligible, from which one obtains pH=1.75.


How does voltage and resistance affect an electrolysis? At school we've learned about calculating the voltage needed to perform an electrolysis. The current is how many electrons pass per time. The formula is given $$I = \frac UR$$

This means that given an electrolysis with $U=\pu{4.0 V}$ we get $$I = \frac{\pu{4.0 V}}{R}$$

If I am correct, what would decide how fast the electrolysis will go, is the resistance. Is the resistance fixed, or can I do anything to lower it? How would increasing the voltage affect the electrolysis? Would it just make it go faster (because $I$ is increased)?

Voltage determines the kind of chemistry at each electrode.  Gold is purified by plating the anode to the cathode at minimum voltage for the half-reaction, so impurities either do not dissolve and do not plate out.  Copper electrorefining electrode mud contains valuable trace elements in concentrated from.

Current determines the amount of chemistry, ~96,500 coulombs/mole of electrons.

Cell internal resistance converts input energy, power multipleid by time, $\ce{I^2Rt}$, into useless heat.  You then want small electrode spacing and good electrolyte stirring.  You then need enough voltage to do the chemistry, plus more to compensate for resistance losses, but not so much voltage that undesirable chemistries are enabled.  Does cell resistance increase (ohmic) or decrease with temperature?


How do I calculate the isoelectric point of amino acids, each of which has more than two values of pKa? For most amino acids, the $\mathrm{pI}$ is simply the arithmetic mean of the amino and carboxyl $\mathrm pK_\mathrm a$s. However, for tyrosine and cysteine, which have more than one $\mathrm pK_\mathrm a$ value, this rule of thumb doesn't apply.
I see that for tyrosine, it's the $\mathrm pK_\mathrm a$s of the carboxyl and amino groups that are averaged, but for cysteine it's those of the carboxyl group and the side chain.
I haven't been able to find an explanation of why this is the case, or what the reasoning behind the calculations is?

Since the $\mathrm{pI}$ is the $\mathrm{pH}$ at which the amino acid has no overall net charge, you need to average the $\mathrm pK_\mathrm a$ values relevant to the protonation/deprotonation of the form with no net charge. Here are the acid-base equilibria for tyrosine:

The form with no net charge is in red (+1 and -1 cancel out to give no net charge). It is the $\mathrm pK_\mathrm a$ values on either side of this form (in blue) that matter, hence the $\mathrm{pI}$ of tyrosine is $5.66$ (the average of $2.20$ and $9.11$).

It just so happens that $2.20$ is the carboxyl $\mathrm pK_\mathrm a$ and $9.11$ is the amino $\mathrm pK_\mathrm a$. If the side chain $\mathrm pK_\mathrm a$ were lower than $9.11$, then you should average the carboxyl and side chain $\mathrm pK_\mathrm a$'s instead.

The same logic applies to cysteine (look up the $\mathrm pK_\mathrm a$ values and draw out the differently protonated forms). You'll find that since the side chain has a lower $\mathrm pK_\mathrm a$ than the amino group, you average the carboxyl and the side chain $\mathrm pK_\mathrm a$'s.

This procedure can of course be extended to the amino acids with acidic side chains (aspartic acid; glutamic acid) and those with basic side chains (lysine; arginine; histidine).


History of the word copperas I searched for copperas on google and it says ferrous sulphate. My dictionary shows that the word copperas is derived from copper.

I was just wondering what has ferrous sulphate to do with copper

You are on the right track.

The Brockhaus Konversationslexikon (Brockhaus Encyclopedia, 14th edition, 1894-1894)

explains the German term Kupferwasser (copper water) as

  veraltete Bezeichnung für unreinen (kupferhaltigen) Eisenvitriol

which means outdated name for impure (copper-containing) iron vitriol.

Note that iron vitriol is an old term for iron(II) sulfate,  such as vitriolic oil is an outdated name for sulfuric acid.

If you take a look at English wikipedia page on iron(II)sulfate, you'll see that one of the common minerals is a heptahydrate, melanterite. If you follow the link, you will recognize that one of the associated minerals is chalcanthite, which is $\ce{CuSO4·5H2O}$.

To cut the long story short, the term copperas stems from the fact that natural iron(II)sulfate was often found to be associated and maybe contaminated with copper(II)sulfate.


Is there such a thing as a "minimal soap" molecule? Wikipedia's Soap gives sodium stearate as an example of soap, and apparently I've been eating it:

  Sodium stearate is the sodium salt of stearic acid. This white solid is the most common soap. It is found in many types of solid deodorants, rubbers, latex paints, and inks. It is also a component of some food additives and food flavorings.

What would be the smallest or simplest molecule that we could reasonably call a soap? Perhaps a functional definition would be that it could perform some of the functions of soap in the same way that soap does.

In Why does bleach feel slippery? and its follow-up Is it known for sure that bases feel slippery because of the production of soap/surfactant? the saponification of other existing molecules is discussed, and I'm not looking for that here.

Source

It boils down to the definition of soap. Wikipedia defines a soap as the salt of a fatty acid.
IUPAC claims the smallest fatty acid can be considered to have 4 carbons.
Therefore the simplest soap molecule would be a (generally sodium) salt of butyric (butanoic) acid, i.e. sodium butyrate.

Now apart from the chemical definition, a soap must adhere to its function in order to be defined as such. Therefore, before calling something a soap, we would have to know if the substance does in fact act as a surfactant in a oil-water surface (reducing the interfacial tension) - but that will depend on the nature of the oil in question and the purity of the water. Only for a well defined system we can then conclude that such molecular salt is acting as a soap in it.


Why is N2 enthalpy zero, entropy 191.5, yet G is listed as zero and not [-T*0.1915]? In standard tables, $\ce{N2}$ at STP has an enthalpy formation ($\Delta H_\mathrm f$) listed as $0\ \mathrm{kJ/mol}$, entropy ($S$) as $0.1915\ \mathrm{kJ/K}$, and free energy of formation ($\Delta G_\mathrm f$) as $0\ \mathrm{kJ/mol}$.

If we use the std eq.: $\Delta G = \Delta H-T\cdot\Delta S$, I don't get zero with those numbers, obviously. (assuming STP and all that) But, $\Delta G$ is zero for nitrogen?

My initial thought is that $0.1915\ \mathrm{kJ/K}$ is not $\Delta S$ as required by the equation, but just $S$, so I need to figure out $\Delta S$ for $\ce{N2}$ and use that instead. Okay, that would make sense, but it doesn't follow for this example, which I see everywhere:

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

$\Delta H$ is calculated for the reaction using the appropriate $\Delta H_\mathrm f$

$\Delta S$ is calculated using the appropriate $S$ values (So $0.1915\ \mathrm{kJ/K}$ for $\ce{N2}$, not $0$)

Then they use these reaction $H$ and $S$ to get $\Delta G$

However, if I just use the $\Delta G$ values found in databases like the CRC Handbook (so $0$ for $\ce{N2}$), I don't get the same answer using $\Delta G_\mathrm r=\Delta G_\text{products}-\Delta G_\text{reactants}$

This is driving me nuts, I know I must be missing something simple here.

Enthalpies of formation and Free energies of formation are expressed relative to the standard state of 25 C and 1 bar.  So, in the standard state, they are taken to be zero.  The entropy value you cited is the absolute entropy, relative to absolute zero.  If you wanted the entropy of formation of nitrogen relative to the standard state of 25C and 1 bar, that would be 0.


Question on notation Here, when they write $v/(\mathrm{mol\ s^{-1}\ (kg\ \text{protein})^{-1}})$ is $0.30$ for the first row first column, do they mean that the value is $0.30\ \mathrm{mol\ s^{-1}\ (kg\ \text{protein})^{-1}}$ or $0.30/(\mathrm{mol\ s^{-1}\ (kg\ \text{protein})^{-1}})$?

The expression
$$v/(\mathrm{mol\ s^{-1}\ kg^{-1}})=30$$
means
$$v=30\ \mathrm{mol\ s^{-1}\ kg^{-1}}$$

This notation is used in the table, so that the entries are all just numbers without any unit symbols.

For an explanation, I copy the following section from my meta answer, which is mainly a collection of rules and examples taken from various standards and guides (see the meta answer for references).

  Unit symbols are mathematical entities and not abbreviations. In expressing the value of a quantity as the product of a numerical value and a unit, both the numerical value and the unit may be treated by the ordinary rules of algebra.

  (…)

  In expressing the value of a quantity as the product of a numerical value and a unit, both the numerical value and the unit may be treated by the ordinary rules of algebra.

    the equation $T=293\ \mathrm K$ may equally be written $T/\mathrm K=293$

  It is often convenient to write the quotient of a quantity and a unit in this way for the heading of a column in a table, so that the entries in the table are all simply numbers.

    $$\begin{array}{ll}
\hline
T/\mathrm K & p/\mathrm{kPa} \\
\hline
300.00&3.5368\\
310.00&6.2311\\
320.00&10.546\\
330.00&17.213\\
\hline
\end{array}$$

  The axes of a graph may also be labelled in this way, so that the tick marks are labelled only with numbers.

    do not write “mass [kg]” for “mass in kilogram”

Also note that expressions for units shall contain nothing else than unit symbols and mathematical symbols. Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted. Therefore the use of the unit “$\mathrm{mol\ s^{-1}\ kg^{-1}\ \text{protein}}$” as written in the question is not permissible. The correct unit is $\mathrm{mol\ s^{-1}\ kg^{-1}}$.


How to find the number of atomic microstates for a given electronic configuration? Given some electron configuration, I know the following formula can be used to determine the number of microstates:

$$\text{# mircostates} = \frac{(\text{# electron positions})!}{((\text{# electrons})!(\text{# positions} - \text{# electrons})!)}$$

So for 2 electrons in a $\mathrm{p}$ orbital it would be $6!/(2!4!) = 15$, or for two electrons in a $\mathrm{d}$ orbital it would be $10!/(2!8!) = 45$.

I'm a little confused how to apply this for a situation where all the electrons are not in the same $l$ level.

For a configuration like one electron in $\mathrm{s}$ and one in $\mathrm{d}$ would the number of microstates be the product of the above formula for each $l$ level?

$$\frac{2!}{1!1!}\cdot \frac{10!}{1!9!} = 20$$

Or for one electron in $\mathrm{s}$ and one in $\mathrm{f}$:

$$\frac{2!}{1!1!}\cdot \frac{14!}{1!13!} = 28$$

I assumed it would work like this because the formula reminds me of multiplicity, so it would make sense for it to be a product. But it also occurred to me that the 2 $l$ levels could be combined for the formula, so for a $\mathrm{(s^1)(d^1)}$ electron configuration, the number of positions would be $$\frac{(10+2)!}{2!(10+2-2)!} = \frac{12!}{2!10!}= 66,$$ which is must bigger than what I had before.

Can anyone clear up how to apply this formula to a case like this?

For clarification, the exact specification in the problem was:

  Imagine a $\mathrm{s^1\, f^1}$ electron system (that is, 1 s electron and 1 f electron; no p or d electrons).  How many total microstates exist for this system?

I think this means that the one $\mathrm{s}$ and one $\mathrm{f}$ electrons are each confined to their $l$ level, which is why I thought the multiplicity of microstates would be the product of the above formula for each, but I'm not sure.

You really have to calculate out things exactly, and particularly carefully if electrons are in the same orbital, e.g. $p^2$, when the Pauli principle has to be checked. Its a bit tedious but it can be very quick with a bit of practice especially if electrons are in different orbitals.
You start by calculating the spin, orbital and total of all angular momentum. Each electron has spin quantum number s = 1/2 and magnetic quantum number $m_s$ = +-1/2. Orbitals have angular momentum of s=0, p=1, d=2, f=3 etc units.
The total spin angular momentum is the series of values $S=|s1+s2|...|s1-s2| $ where s1 and s2 =+1/2 are the quantum numbers of each electron. The total angular momentum is   $L=|l1+l2|...|l1-l2| $  and total angular momentum $J=|L+S|...|L-S| $
Each of these are called Clebsch-Gordon series. The way to calculate them is to calculate the maximum and minimum values and make intervening values separated by 1. Note that sometimes there is only one value and all values are positive. e.g. J=5/3, 3/2, 1/2 or J could be 0 or 3 etc.
The term symbol has the form $^{2S+1}L_J$.
The super-prefix is the spin multiplicity, for spin angular momentum $S$ this is  $2S+1$ or in general $2X+1$ for angular momentum X.
 In your question $s^1f^1$ the total spin using the series above is  S = 0 and 1. The spin multiplicity is thus 1 and 3, singlets and triplets terms only. The total orbital angular momentum L = 3 since the s orbital has zero angular momentum. Thus the terms so far are
$^1F$ and $^3F$ The number of microstates (multiplicity) in each term is $(2S+1)(2L+1)$. This is 7 for the $^1F$ and 3.7=21 for the $^3F$ making 28 in total. These states are also degenerate in energy. You should calculate that $s^1d^1$ has 20 microstates.

The total angular momentum J series associated with S=1 is $J=|3+1|.... |3-1|$ or 4, 3 and 2 and with S=0 , J=3.  Thus only triplet states have J = 4 and 2 and both singlet and triplets J=3. The terms symbols for the levels are
$^1F_3$, $^3F_4$, $^3F_3$,  $^3F_2$
 The total multiplicity from the J values is is 7+9+7+5 = 28 the same as from terms.

When the electrons are in the same orbital then the only secure way of calculating terms is to list all the individual quantum numbers in a table, with columns $m_{l1},m_{l2},m_{s1},m_{s2}$, S and L and remove those for which the Pauli principle removes a row. (See table in Steinfeld 'Molecules & Radiation' chapter 2 and Engel & Reid, 'Physical Chemistry' chapter 22). In particular you have to look out for when two rows have the same set of quantum numbers but in a different order; for example $m_{s1}$=0, $m_{s2}$=1 in one row and $m_{s1}$=1 and $m_{s2}$=0 in another but the $m_l$ values are the same, and vice versa. Doing this you should find, for example that $p^2$ has 15 terms, as $^1D$, $^3P$ and $^1S$ Hope this rather long answer helps you understand things.


Isn't 11-trans-retinal more stable than 11-cis-retinal? 11-cis-retinal can be found as a part of rhodopsin and it is the non-excited form of retinal. By absorbing light, retinal goes to a higher energy state and converts to 11-trans-retinal. At least I understand it so.
If all above is correct, I just don't get one thing. Shouldn't the cis form be less stable than trans (due to steric hindrance)?

Cis-retinal is less stable than the trans isomer.  The carotenoids ingested by the body are all trans.  Enzymes in the body convert them to trans-retinoic acid, then turn them into a fatty ester which is then converted by the enzyme retinoid isomerohydrolase to cis-retinal.  The higher energy state of the cis isomer makes it prone to photoisomerization back to the lower energy trans isomer.


Entropy and enthalpy change on mixing liquids Question:

  A concentrated solution of copper sulphate, which is dark blue in colour, is mixed at room temperature with a dilute solution of copper sulphate, which is light blue. For this process:

  Entropy change is positive, but enthalpy change is negative.
  Entropy change and enthalpy changes are both positive.
  Entropy change is positive and enthalpy does not change.
  Entropy change is negative and enthalpy change is positive.

The final conc. of the mixture would be less than the initial concentration of the concentration $\ce{CuSO4}$ given.
Therefore  I think the entropy change is positive. But, what about enthalpy?  What factors should I look into to find answer to these types of questions?

This is difficult, because we do not know what is your current understanding and at what “level” of approximation you expect the question to be answered.

At a undergraduate level, or in the absence of any details (or thermochemical data or access to thermochemical tables), I would go with option (3).

Enthalpy: we are not given any indication at what is the enthalpy of mixing. In the simplest approximation, the enthalpy of mixing is zero, as in an ideal solution. It would correspond to a regime where the enthalpy of dissolution ($\Delta H_{\text{sol}}$) does not depend on the concentration of the solution. It's probably a good approximation, until you reach high concentrations.
Entropy: the entropy of mixture is positive, i.e. the process of mixing the two solutions is spontaneous. It can be explained by this reasoning.


Diatomic halogens - bond strength Why is the $\ce{F-F}$ bond relatively weak? And why does bond strength decrease from $\ce{Cl-Cl}$ to $\ce{I-I}$?

Does this have anything to do with "lone-pair" repulsion on the rather small $\ce{F-F}$ molecule? I haven't asked my professor but I'm suspecting this is going to the first thing on his mind. I'm sure that many of you here will have rebuttals to this argument.

On the other hand does $\ce{F2}$ even have lone pairs? There are no non-bonding orbitals populated on the molecular orbital diagram of fluorine. I wouldn't expect there to be any non-bonding orbitals populated either, as this is a simple diatomic we're dealing with and the atoms involved are identical ... so each atom's orbital should be able to interact with the other atom's orbital effectively.

I see as I write the answer, that many comments are spot-on.

Whether or not $\ce{F2}$ has lone pairs depends on your view of molecular orbitals and valence bond (or Lewis) descriptions of bonding.

Covalent radii for fluorine compounds are usually around 0.64 Å for F, but yet the F-F bond distance is 1.43 Å.

If we go for the Lewis or valence bond view, we'd argue that the large number of lone pairs on a small atom will be highly repulsive.

If I go for the molecular orbital view, I'd point out that in reality, the anti bonding orbitals $\pi^*$ and $\sigma^*$ are higher in energy than their bonding counterparts, particularly in the 2nd row, so the F-F molecule will be weaker than a simple single bond.

There's a really nice article by Peter Politzer in JACS that I like to use when discussing this "anomaly": "Anomalous properties of fluorine" J. Am. Chem. Soc., 1969, 91 (23), pp 6235–6237

  It appears rather that the anomaly involves the fluorine atom, in whatever circumstances, and its interaction with an external electron which enters its outer shell. [...] In any case, it is suggested that because of the exceptionally small size of the fluorine atom (see Table I), the entering electron feels unusually large forces of attraction from the nucleus and repulsion from the electrons already associated with the atom. It appears that the magnitude of the anomaly in the repulsive forces is greater than in the attractive forces.

In other words, regardless of the MO or VB view, the issue is one of electrostatic repulsion.


Melting point of alkaline earth metal carbonates The melting point of carbonates of group 2 elements is (From Wikipedia)

$\ce{BeCO3}$ : 54
$\ce{MgCO3}$ : 540
$\ce{CaCO3}$ : 1339
$\ce{SrCO3}$ : 1494
$\ce{BaCO3}$ : 811

A similar trend is followed by alkali metal carbonates (from Wikipedia). But in the inorganic chemistry book by Shriver and Atkins, this is given

Is there a difference between decomposition temperature and melting point? Or is the values given in the book wrong?

Melting and decomposition are different things, as you guessed in your question.

Melting:

Something can melt, and then re-solidify, undergoing a physical but not chemical change.

A simple example would be ice, but it is also true for many chemicals. For example tert-butanol is a solid at room temperature (at least in a well air-conditioned lab), but melts readily at slightly elevated temperatures without any degradation. Dimethylsulfoxide is another example of a chemical which often solidifies in a cool lab, but can be melted back down with only the heat of your hand.

Decomposition:

Decomposition is a different thing, whereby the material undergoes a chemical change (usually in addition to physical change). This is often seen in laboratory settings when trying to measure the melting point of molecules with high melting points, such that at the temperatures needed to melt the solid, side reactions are often observed, destroying the material. For this reason, in the literature, you will see melting points reported as  mp. (decomp.), i.e. the temperature at which your molecule decomposed rather than melted.


What would be the name of this compound? Although I assume this compound is likely unable to be formed, my chemistry teacher put it up on the whiteboard to show the extent of what you can name using IUPAC nomenclature, out of interest I tried to name it and got hexcyclo[2.2.0.0.0]cyclohexane, is this correct or am I mistaken?

I think it is Cubane ($\ce{C8H8}$), a synthetic hydrocarbon molecule (Wikipedia), which has synthesized using following scheme:

I am confident about that because your teacher had said that there wasn't a central carbon atom (assumingly). See following illstration:


Why Gold and Silver react minimally with atmosphere? Gold and Silver atoms interact with one another to form reasonably strong metallic bonds, but surprisingly they have almost no chemical interaction with atmosphere. How can we explain it based on the atomic structure of Gold and Silver. What is so unique about its chemical structure that is not present in other transition metals that makes them noble (non-reactive)?

Copper, silver, gold. What is the energy of the highest "s" orbital versus the highest "d" orbital?  Gold is a very heavy nucleus.  An "s" orbital has non-zero probability at its nucleus.  Special relativity effects make its "s" electrons more massive, bringing them into the atom.

A silver atom is a soft Lewis acid.  It reacts well with large soft Lewis bases.  Silver quickly tarnishes with sulfide.  Gold is not chemically inert at all to even mild oxidation if the product cation is coordinated by ligands.  Salt water plus air will tarnish gold.  Dilute cyanide plus air will dissolve it (e.g., heap leaching of gold ore).

Thermodynamics proposes, kinetics disposes.  Aluminum is tremendously reactive toward air and water.  Anything that permanently breaches its adherent oxide coating (e.g., abrasion plus amalgamation) sets it off.


Proper disposal of calcium hydride/calcium hydroxide I had probably 2-3 grams of calcium hydride leftover in a schlenk flask from predrying a distillation of benzyl chloride. For the disposal, I diluted the flask contents in about 30mL isopropanol and then, over a water bath (flask open but under heavy stream of argon), added distilled water every few minutes until well after the bubbling had stopped, which I assumed was exothermic evolution of hydrogen gas. I also thought any leftover benzyl chloride would just make trace toluene, benzyl hydroxide, and calcium chloride.

I then figured the milky-white solution was okay to pour down the sink with lots of water running and removed about half of it in this manner, although I noticed a brief but strong smell from the vapors as I did so and stopped. I checked and noticed some sites place it into a basic waste container. Was it a mistake to pour down the sink? I thought limewater would be harmless with enough water. Were there any other errors in this process? I may have to repeat this in the future with calcium hydride again or phosphorous pentoxide (seen various methods on P$_2$O$_4$, but one mentioned sink disposal as well).

The safest way to get rid of calcium hydride is to pour over ice. The water quenches the reagent itself, with the ice helping to control the exotherm (if you just add water, a lot of heat and gas is liberated, having a tendency to spray hot chemicals over the whole hood).

Once everythings quenched, it's helpful to add some hydrochloric acid before disposing of down the sink to prevent it clogging the drains.

Note that whilst clean calcium hydride is safe (once quenched) to be disposed of via the normal drainage, you must ensure any chemicals are washed off first (hexane works well), especially odorous things such as amines which you may have been drying with the calcium hydride.


Determine the rate equation (intermediate reaction) I was helping some students today, quite successfully, until I came across this problem:

$$\begin{align}
\ce{A &<=>[$k_+$][$k_-$] B + C} \\[5pt]
\ce{C + D &->[$k_1$] E}
\end{align}$$

The first equation is a fast process  with forward and reverse rate constants $k_+$ and $k_-$, respectively.
The second process is slow, with rate constant $k_1$.

The objective is to write down a rate law for the resulting equation:

$$\ce{A + D -> B + E}$$

The technique that had been working was to write down the rate for the slow process...

$$r = k_1[\ce{C}][\ce{D}]$$

Since $\ce{C}$ is a product in the fast reaction, which is an equilibrium equation, we also have:

$$k_+[\ce{A}]=k_-[\ce{B}][\ce{C}]$$

This is where things went awry, because the rate law that we seek is a function of $[\ce{A}]$ and $[\ce{D}]$ only. Considering the stoichiometry, it seemed appropriate to put $[\ce{B}]=[\ce{C}]$. This leads to the substitution $[\ce{C}]=(\frac{k_+}{k_-})^{1/2}[\ce{A}]^{1/2}$. The interface has declared the result to be invalid, so it must be an illegal maneuver. How, then, do I eliminate $[\ce{B}]$ and $[\ce{C}]$ and come up with the desired reaction rate law?

You can't write $[\ce{B}] = [\ce{C}]$ because  $\ce{C}$ is consumed in the second step but $\ce{B}$ isn't. However, you were pretty much on the right track.

Since $\ce{E}$ is produced from $\ce{D}$, you can write $[\ce{E}] = [\ce{D}]_i - [\ce{D}]$ where $[\ce{X}]$ denotes the concentration of $\ce{X}$ at a certain point in time and $[\ce{X}]_i$ denotes the initial concentration of $\ce{X}$. This is assuming that the initial concentration of $\ce{E}$ is $0$, of course.

On top of this, $\ce{C}$ is either produced from $\ce{A}$, or consumed in the production of $\ce{E}$. Therefore you can write $[\ce{C}] = [\ce{A}]_i - [\ce{A}] - [\ce{E}] = [\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]$. (There is the reverse step in which $\ce{C}$ produces $\ce{A}$, but the equation still holds, since you can just consider this to be "negative" production of $\ce{C}$ from $\ce{A}$.)

Your rate law is therefore: $r = k_1[\ce{D}]\big([\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]\big)$. The initial concentrations are merely constants, so you could view this as being of the form $r = k_a[\ce{D}] + k_b[\ce{A}][\ce{D}] + k_c[\ce{D}]^2$. I can't think of any way to simplify this (the steady-state approximation is not very valid here since $\ce{C}$ is produced quickly and consumed slowly, and in any case it does not yield a nicer expression). I imagine that one could use some form of an integrated rate law to find expressions for $[\ce{A}]$ and/or $[\ce{D}]$, but it would just make things even uglier.


POI is a bent molecule similar to water. How are the two molecules different it terms of bond angles and what causes this? phosphorus oxyiodide ($\ce{POI}$) has the VSPER AXE formula of $\ce{AX2E}$ while water, $\ce{H2O}$ has the AXE code of $\ce{AX2E2}$. According to my work sheet $\ce{POI}$ should have a $120^\circ$ bond angle while $\ce{H2O}$ should have a bond angle of $104.5^\circ$.

Does this mean that due to the extra lone pair it creates less resistance resulting in a greater angle? Also, does it having three different elements in the equation do anything?

Does this mean that:

    Due to the extra lone pair $\color{red}{\text{it}}$ $\color{limegreen}{\text{creates less resistance}}$ resulting in $\color{blue}{\text{a greater angle}}$?

More like:

Due to the extra lone pair $\color{red}{\text{in } \ce{H2O} \text{ the pendant atoms}}$ $\color{limegreen}{\text{experience more repulsion}}$ resulting in $\color{blue}{\text{a smaller angle}}$ $\color{magenta}{\text{compared to }\ce{POI}}$?

  Also, does it having three different elements in the equation do anything?

No, I think the three different elements in the question were there to evaluate if you were:

estute to the fact that oxygen forms a double bond thus
phosphorous has 1 unbonded pair of electrons (not 2) and
if the double bond would hang you up in obtaining an answer.