What is the pH of a 5M solution of hydrochloric acid? What is the $\mathrm{pH}$ of a $\pu{5M}$ solution of hydrochloric acid?

So in order to solve this, apparently all you need to do is plug it into the equation:

$$\mathrm{pH} = -\log[\ce{H3O+}]$$

where the concentration of $\ce{H3O+}$ is $\pu{5M}$. But why can we assume this? Don’t we need to create an ice table from $\ce{HCl + H2O-> H3O+ + Cl-}$ and use the $K_\text{a}$ value of $\ce{HCl}$ to solve this?

As explained in pH Paradoxes: Demonstrating That It Is Not True
That pH ≡ -log[H+] Journal of Chemical Education Vol. 83 pages 752-757.

pH = -log (hydrogen ion activity).

and "pH of 7.6 M HCl is about -1.85 (not -0.88)"

So how far off is -log (5) = -0.69 from the real answer?

From this table of activity coefficients 5m HCl has an activity coefficient of 2.38, so activity is 11.9, which yields pH = -1.1.

Unfortunately the table is in terms of molality rather than molarity.  5M HCl is about 5.5m, and activity coefficient is dramaticly increasing with concentration (see table I of the pH paradox reference) so the true pH deviates somewhat further from -0.69, but pH = -1.1 is a reasonable estimate.

Just to give a quick and logical answer, if you look at the following acids:

$\ce{HCl}$
$\ce{HBr}$
$\ce{HI}$
$\ce{H2SO4}$ - first dissociation only
$\ce{HClO4}$
$\ce{HClO3}$
$\ce{HNO3}$

These acids above are Strong Acids, meaning that they completely dissociate, so for that purpose there is no need of a certain K value, since none of this is an equilibrium reaction.

  Just a forward reaction. $\ce{->}$ not $\ce{<->}$

NOTE: This is usually for general purposes. Not to be super accurate.


Why is the carbon-fluorine bond stronger than the other haloalkanes? What effect makes  C-F bond  stronger than  other halo-alkanes?

From wikipedia article of carbon-fluorine bond,

  The high electronegativity of fluorine (4.0 for F vs. 2.5 for carbon)
  gives the carbon–fluorine bond a significant polarity/dipole moment.
  The electron density is concentrated around the fluorine, leaving the
  carbon relatively electron poor. This introduces ionic character to
  the bond through partial charges ($\ce{C^{δ+}—F^{δ−}}$). The partial
  charges on the fluorine and carbon are attractive, contributing to the
  unusual bond strength of the carbon–fluorine bond. The bond is
  labeled as "the strongest in organic chemistry," because fluorine
  forms the strongest single bond to carbon. Carbon–fluorine bonds can
  have a bond dissociation energy (BDE) of up to 544 kJ/mol. (This is why) The BDE
  (strength of the bond) is higher than other carbon–halogen and
  carbon–hydrogen bonds. For example, the molecule represented by
  $\ce{CH3X}$ has a BDE of 115 kcal/mol for carbon–fluorine while values
  of 104.9, 83.7, 72.1, and 57.6 kcal/mol represent $\ce{C–X}$ bonds
  to hydrogen, chlorine, bromine, and iodine, respectively.


Extracting urea from urine What is a process for the extraction of urea from urine? These are the steps I have come up with:

Add 2-propanol to get rid of any bacteria present. (The 2-propanol likely contained some sodium chloride.)
Precipitate out the proteins and neutralize any acids present using sodium carbonate. (At least, that is what appears to happen upon addition.)
Crystallize out the sodium and potassium chlorides, as urea is much more soluble than those salts.

At the second step, three layers formed: a top clear yellow layer (which I assume was 2-propanol), a middle very cloudy yellow layer (which quickly disappeared into the bottom layer) and a bottom slightly cloudy yellow layer (which I assume was water). The bottom two layers contained particles floating around.

Then, I assume that the water layer remaining contains mostly urea. However, how do I get rid of everything else (urobilin, etc.)?

The end goal is to generate ammonia from the urea by adding some calcium hydroxide. However, it would also be nice to get rid of other impurities which do not impede this reaction.

First, from Boundless.com (Chemical Composition of Urine), "Urine is an aqueous solution of greater than 95% water, with the remaining constituents, in order of decreasing concentration urea 9.3 g/L, chloride 1.87 g/L, sodium 1.17 g/L, potassium 0.750 g/L, creatinine 0.670 g/L and other dissolved ions, inorganic and organic compounds (proteins, hormones, metabolites)."

Urine color is generally a result of the degradation of the heme molecule from blood into the urinary products, called bilirubin, urobilinogen, urochrome, urobilin and uroerythrin. In essence, they are all tetrapyrroles amounting to something on the order of 1mg per 100ml (or roughly 0.001% of urine)... so this is a very minor part of urine. Also, it should be noted that over 3,000 chemicals have been found in urine, so there's something to consider.

Activated carbon filtration might be able to be used to preferentially remove the tetrapyrroles and some other organic molecules while allowing urea to pass (for the most part).

There are various forms of activated carbon (of course), and so the amount of urea absorption (by carbon) will be variable; the adsorption capacity of the activated carbon for urea will (likely) be different for activated carbon from different sources. Absorbance of urea from one report was about 1.69 mg urea/g carbon, where the urea concentration was 100 mg/L (see picture below). At greater concentrations (like 930mg urea per liter), lesser amounts of acivated carbon will become saturated; in other words the activated carbon will not be able to absorb more urea, but it might be able to effectively remove the tetrapyrroles or some other minor stuff.

But I think that the first thing to do is deal with the concentration. Urea is only about 0.93% of urine. So you only have about 0.93g urea per 100ml of urine.

To concentrate urea: boil off the water (which will also kill bacteria). Urea doesn't boil. The melting point for pure urea is 132.7 °C and urea will begin to decompose at around 150 °C. The residue (which should look creamy) will have a relatively high urea concentration which can then be extracted with (warm/hot 60 °C) ethanol or diacetone alcohol. Salt and protein are significantly less soluble in alcohol(s), so they should hopefully be (mostly) removed at this step.

The solubility of urea in 95% alcohol (100 g/L), absolute ethanol (50 g/L), and methanol (166 g/L) at 20 °C (room temperature) sounds pretty fair. And from TECHNICAL MEMORANDUM No. 13/M/67 (Investigations of Crystal Modification of Urea) "in the case of diacetone alcohol the solubility [of urea] ranged from 2 per cent at 20 °C to about 10 per cent at 100 °C, offering possible development of a process yielding free-flowing crystalline urea."

The solubility of salt (at 25 °C) in ethanol (0.65 g/L) and methanol (14.9 g/L) is significantly lower.

Now, the solubility of a hodgepodge of stuff including tetrapyrroles is something that you will have to experiment with. But, many similar compounds are readily soluble in acetone and tetrachloroethylene (often had as brake cleaner). Urea should not be highly soluble in these two solvents, but I'm a afraid I can't find a reliable solubility listed online, and I happen to be out of acetone right now. Cleaning an alcoholic extract of boiled urine with tetrachloroethylene might require using a separatory funnel.

On a historical note, (around 1727) Herman Boerhaave was able to extract urea from urine by working with the concentated-by-boiling residue (having the consistency of "fresh cream"). After filter pressing the residue, he allowed it to dry for a year... until he obtained a solid mass with an unknown oily substance (on top).  He discarded the oil and rinsed the crystals (filtride) with cold water to remove salt, leaving urea. He then recrystallized the urea from hot water. Previously (in 1669), Hennig Brand discovered the philosophers stone (phosphorus) by cooking the bejeezus out of (lots of) urine.

Personally, instead of waiting a year, I would take a chance on extracting urea from the cream. Also, I read today that boiling urine causes irreparable stink-damage to a home; so get a hot plate and an exension cord and boil it outside. Another recipe (as if I had nitric acid to spare), calls for reacting concentrated urea (say 100g in 100ml water) to make urea nitrate. The addition of nitric acid is exothermic and urea nitrate is explosive, so make sure the solution doesn't boil off (ice bath)... and urea nitrate is precipitated in the freezer. Then after you have clean urea nitrate, you can neutralize the acid under alcohol to recover urea. I would reserve this (nitric acid) method for the final work up, using relatively clean urea.

Also, here's a couple of pay-walled articles that might be useful to you if you really want to do this: Solubility of urea in water-alcohol mixtures and Urea in the history of organic chemistry: Isolation from natural sources

Urea is soluble in alcohol and methanol while the salt isn’t, so you simply boil the urine to reduce its volume, add methanol, separate the methanol layer, add $\ce{NaOH}$ and a bit of $\ce{H2SO4}$ or $\ce{HCl}$ and evaporate the solvent. There should be urea pure enough for anything you want to do …


Water: What to use H3O+ or H+? Why can we use $\ce{H+}$ and $\ce{H3O+}$ interchangeably?

I have seen in many places that first the reaction is written including Hydronium ion but then in bracket its written that we can write it as $\ce{H+}$ for simplicity. I am not clear with this.
$$\ce{H2O +H2O <=> H3O+ + OH-}$$
OR
$$\ce{H2O(l) <=> H+(aq) + OH- (aq)}$$

Since your solvent is itself water, it makes no difference whether you use $\ce{H+}$ or $\ce{H3O+}$.
$\ce{H3O+}$ is basically the hydrated form of $\ce{H+}$. If you know, the oxygen atom in water contains two lone pairs. When it donates one of the lone pairs to the hydrogen atom which doesn't have any electrons, you get $\ce{H3O+}$.
So,
$\ce{H3O+}$ is not $\ce{H+}$
$\ce{H3O+}$ is $\ce{H+(aq)}$
This means that the aqueous form of $\ce{H+}$ is represented as $\ce{H3O+}$

In all cases, acids yield protons ( or hydronium ions H3O+) and bases yield OH- (hydroxide) ions in aqueous solutions.

The H3O+ ion is considered to be the same as the H+ ion as it is the H+ ion joined to a water molecule. The proton cannot exist in aqueous solution, due to its positive charge it is attracted to the electrons on water molecules and the symbol H3O+ is used to represent this transfer.

The equation can be written as:

H+ + H2O(l) → H3O+(aq).

This is hydrolysis as it is involving water as a reactant.

Consider the first equation in the question , the ionisation equation of water:

H2O(l) + H2O(l)→H3O+(aq) + OH-(aq)

The  H3O+ is the conjugate acid of H2O. So H3O+ is used as a shorthand for a proton in aqueous solution. In a non-aqueous solution the proton would form a different structure.

The second equation:

H2O(l) → H+(aq) + OH-(aq)

Shows that H2O is made up of equal parts H+ and OH- ions and is amphoteric (can be an acid or a base) having a deprotonated form  (OH-). The ionic component is at a very low concentration and a water molecule is generally considered covalent with a dipole moment favouring a slight positive charge.

The H3O+ ion concentration in pure water at 25° C is 10^-7 dm^-3. This can be written as:

[H3O+] = 10^-7

where the symbol [ ] means the "molarity of" (units in moles dm^-3).

The number of H3O+ and OH- ions formed by the ionisation of pure water must be equal ( from the equation):

[H3O+] = [OH-] = 10^-7).

This shows that pure water is neither acidic or basic, it is neutral. The product of [H3O+] = [OH-] is the ionic product of water.

[H3O+][OH-]=10^-7 × 10^-7 = 10^-14

shows that in aqueous (water) solutions, whether acidic, basic or neutral,  the product of the ion concentrations equals 10^-14.

Acidic solutions contain more H3O+ ions than OH- ions. For basic solutions it is the reverse.

Therefore a water solution is :
Neutral when [H3O+]= 10^-7.
Acidic when [H3O+] > 10^-7.
Basic when [H3O+] < 10^-7.


Two different electronic configurations for cerium Depending on the textbook there are two different electronic configurations stated for cerium. On the one hand

$$[\ce{Xe}]\mathrm{4f^15d^16s^2}\quad$$

and on the other hand

$$[\ce{Xe}]\mathrm{4f^26s^2}$$

Are both configurations possible? If not, why so?

The filling of electrons follows Aufbau principle. However, introduction of $\mathrm{d}$ and $\mathrm{f}$ orbitals and resulting poor shielding affects predicted energy order, and it consists last 2 or three shells incompletely filled. The energy associated with this subshells is nearly same, and one of the subshell of outer shell is bigger in size than subshell of inner shell. So, to perform minimum repulsion between electrons, one or two electrons jump to a bigger subshell (with similar energy) so this exceptions occur.
In case of cerium, energy of $\mathrm{4f}$ and $\mathrm{5d}$ subshells are quite similar and $\mathrm{5d}$ subshell is bigger than $\mathrm{4f}$ subshell. So 1 electron in $\mathrm{4f}$ jump to $\mathrm{5d}$ subshell.

The configuration of cerium is
Ce -> 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f1 5d1
https://en.wikipedia.org/wiki/Cerium

The filling of electrons follows Aufbau principle nicely till Z<20, however, introduction of d and f orbitals and resulting poor shielding affects predicted energy order. Lanthanoids have their atomic configuration with a variable occupancy of 4f energy level and a common 6s2 level.The energy level of the 5d sub-shell increases than that of the 4f sub-shell moving across period(57–71). As a result of the low energy of 4f sub-shell electrons occupy these sub-shells before occupying the 5d sub-shell. As we progress from atomic number 57 to 71, an electron is added in 4f energy level with increase in atomic number with the exception of some metals where the electron occupies a 5d energy level.

In case of Cerium, energy of 4f and 5d subshells are quite similar and thus explains the electronic configuration stated above.


Physical significance of magnitude of ΔG (change in Gibbs free energy) of a reaction In thermodynamics, we know that if $\Delta G$ of a reaction is negative, the reaction is spontaneous.
Suppose we are given with two spontaneous reactions ($\Delta G\lt0$) with corresponding magnitudes of $\Delta G$. Can we extract any other information from these magnitudes?

It is only a rule of thumb that a negative value for $\Delta G^0$ of a reaction means that a reaction is spontaneous.  If you start out with pure reactants (and no products present), all reactions are spontaneous, irrespective of the $\Delta G^0$.  A negative value for $\Delta G^0$ only signifies that the equilibrium constant is greater than 1.

The change in the Gibbs free energy for a process at a constant temp. and pressure is equal to how much non-expansion work it can do.


Which is more acidic HOF or HOCl? My attempt to the question:
We know that more polar the $\ce{O-H}$ bond, the more acidic is the compound. According to electronegativity, the $\ce{O-H}$ bond of $\ce{HOF}$ should be more polar than $\ce{O-H}$ bond of $\ce{HOCl}$ since electronegativity order is $\ce{F} > \ce{Cl}$. But we should not miss another factor! That is the backbonding of  2p electrons (lone pairs) of oxygen with the 3d vacant orbital of chlorine. This is not possible in $\ce{HOF}$ due to the absence of vacant orbitals. Backbonding reduces (maybe, I do not know) electron density in oxygen and therefore polarity of $\ce{O-H}$ bond increases.

If we compare the HOF and HOCl
1) In HOF , the oxygen atom cannot do backbonding since fluorine has no vacant orbital while in HOCl it can do hydrogen bonding it can do backbonding which reduces electron density on oxygen by a slight extent( Remember 2pπ-3pπ backbonding is not very much effective due to size difference of 2p and 3p orbitals) . But yeah this factor makes HOCl bond more polar .
But there are two more important factors-:

2[)INDUCTIVE EFFECT]-  among oxygen and fluorine , fluorine is more electronegative , so it partially pulls the sigma bonded electrons towards itself giving a  delta+ charge to oxygen .
While in HOCl , O is more electronegative than Chlorine so the opposite mechanism happens and O gets a slight negative charge.
So according to this HOF has increased bond polarity than HOCl.
3) This factor is the most important and sufficient to answer almost any acid comparison . Compare the stability of the ion formed after releasing H+ ( formally called as the conjugate base of the acid).
So the question left is simply the comparison between OF- and OCl- ,
 The first two factors again work here but another important factor is the 2p-2p electron repulsions( due to smalls ize of sigma bond the electrons in the orbitals of O and F repel each other. This factor  in itself is quiet dominating) present in OF- makes OF- less stable than OCl-

So according to me HOCl must be more acidic than HOF.

Another important thing i wanted to draw attention to is that HOF is not the common acid that can be kept in laboratory. It is quite reactive and reacts vigorously even at 0°C , measuring its acidity is quite diffcult . I also searched online but couldn't find any data concerning Ka values for ionisation of HOF.
I hope this helps.

According to me, $\ce{HOF}$ is more acidic than $\ce{HOCl}$.
As the $\ce{O-H}$ bond length decreases, the acidic nature of the compound increases.
In $\ce{HOF}$ , the $\ce{O-H}$ bond length is $\pu{96 pm}$ while in $\ce{HOCl}$, the $\ce{O-H}$ bond length is $\pu{97 pm}$ and so $\ce{HOF}$ is more acidic than $\ce{HOCl}$.


Finding molecular formula with limited information given I have recently stumbled upon a question I can't seem to solve :

  Chlorine can also be used to synthesise a bromine-containing oxoanion that is mainly used in continuous or batch dyeing processes. When chlorine is passed through an aqueous solution containing potassium hydroxide and potassium bromide, the bromide ions are oxidised to the bromine-containing oxoanion. Careful addition of aqueous lead nitrate to the resulting solution precipitates 4.17g of $PbCl_2$. When this precipitate is filtered off and the resulting filtrate is evaporated, 0.835g of a white crystalline solid is obtained. The white solid has the following composition by mass : $K$, 23.4%; $Br$, 47.8%; $O$, 28.8%. $\mathbb \ Determine\ the\ formula\ of\ the\ white\ crystalline\ solid.$

I attempted by first finding the empirical formula of the solid using by dividing the percentage compositions by their respective elements' mass number, and ended up getting $KBrO_3$ as the empirical formula, which would make my second step to find $n$, where the molecular formula of the white crystalline solid is $K_nBr_nO_{3n}$. However, how can I go about doing this without being given how many moles of the white crystalline solid was precipitated? I can't seem to construct the equations that could potentially lead me to the mole ratio between $PbCl_2$ and $K_nBr_nO_{3n}$. If possible, can anyone tell me how to get a better grip on forming equations in scenarios such as this? I have been struggling in this topic for a while and this is likely to be one of the key reasons why. Sorry for the lengthy question!

Ionic solids like NaCl, KOH, KBr, PbCl2 and KBrO3 have ratios of individual elements, but the "molecule", if you could use that term to describe it, would be the whole crystal, because all the ions are attracted to nearby ions, ad infinitum.

The white solid could be written as "K23.4% Br47.8% O28.8%", and every time this experiment is run, the same product with the same composition will be obtained. And it will behave exactly the same as the commercial product called potassium bromate with a label saying KBrO3. We prefer the simpler formula.

The numbers given for the weight of PbCl2 and the product KBrO3 do not define a specific reaction, because other reaction products and excess reactants are not specified. And to be realistic, it seems that the filtrate was not evaporated to dryness, because all these other components are not obtained. It seems that this question is posed not just to get you to divide 3 percentages by 3 atomic weights, but to decide what numbers are important to answering the question. The mole ratio between PbCl2 and KBrO3 is unimportant (it's 3) - you got the answer at KBrO3 and should stop there.

Remember this riddle: What has 4 legs and a tail, barks like a dog and concrete?
Answer: A dog. (I just threw in the concrete to make it hard.)

A common test technique of teachers is to throw in extra information to see if you know how to select the correct information to use.


What happens to acids/bases left out overnight? I conducted a titration experiment a few days back and was unable to neutralize the solution in time, so I had to leave the solutions of HCl and NaOH out overnight in order to use them the next day. Also, one of my partners blew into the buret to get a drop of NaOH out. At the end of the experiment, we found that we required 6-7 mL more base to neutralize the HCl solution; another group who was forced to leave their solutions overnight used 3 mL more even than we did.

My hypothesis is that CO2 from the air formed carbonic acid in both solutions, mainly due to the instance where my partner blew into the buret, and thus the acidity of both solutions required that we use more base than we would have originally. As far as I've researched, this seems to be the case. Just to be sure, though: is this a correct assumption? If not, what could be the cause of this discrepancy?

First of all, blowing into the buret seems like a less-than-safe thing to do. $\ce{NaOH}$ could easily fly out into his/her mouth and nose.

I would expect that your acid and base solutions were already at the equilibrium concentration of $\ce{CO_2}$ when you got them, since they were exposed to air. I think that dissolving large amount of $\ce{CO2}$ into solution at atmospheric pressure is no easy task. I just don't think small amounts of $\ce{CO2}$ would account for the difference.

When you say "left them out" do you mean uncovered? If so, there's probably a fair amount of evaporation happening. The analyte solution could be affected depending on your experiment, and the concentration of your titrant would go up, which means you'd actually need less of it to reach the end-point.

When acquiring both the Acid $\ce{HCl}$ and the base $\ce{NaOH}$, your teacher should have given you a certain amount of either one to start with. Using the data of the given quantity you can acquire the necessary amount of base/acid needed to neutralize the other using the concentration and volume. This is known as the theoretical value needed to carry out the neutralization reaction $\cdots$ and this is probably where from my understanding you misunderstood why you needed more base. Your titration was not wrong as the experimental value (value gotten when experiment is carried out) is usually higher than the theoretical value meaning if 23 ml was required but you used 28 ml, its fine. However the real answer to your question lies in paralax errors (known as reading errors), human errors and friction along the inside of the burret, which causes the discrepancy. Like for example if you wanted to get 200 mL of $\ce{HCl}$ .. you might of gotten 200.5 mL where as the other group got 201 or 202 mL etc.


What is the empirical formula of the polyethylene? Polyethylene is a polymer of only carbon and hydrogen.  If 2.300 g of the polymer is burned in the oxygen it produces 2.955 g H2O and 7.217 g CO2.  What is the empirical formula of the polyethylene?

So my method for this problem is
1) find the mass of carbon from CO2
2) find mass of H from H20
3) Find mass of O by subtracting
4) cont. answer with formula from % composition

Therefore:

(7.217 g x 1mole x 1 mole Carbon x 12 grams) /  44g CO2 = 1.96 g
(2.955 g x 1mole x 2 mole Hydrogen x 1g)     /  18g H2O = .328 g

1.96g / 12g = 0.163
.328  / 16g = .0205 (smallest #)

.163 / .0205 = 8 Carbon
.0205 / .0205 = 1 Hydrogen

The answer is CH2 :/, but I am having a hard time finding out where I slipped :(

Just some comments: First, it is poor practice to use constants with so few significant figures that the constants contribute to the error of the results. H has a mol mass of 1.00797, C 12.011 and O 15.9994 not using 1.008, 12.001, and 15.999 is sloppy/lazy. I got a mass of C as 1.969686 (ignoring significant figures "inside" a calculation is better than trying to manage them; apply significance to the results after you're done number crunching. You got a mass of 1.96. One of us is very wrong. A 2% error 1.96 vs 1.97 FROM CALCULATION is really inexcusable (unless you're doing it in your head, or need it to keep your spaceship from crashing in the next 2 minutes and are doing it on a wall with your own blood...). Next, your "calculation":(7.217 g x 1mole x 1 mole Carbon x 12 grams) / 44g CO2 is RUBBISH. The use of the unnecessary parentheses suggest you don't understand when to use them, but that is a minor issue. Dimensional analysis (working with the units of measure) of that calculation results in your 'final' units to be (grams² x moles x moles C ÷ grams CO2 )
This indicates a profound misunderstanding of what you are doing.
In a mole of CO2 there is 1 mole of C and two moles of O. You probably know this. That means there are 12.011 grams of C and 31.998 grams of O in one mole (44.009 grams) of CO2. The calculation is 12.011 (grams C)/(12.011 +31.998)(grams total) = 0.2729 (g C/g total) and there's nothing wrong with claiming this as 0.2729 without any units (g/g "cancel") and the ratio can be used without having to continue to manage its units making further calculations easier. It is true that 7 g of CO2 can be multiplied by 0.2729 to get 1.97 g of C. (note that if we carry through the units of the ratio 0.27 gC/gTot that the dimensional analysis gives us gTot (same as gCO2) x gC/gTot which gives us units of gC → showing that we've done it right (in other words, it can be useful to carry the units through calculations). You should NOTE that "moles" don't come into this, although arguably you could claim that it isn't 12.011 gC; it's 12.011 g C per mole (or per mole C). NONE of this justifies what you did in multiplying (not dividing!) by moles! Wrong, wrong, wrong. Doing the  calculation for the H, it is 0.33068 grams of H another HUGE mistake you made (0.3307 g of H is 0.3281 moles of H, which is, I suppose where you screwed up).
Think of dimensional analysis as a check that your calculation makes sense. What you did makes no sense at all, imho. But don't get discouraged; the process of understanding what you're doing, rather than throwing things at the wall takes practice time.


Are there other carbon free gasses that can be used for energy transport, apart from hydrogen? Various gasses are currently used to transport and store energy for heating and electricity generation.

Most of them contain carbon:
Methane, propane, butane, carbon monoxide and natural gas.

It could be of advantage to use gasses that do not contain carbon.
One example for that is hydrogen.

Are there other gases that theoretically could be used, if they could easily be created?
I assume most can not practically used because they are not trivial to generate, but I like to ignore that, and assume they can be created.

Are there other gases that could be used, in principle, as energy carrier?

Yes, ammonia has been cited.

Per a source, 'Ammonia as a Hydrogen Source for Fuel Cells: A Review', to quote:

  Like hydrogen, ammonia is carbon free and can be produced from any energy resource. However there are also some significant advantages in terms of storage and transport.

Boranes like $B_2H_6$ or $B_6H_{10}$ are also a possibility. Probably too expensive !


Why do Hydrogen and Helium have molar volumes higher than an ideal gas? In my chemistry textbook, there is a table of the molar volumes for different gases. Most of them are below 22.42 l/mol, the ideal gas molar volume, but I noticed that hydrogen and helium do not. Hydrogen is 22.433 l/mol and helium is 22.434 l/mol. Why is this? What makes these particles, with mass, have higher molar volumes than the basically mass-less particles of an ideal gas? And why does the molar volume increase when going from Hydrogen to Helium?

As hydrogen and helium's molecular weight are too less, the intermolecular attractions are also too less. So the a/V^2 is negligible.  So the v.d.o equation becomes P (V - nb) =RT . By which, it can be said as the pressure of these elements are too less and from Boyle's law , the volume is higher.


Can Carbon Form bonds without Hybridization? Carbon has two electrons in its p orbital which should be able to form bonds, are there any examples in which this occurs instead of carbon hybridizing before bonding?

Hybridisation is a mathematical concept. It is one way of deriving chemical bonds from elements but not the only way. A different one (that computational chemistry typically uses) would be to just plant atoms at certain positions in space, and then mix their unhybridised atomic orbitals with each other to see what comes out. One can then back-calculate this to localise the molecule-spanning orbitals to single bonds and thereby deduce a hybridisation that would have worked to build up the molecule. Take the following example of methane (taken from Professor Klüfers’ internet scriptum for basic and inorganic chemistry at the university of Munich):

On the left you have four hydrogen s-orbitals, on the right you have the orbitals of carbon. As you see, there is no hybridisation assumed at all and the ground state is taken to be a fully populated $\mathrm{2s}$ orbital and three $\mathrm{2p}$ orbitals populated with two electrons amoung them for a total of $\mathrm{2s^2~2p^2}$. The hybrid orbitals are gained from symmetry considerations (in a tetrahedron, the p-orbitals transform as $\mathrm{t_1}$) and hydrogen orbitals were selected as was seen fit. The electrons are added a posteriori. Spin conservation would have produced a twice-excited state with very high total energy, so immediate spin pairing is assumed.

To localise the bonds, linear combine all four orbitals. If you take every orbital with factor $+1$, you arrive at the $\ce{C-H}$ bond to the upper right hydrogen. By shifting around a factor of $-1$ among the three $\mathrm{t_1}$ molecular orbitals during linear combination, you can address every other $\ce{C-H}$ bond. The resulting bonding orbital looks like n $\mathrm{sp^3}$ orbital mixed with a hydrogen $\mathrm{s}$ orbital.

Hybridization isn't actually an operation carried out on the orbitals. It is an operation carried out on the mathematical functions that define the orbitals (refer chapter 1 of A Guidebook to Organic Reaction Mechanisms by Peter Sykes). As @Ivan Neretin said, Carbon, like all other period 2 elements, wants to attain octet badly. It does this by promoting two 2s electrons to 2p orbitals. The energy required for this is much less than that released during formation of two extra bonds (for eg.: formation of 2 C-H bonds in methane would release 97kcal/mol).

Now, carbon atom could form 3 directed bonds, which are mutually perpendicular, using three 2p orbitals and one non-directed bond using 2s orbital. Whereas in fact, it is known that the 4 C-H bonds are disposed at an angle of 109°28'. This can be accounted for if we use the concept of hybridization in functions defining the orbitals.


When are Kp and Kc values equal? The picture shows the correct answers to this question.

I'm not sure why these answers are correct. $K_p$ only counts with gases, while $K_c$ only counts with aqueous solution + gases. I assumed that any reactions with only gases, or a mix of gases and solids, will have an equal $K_p$ and $K_c$ value, but that is clearly not the case.

Disclaimer: The following information has been extracted from my own Chemistry lecture notes which was built on contributions from many other general chemistry textbooks.  It is a stereotypical example of relating $K_c$ and $K_p$.

$K_c$ and $K_p$ relationship

For a simple reaction involving gas phase substances

$$\ce{aA(g) <=> bB(g)}$$

The equilibrium constant can be written in terms of molar concentrations or partial pressures as

$$K_c = \frac{[\ce{B}]^b}{[\ce{A}]^a} \qquad\text{or}\qquad K_p = \frac{(P_\ce{B})^b}{(P_\ce{A})^a}$$

Assuming that the gas behaves ideally (a common and convenient assumption for general cases), we can write the ideal gas law for each gas such that

$$ PV = nRT \quad\to\quad P_\ce{A}V = n_\ce{A}RT \quad\to\quad P_\ce{A} = \left(\frac{n_\ce{A}}{V}\right)RT$$

$$ PV = nRT \quad\to\quad P_\ce{B}V = n_\ce{B}RT \quad\to\quad P_\ce{B} = \left(\frac{n_\ce{B}}{V}\right)RT$$

We note that
$$ \frac{n_A}{V} \equiv M $$

where $M$ is simply molarity (moles/unit volume).  Since $[\ce{X}] = M_x$, we get

$$P_\ce{A} = [\ce{A}]RT \qquad\text{and}\qquad P_\ce{B} = [\ce{B}]RT$$

Substitute these terms into the $K_p$ equation to give

$$K_p = \frac{(P_\ce{B})^b}{(P_\ce{A})^a} = \frac{([\ce{B}]RT)^b}{([\ce{A}]RT)^a}$$

and recognize that

$$\frac{[\ce{B}]^b}{[\ce{A}]^a} = K_c$$

Plug this relationship into the $K_p$ expression to give

$$K_p = K_c (RT)^{b-a} \quad\to\quad K_p = K_c (RT)^{\Delta n}$$

Where $\Delta n$ is simply the difference of the sum of the stoichiometric coefficients of the products and reactants from a balanced chemical equation like so

$$\Delta n = \sum (\mathrm{products}) - \sum (\mathrm{reactants})$$

It should be clear from this that the only way $K_c = K_p$ within the context of an ideal gas is when $\Delta n = 0$.

$$\begin{align*}
K_p = K_c(RT)^{\Delta n} \quad\to\quad K_p &= K_c(RT)^{0} \\
&= K_c (1) \\
&= K_c
\end{align*}$$

By visual inspection of your image, it is clear why these are the correct answers based upon the derived relationship offered here.

Kc=Kp when the mols of gas on the left side of the equation is equal to the number of mols of gas on the right side of the equation.


Why does soap reduce surface tension of water One explanation I read is that soap's long non-polar end will seek the air at the interface between water and air, and that the ionic end will form ionic-dipole bonds with the water. In the process of moving to the surface, the soap molecules force apart the water molecules, and hence the water molecules no longer have hydrogen bonds with each other, and the surface tension is weakened. However, wouldn't the ionic heads of the soap molecules just form stronger ionic-dipole bonds with the water, increasing inter molecular forces, and therefore increasing surface tension? What is the true explanation for why soap reduces surface tension of water?

Apoorv did not give the right explanation, because this would imply, that surface tension goes especially down, when micelles are formed, but the opposite is the case. Compare e.g.: first chapter of The Colloidal Domain by D.Fennell Evans and Håkan Wennerström.

First thing to note is that pure water has a higher surface tension than hydrocarbon chains which is due to the strong dipole dipole interaction.

If there is a very dilute solution of surfactant in water no micelles are formed but you have monomers "floating around". In contrast to ions in water they are not homogenously distributed, but they concentrate especially on the surface where their polar head is directed to the water and their hydrophobic chain is directed to the air. In the end you substituted water molecules with hydrocarbon chains which leads to a lower surface tension.

If you increase the concentration of surfactant, they start to form micelles in the water (and not on the surface). So after reaching this critical micelle concentration (called CMC) the surface tension stays quite constant. This constant behaviour is used experimentally to estimate the CMC.

See the figure below Equation 11:
http://infohost.nmt.edu/~jaltig/SurfaceTension.pdf

Please note that Apoorv described some kind of reversed micelle. The normal way is that they stick their polar head to the water and have their hydrocarbon chain inside the micelle.

In the picture the white group is the polar head:
https://en.wikipedia.org/wiki/Micelle#/media/File:Phospholipids_aqueous_solution_structures.svg

The polar end of the surfactant is attracted to the polar water molecule. When enough surfactant molecules attach to a water molecule, it gets covered in surfactant and forms a unit. These units have a weaker force of attraction because only the non polar head of the surfactant is exposed thereby lowering the surface tension.


Why does cinnamaldehyde pass the ferric chloride test? I was given some data to determine what the unknown compound was and figured out that it was cinnamaldehyde. After reviewing the classification test given to me by my professor, I was a bit confused as to why it would pass the ferric chloride test as well as pass a Lucas test but yet show no reaction.

Can someone explain to me why that is possible?

Also, in terms of solubility, why would cinnamaldehyde be soluble in sulfuric acid only and not both sulfuric acid and $\ce{NaHCO3}?$

I think the cinnamaldehyde gives $\ce{FeCl3}$ test due the similar reason why acetic acid can give $\ce{FeCl3}$ test as $\ce{Fe^3+}$ can act as oxidizing agent.

As in case of acetic acid $\ce{Fe(CH3COO)3}$ (red colour) is formed, a similar complex may be produced.


How many valence electrons does a copper atom have? The valence electron configuration of $\ce{Cu}$ is $\ce{4s^1 3d^{10}}$. According to this the answer is $11$, but shouldn't it be $1$ since the d shell is already filled up?

Yes, copper only has 1 valence electron. Remember: valence electrons only include the electrons in the highest energy (n) shell.


On the origin of Gibbs Free Energy It is intuitive to me that for a spontaneous process involving any system i.e. one occurring without constant human interaction $$ \text{d}S \geq 0 $$ as I can observe increasing energy dispersal in everyday life, for example my tea turning cold.

What is not intuitive to me is that for all systems $$ \text{d}S \geq 0  ⇔\text{d}G = \text{d}H-T\text{d}S-S\text{d}T  \leq 0 $$

Why is this? I imagine that the explanation requires an understanding behind the origin of the definition of Gibbs free energy, G = H - TS.

The Gibbs energy has not been derived this way. It came from a thought process. About 100 years ago, Gibbs was puzzled by the very existence of spontaneous endothermic reactions. In mechanics, objects always fall down. They never "fall up" spontaneously. They never gets more energy spontaneously. In exothermic reaction, the matter looses spontaneously energy, like a falling object. In endothermic reactions, the atoms and molecules receive energy from the surroundings. They increase their energy. It is surprising. So Internal energy U and Enthalpy H are not equivalent to the gravitational energy. A spontaneous reaction can happen in chemistry with $\Delta$H <$0$ ou >$0$. As a consequence, H or U are not the good criteria to foresee whether a chemical reaction will spontaneously happen.

So Gibbs was thinking : What is the energy that always gets out of a system of molecules and atoms in a spontaneous reaction, even when the reaction is endothermic ? The answer is of course : the electric energy. A galvanic cell can only produce energy, whatever its heat effet. A working galvanic cell may become hotter or cooler, but it can only deliver electric energy. The amount of electric energy delivered by a cell is defined by the difference of a new potential energy, called Gibbs energy $G$, and defined so that $\Delta$$G$ is always negative in a cell. $\Delta$$G$ is obtained form the expression : $\Delta$$G$ = - $zEF$, where $E$ is the voltage of the cell, $F$ is the Faraday, and $z$ is the number of electrons exchanged in the reaction inside the cell. Furthermore, $\Delta$$G$ is defined and calculated in the same way as $\Delta$$H$. Gibbs energies of formation exist like enthalpies of formation.

For example, one may write : $\Delta$$G_{react} = \Sigma$$G_{fin}$-$\Sigma$$G_{in}$ and so on. You probably know the rest to the theory. And for example the relationship between Gibbs energy and the equilibrium constant.

Now if you plot $\Delta$$H$ and $\Delta$$G$ of a given chemical reaction versus $T$. you will notice that $\Delta$$H$ practically does not change much with $T$. But the voltage $E$ and of course $\Delta$$G$ do change much with $T$. The line $H$ vs. $T$ is nearly horizontal. But the line $G$ vs $T$ is inclined. The strange result of this drawing is the fact that both lines are joining one another at $T$ = $0$ $K$. So the difference of these energies is of course : $H-G$ = $T\Delta S$, where $\Delta S$ is the slope of the $\Delta G$ curve.


Why is Ag+ a soft acid? A soft species is typically large, polarizable and has a low charge (magnitude). The silver cation is not highly charged (this supports softness) however, it is anomalously small due to high nuclear charge and poor shielding by the d-orbitals. Consequently, shouldn't it also be less polarizable?

It does have a tendency to form bonds that are significantly covalent in nature which backs up the assertion that $\ce{Ag^+}$ is soft; however, I don't understand why.

According to HSAB theory, Soft Acids have:

large atomic/ionic radius;

low or zero oxidation state bonding;

high polarizability;

low electronegativity.

The ion $\ce{Ag+}$ satisfies all the aforementioned properties.
I don't agree with you when you say it is anomalously small due to high nuclear charge and poor shielding by the d-orbitals. The effective nuclear charge has nothing to do here.

According to http://www.webelements.com/silver/atom_sizes.html: The radius of this ion is $114\ \mathrm{pm}$ when it's in 4-coordinate, tetrahedral geometry and it's $116\ \mathrm{pm}$ when  it's 4-coordinate, square-planar geometry. Please just compare these radii with the radius of  $\ce{Al^3+}$ ion : $53.5\ \mathrm{pm}$ which is considered as a hard acid.
I hope it's clear now!

probably it is related to its filled $\mathrm{d}$ orbitals. Filled or nearly filled orbital elements counted as soft acids, and $\ce{Ag+}$ has totally filled $\mathrm{d}$ orbitals, and it can donate to ligands, making ligand-$\pi$ bonding, which causes stronger interaction.


Assay of stainless steel I frequently need to discriminate between 304 and 316 stainless steel which cannot be done by eye.

304 SS has 18% chromium and 8% nickel

316 SS has 16% chromium, 10% nickel and 2% molybdenum

How can I do a chemical assay to discriminate between them and ideally identify their precise composition to 1% or better accuracy? I assume the basic procedure is to create a solution, then do a titration of some kind.

If the first step in your answer is... "dissolve the sample in acid," be aware that stainless steel is pretty resistant to most acids and I need a procedure that will work within 24 hours.

Also, please no answers like "use an xrf machine" or some surface detection technology. I need a chemical process that assays the whole sample, not just its surface.

Digest a sample of the metal in aqua regia, I have found that the gauze in a kitchen sieve can be digested in a very short time with a mixture of 16 M nitric acid and concentrated hydrochloric acid.

Next you should dilute it with water and filter it to remove any particles.

I would use either flame AAS or ICPOES to measure the metals after dilution to a suitable concentration in plastic ware. I would want to measure the Fe, Ni, Cr and Mo in the sample. The Fe to Ni and Fe to Cr ratio may help you distingish between a lot of grades of stainless steel.

One alternative for iron is to use 2,2'-bipy and hydroxylamine to form the deep red iron(II) tris bipy complex.

This can be measured with a UV / vis machine in a 1 cm cuvettes, you will have to make a series of solutions in volumetric flasks. You make 0, 1, 2, 5, 10 and 20 ppm iron by adding a 1000 ppm standard to the flasks, then add a solution of 2,2'-bipy in water, the hydroxylamine and then dilute to the line. Mix them well and then measure the red colour with the UV / vis machine.

I am sure that there are colourmetric methods for Mo and Cr but sadly I do not know them.

Asking this question, I assume you are not a chemist, and neither have a lab or a chemist available in the place you want to do this, right?

Buy yourself a handheld XRF analyser. They look like a toy pistol, come nicely coloured and with touchscreen displays, and there are a dozen or so companies building them. Google gives you many hits, although no prizes. I'll guess you can get one in the 2-3000 € region? Definitely cheaper than hiring a chemist, or starting to fiddle around with acids and other stuff yourself, if you need to do it on a daily basis, and professionally.

(The machines on wikipedia https://en.wikipedia.org/wiki/X-ray_fluorescence look very intimidating, google has much nicer pics https://www.google.com/search?q=handheld+xrf . ;-)


Comparing the Lattice energy of different compounds I am confused about some concepts related to Lattice energy.

First of all,how to compare lattice energy between $2$ compounds belonging to different groups and periods?

1.Between $\ce{AlF3}$ and $\ce{MgO}$ (In $\ce{AlF3}$ product of ionic charge is $3\times1$ and in MgO it is $2\times 2$. So,according to me $\ce{MgO}$ should have higher lattice energy but the opposite is true.

2.Between $\ce{KF}$ and $\ce{LiF}$ (According to Fajan's rule $\ce{KF}$ is more ionic than $\ce{LiF}$ and lattice energy depends directly on ionic character).But, value of lattice energy of $\ce{KF}$ is $\pu{808 kJ mol-1}$ and $\ce{LiF}$ is $\pu{1030 kJ mol-1}$. How is this possible?

So,can you please explain me how to compare lattice energies of different compounds and which factor is more dominating size of cation and anion or charge on ions.

You might want to look at

https://scilearn.sydney.edu.au/fychemistry/prelab/e12.shtml

and the related page

https://scilearn.sydney.edu.au/fychemistry/calculators/lattice_energy.shtml

which has a nice lattice energy calculator that allows you to play with the parameters and see how the lattice energy varies, but I'll try to summarise those the argument below.

Here we are dealing with the ionic model - everything is totally ionic, there is total charge separation, all binding is electrostatic. Thus our system consists of a set of point charges. We can write the energy of such a system as

$$
E=N_AM \frac{z_1 z_2 e^2}{4 \pi \epsilon_0 r}
$$

where $N_A$ is the Avogardro constant, $z_1$ the charge on the first ion, $z_2$ the charge on the second, r is the closest separation of the ions and M the Madelung constant. Now note the Madelung constant only depends upon the structure of the crystal - all Sodium Chloride type crystals, whether it be NaCl, KF, or FeO, all use the same value off M. Looking at the formula we can see that FOR A GIVEN STRUCTURE (i.e. with the same M)

High charges on the ions mean high lattice energy
Small separation means high lattice energy

Thus point 2 addresses point 2 in your question, K is bigger than Li, hence the separation is bigger in KF than LiF, hence KF has a lower lattice energy than LiF. The first point explains why MgO has a higher lattice energy than NaF. However it doesn't cover your first example, as $\ce{AlF3}$ and MgO have different structures (and actually I don't think it is a very good question).

Comparing different structures is very difficult. About the best we can do is the electrostatic term from the Kapustinskii equation (https://en.wikipedia.org/wiki/Kapustinskii_equation)
$$
U_{L}={K}\cdot {\frac {\nu \cdot |z^{+}|\cdot |z^{-}|}{r^{+}+r^{-}}}
$$
This is an approximation to the above. Here K is a constant which is INDEPENDENT of structure and $\nu$ is the number of ions in the empirical formula, 2 for MgO, 4 for $\ce{AlF_3}$. Thus assuming ${r^{+}+r^{-}}$ is the same for both of these $\ce{AlF3}$ has a bigger lattice energy because $\nu\cdot |z^{+}|\cdot |z^{-}|$ is bigger - For $\ce{AlF3}$ it is 4*3*1=12, while for MgO it is 2*2*2=8. What this is really saying is that the $\ce{AlF3}$ structure has a much larger Madelung constant than the NaCl one, enough to overcome the charge differences, and thus it has a higher lattice energy.

in AlF3  in case of each AlF bond,  charge of Al =3
=> product of charges in case of each bond=3
=>sum of products of charges =3x3=9>4
so, AlF3 has lattice energy more than MgO.


Why do other sugars melt whereas sucrose decomposes? Sucrose decomposes at 186 °C. It doesn’t melt but when it reaches decomposition temperature, sucrose decomposes into fructose and glucose.

After cooling and hardening it is no longer sucrose. Am I right?

You're mostly right.  Sucrose is a disaccharide composed of glucose and fructose.

It is a crystalline material.  Heating sucrose results in a complex thermal process that involves both melting and decomposition.  Depending how fast you heat it, you will observe different things.  Basically, sucrose will start to decompose when it starts to melt.  If there is some water present during the heating process, then a reaction known as sucrose inversion will occur and the sucrose will break down into its components, one molecule of glucose and one molecule of fructose.

However if dry sucrose is heated to its melting \ decomposition point, then more complex things happen.  Significant amounts of glucose are initially detected.  Continued heating leads to "caramelization".  Caramelization is a condensation process where water is eliminated from sucrose; the initial loss of water is followed by isomerization and polymerization.  Over 4,000 products are thought to be formed during caramelization.  When this complex mixture is cooled, it is no longer sucrose.

when you melt it you break the bonds and if you break the bonds then it is not sucrose So what you are saying if it is melted at the right temperature for the right amount of time with the right amount of liquid then it will melt but it has to be anything but monosaccharides or disaccharides so no sugar in the liquid except for the sucrose


Propanoic acid reaction with potassium permanganate I was conducting a lab where we have to identify/justify how functional groups affect the oxidation of a molecule.

I used propene, propanone, and propanoic acid for this reaction. From what I have learned in class, ketones and carboxylic acids cannot get oxidized. However, in my experiment the ketone somehow is not getting oxidized while the propanoic acid is. I tried using butanoic acid just to be sure but it turns out that it gets oxidized as well. I have also made sure that there are no contaminants. I used 4 ml of 100% propanoic acid with 0.003 mol of $\ce{KMnO4}$.

Can you detail the concentrations of both the acid as well as $\ce{KMnO4}$ and also the time it took to de-colourise?

I tried an experiment however, I only have Glacial Acetic Acid, which didn't seem to react ( as the smell was still prominent).

I can suggest you a simple trick to analyse whether it is decomposing. Smell Propionic acid before adding $\ce{KMnO4}$. Then smell it after the colour of $\ce{KMnO4}$ is gone. Keep adding $\ce{KMnO4}$ till the pink colour stay(similar to titration). Then smell again. If the smell is gone then the carboxylic acid is oxidised.

There are 2 possible oxidation products of carboxylic acid:

Ketone
Carbon Dioxide

Both can be pretty obvious. Ketone will smell differently than the acid(more like nail polish remover) and Carbon Dioxide should be visible because of effervescence.

Could it be that there is an isohypsic (non redox) reaction that changes $\ce{KMnO4}$ into another $\ce{Mn(VII)}$ compound? According to wikipedia, it has been known to happen with other acids.

It would be nice to know if there were other methods available to analyze the reaction mixture (such as TLC).


How does one determine the probability of H⁺ to H₂O in pure water? The ratio of $\ce{H+}$ to $\ce{H2O}$ is something around $10^{-8}$. How did people determine that?

You cannot use the $K_\mathrm{w}$, $\mathrm{p}\ce{H}$, $\ce{[H+]}$ ion concentration value to answer because you get all that from the probability.                                                                   Is there any independent way to find the probability. Please elaborate on the steps.
I do not want to know why the ionic product is what it is. how was the ionic product found ?

The answer to this question was surprisingly difficult to find because all modern ways to measure pH use reference solutions of known pH and thus rely on prior knowledge of the $H^+$ activity of said reference solutions.

Finally I found this on wikipedia:

  it is possible to measure the concentration of hydrogen ions directly, if the electrode is calibrated in terms of hydrogen ion concentrations. One way to do this, which has been used extensively, is to titrate a solution of known concentration of a strong acid with a solution of known concentration of strong alkaline in the presence of a relatively high concentration of background electrolyte. Since the concentrations of acid and alkaline are known, it is easy to calculate the concentration of hydrogen ions so that the measured potential can be correlated with concentrations. The calibration is usually carried out using a Gran plot.[9] The calibration yields a value for the standard electrode potential, $E^0$, and a slope factor, f, so that the Nernst equation in the form
  $$E = E^0 + f\frac{2.303RT}{F} \log[\mbox{H}^+]$$
  can be used to derive hydrogen ion concentrations from experimental measurements of E. The slope factor, f, is usually slightly less than one. A slope factor of less than 0.95 indicates that the electrode is not functioning correctly. The presence of background electrolyte ensures that the hydrogen ion activity coefficient is effectively constant during the titration. As it is constant, its value can be set to one by defining the standard state as being the solution containing the background electrolyte. Thus, the effect of using this procedure is to make activity equal to the numerical value of concentration.

This means that the precise cocentration of $H^+$ is actually unknown what is known is it's chemical activity which has been defined as equal to the concentration for a certain standard solution.

You are, I believe, referring to the auto-dissociation reaction of water:

$$2\mathrm{H_2O}\text(l) \leftrightarrow \mathrm{H_3O^+}\text(aq)+\mathrm{OH^-}\text(aq)$$

This equilibrium reaction obeys the following equation (the square bracketed quantities are concentrations):

$$K_W=[\mathrm{H_3O^+}][\mathrm{OH^-}]=10^{-14}\:\mathrm{mol^2/L^2}$$

At $pH=7.00$ and in pure water this means:

$$[\mathrm{H_3O^+}]=[\mathrm{OH^-}]=10^{-7}\:\mathrm{mol/L}$$

$$pH = -\log[\mathrm{H_3O^+}]=7.00$$

The concentration of water in water is $55\:\mathrm{mol/L}$, so the ratio of $\mathrm{H_3O^+}$ to $\mathrm{H_2O}$ in pure and neutral water is about: $$1.8 \times 10^{-9}$$

Naked (non-solvated) protons $\mathrm{H^+}$ do basically not exist in water because $\mathrm{H_2O}$ is such a hard Lewis acid (nucleophile), which immediately bonds to any free protons.

The $pH$ of water is determined by means of a $\mathrm{H_3O^+}$-selective electrode, commonly known as a $pH$-meter.


Reversed size dependence in ion solubility My book says that solubility increases with interionic distance, because the attractive forces between ions gets smaller. However, if one of the ions in a binary ionic system is very large compared to the other, the trend is reversed. For example, CsF has greater solubility than CsCl. Why is this?

This can be explained with some crude thermodynamic models. The lattice enthalpy $\Delta H_\mathrm L$ of an alkali metal halide $\ce{MX}$ is given by

$$\Delta H_\mathrm L = \frac{k_1}{r_\ce{M} + r_\ce{X}}$$

and the hydration enthalpies by

$$\Delta H_\mathrm S(\ce{M}) = -\frac{k_2}{r_\ce{M}}\qquad  \qquad \Delta H_\mathrm S(\ce{X}) = -\frac{k_2}{r_\ce{X}}$$

where $k_1$ and $k_2$ are some positive constants and $r_\ce{M}$ and $r_\ce{X}$ are the ionic radii of $\ce{M+}$ and $\ce{X-}$ respectively.

The solubility is therefore a balance between these two quantities: larger (magnitudes of) solvation enthalpies increase solubility, but larger lattice enthalpies reduce solubility. The solubility $S$ is thus related to the difference of these terms. We ignore the entropic contribution, or to be precise, we assume that $\Delta S$ does not vary too much between different salts $\ce{MX}$.

$$S \sim \frac{k_2}{r_\ce{M}} + \frac{k_2}{r_\ce{X}} - \frac{k_1}{r_\ce{M} + r_\ce{X}}$$

Let us now assume that

$$r_\ce{M} = a+b \qquad \qquad r_\ce{X} = a - b$$

where $a,b$ are more positive constants. Here we have implicitly assumed that $r_\ce{M} \geq r_\ce{X}$, but it works the other way round too, so there is no loss of generality. Thus

$$\begin{align}
S &\sim \frac{k_2}{a+b} + \frac{k_2}{a-b} - \frac{k_1}{2a} \\
&= \frac{2ak_2}{a^2-b^2} - \frac{k_1}{2a}
\end{align}$$

This is clearly a minimum when $b = 0$, i.e. $r_\ce{M} = r_\ce{X}$. So, salts with similar cationic and anionic radii are less soluble, and vice versa.

Lattice enthalpy inversely depends on size difference.also lattice enthalpy inversely related to solubility.
Thus as size difference increases solubility increases


Where do the percents in empirical formula questions come from? A typical empirical formula problem reads something like this:

  Determine the empirical formula for a compound that contains 35.98% aluminum and 64.02% sulfur. source

How would those percentages be determined in real life? Spectroscopy?

A percent composition problem would obviously yield percentages, but those start with compound formulas--there'd be no need to determine the empirical formula from the percentages.

How would those percentages be determined in real life? Spectroscopy?

You are right, this analysis comes in the domain of analytical chemistry. The fundamental requirement for formula determination is that your sample must be extremely pure.

There is a whole field. So there are two broad classes of approaches. One is called classical analysis where one would dissolve the sample in an "appropriate" acid or base mixture, and this is followed by target element specific titration or precipitation. This technique results in very reliable data but it is very time consuming

Today most people would use an instrument called the inductively coupled plasma optical emission spectroscopy (ICP-OES), you "srpay" the dissolve sample into a state of matter (plasma) which is as hot as the surface of the Sun. Under those those conditions, compounds break down into their constituent atoms and emit characteristic wavelengths.
For example, if we are interested in knowing how much aluminium is there, we would construct a calibration curve and interpolate the amount of Al in your unknown sample.
The same thing (although less effectively) can be done for sulfur.

Sometimes you need both classical or instrumental analysis.

These percentage are experimental. Suppose you mix 36 g aluminium powder and 64 g sulfur powder, and you heat the mixture. A exothermic reaction will occur. As a final result, a pure compound is produced, which is aluminium sulfide. With any other proportion, the same aluminium sulfide is formed, but some aluminium or some sulfur remains unchanged at the end of the reaction.


pH 10 to pH9 NaOH dilution I have a pH 10 NaOH solution. I need to bring down the pH to 9. So I used the $C_1V_1 = C_2V_2$ approach.

$C_1 = \mathrm{0.0001~mol/L}$

$V1$ is the necessary amount to produce 10 L of pH9

$C_2 = \mathrm{0.00001~mol/L}$

$V_2 = \mathrm{10~L}$

\begin{align}
C_1V_1 &= C_2V_2\\
0.0001~\mathrm{mol/L}\cdot V_1 &= 0.00001~\mathrm{mol/L}\cdot 10\mathrm{L}\\
V_1 &= \frac{(0.00001)(10)}{(0.0001)}\\
V_1 &= 1\mathrm{L}\\
\end{align}

I Applied this calculation in reality and used a pH probe and got a pH of 6.5. I think the drop is too sudden in this type of titration. What should I do?

Generally, you don't make pH of $7\pm2$ by dilution alone. The required dilution is so great that it would leave very little of your compound, and then any tiny impurity would affect the result greatly. Think of a droplet on the pH meter probe left from the previous measurement. Distilled water does not help much. All the same, it contains $\ce{CO2}$ from the air, so its pH is usually about 5.5, not 7 as one might expect.

When people want to have certain pH level more or less reliably, they use buffer solutions. Come to think of it, that's what nature itself does, because our blood is a buffer solution as well.

the dilution increases the pH because the base will be completely sorrounded by water molecules.therefore more of the base will ionise Into constituent ions to produce fewer hydroxide ions than before dilution which increase the PH.


Can we find 100% probability region in an orbital if the position of an electron is limited by speed of light? The square of the wavefunction gives probability density of finding an electron somewhere in the orbital.

The text I'm referring to says that the value of probability density is always higher than zero at any finite distance from the nucleus.

My question is, provided that we knew position of an electron at a give time, could we now draw a boundary surface diagram that encloses a region in which probability of finding the electron is 100% for some point in time in future?

This would be possible because the electron cannot travel faster than the speed of light, so we could draw a sphere of radius $3*10^8$ meters centered at the current known position and say that probability of finding the electron after one second will not be beyond the region enclosed by the sphere.

Or is this question invalid because we can't know the position of an electron ever, not even within a finite uncertainty that would allow us to draw such a sphere?

The (Wave)functional forms for the surface which contains the 100% probability for finding the electron extends to the literal edge of the universe.  In the context of 'wave' it is nonsensical to measure an electron's position, and hence its speed may not be determined.

In the context of particle, an electron, having nonzero mass, may not be accelerated to the speed of light.

The probability density  has some value howsoever small it may be, at any finite distance from the nucleus. It is therefore not possible to draw a boundary surface diagram of rigid size in which the probability of finding the electron is 100 percent.


A few questions about the conditions of the Diels Alder Reactions I'm attempting to extract the diels alder reaction into a generic reaction and am trying to gain a better understanding of it.

Can a conjugated diene react with another conjugated diene via the diels alder reaction? And is the reaction limited to only dienes or can it be expanded to polyenes?

Does anyone happen to have a link as to the details on hetero diels alder reactions and the structures that can undergo such reactions and limitations on chemical species that can react via this reaction.

Yes, conjugated dienes can react with other conjugated dienes in Diels-Alder reactions. A well known example of this is the dimerization of cyclopentadiene. All the standard caveats, restrictions, and other considerations still apply, namely:

The diene must not be rotationally locked in such a way that it cannot adopt an s-cis/synperiplanar conformation. As a logical corollary, cyclic dienes that are already locked into the correct s-cis conformation often have lower activation barriers.
The kinetics of the reaction are accelerated by electron-withdrawing substituents on the dienophile and electron-donating substituents on the diene.
The stereochemistry of the major product is highly substrate-dependent. A preference for the endo isomer exists in cases where the dienophile has unsaturated substituents. In those cases, the transition state is stabilized by an overlap of the $\pi$-electron system between said unsaturated substituents and the $\pi$ orbitals of the newly forming double-bond at the back of the diene. (Just as a side note, Carey & Sundberg refer to this as the "Alder rule" in Advanced Organic Chemistry, but I've never seen that term used anywhere else.)

In principle, there's no reason why two conjugated $\pi$-bonds of a polyene cannot react similarly, provided they satisfy the orbital symmetry and conformational requirements and the reaction kinetics are favorable. Even aromatic molecules can undergo Diels-Alder reactions under the right set of conditions (see, for example, the Diels-Alder reaction between benzene and furan via a benzyne intermediate). There are also numerous instances of well-documented Diels-Alder reactions involving heteroatomic dienophiles, alkynes as dienophiles, polymerization reactions, etc.

If you can get your hands on the aforementioned Carey & Sundberg book, they discuss the Diels-Alder reaction at length (including relatively rigorous explanations of the stereospecifity and reaction kinetics via molecular orbital theory). If you're looking for a comprehensive reference, I doubt there's anything superior to the venerable March's Advanced Organic Chemistry. There's a chapter on Diels-Alder reactions (along with virtually every other organic chemistry reaction known to science), as well as a massive list of accumulated primary literature references if further detail is required.

1.Is an alkyne can be a dienophile in Diels Alder Reaction?Discuss an example and unsaturation in product?
Answer:
An addition reaction that leads to the formation of a ring is called cycloaddition.and the Field Alder reaction is a member of a class of a reaction is called cycloaddition .
The dienophile does not have to be an alkene specifically,the only requirement is that it contain a pi bond. So,It can also be an alkyne.
Example:
The reaction of 1,3-butadiene with ethyne to form 1,4-cyclohexadiene.
Normal Diel-Alder reaction are favoured by the electron donating diene and the electron withdrawing form the dienophile.and in this reaction two new sigma bonds are formed at the expense of two pi bonds.When anthracene is boiled with maleic anhydride in the presence of xylene 9,10dihydroanthro-9-10-endo-alpha-beta succinic anhydride is formed.

Anthracene act as diene.
2)Xylene is used as high boiling temperature solved to proceed Reaction quickly.
Since,the reaction proceeds in a single step cycloaddition reaction. Here, two unsaturated molecules combine to form a cyclic adduct. Thus, diene and the dienophile react to give a cyclohexene derivative.


Why are water droplets shaped like that? With nothing to do, I stared at the droplets that get condensed on the glass panel of my window.

Upon examination, these droplets appear to be in some sorta pattern. There are big droplets as well as small ones that fill the space between the big ones.

The ones in the middle, between rows of small droplets are usually big one.

All these intrigue me to think about what are that factors that contribute to this pattern.

My guess: I think vapor gets condensed on dust particles thus forms those droplets in the picture. But what accounts for the different sizes of the droplets?

From the Wikipedia article on surface tension:

  Surface tension is responsible for the shape of liquid droplets. Although easily deformed, droplets of water tend to be pulled into a spherical shape by the cohesive forces of the surface layer. In the absence of other forces, including gravity, drops of virtually all liquids would be approximately spherical. The spherical shape minimizes the necessary "wall tension" of the surface layer according to Laplace's law.

In short, the more surface tension is, the rounder shapes of water you get. And the opposite goes for gravitational potential energy: The lesser gravitational acceleration results in more spherical droplets of water.

The symbol for surface tension is $\gamma$.

  $\gamma (\ce{H2O}) = 72.8~\mathrm{dyn~cm^{-1}}$ (at $20~\mathrm{^\circ C}$)
   $\gamma (\text{mercury}) = 465~\mathrm{dyn~cm^{-1}}$ (at $20~\mathrm{^\circ C}$)[1]

That's the reason you hardly ever see mercury drops out of their spherical shape.

In short

  The spherical shape minimizes then necessary "wall tension" of the surface layer according to Laplace's law.[2]

Oh and I almost forgot: This great article - Antonin Marchand, Joost H. Weijs, Jacco H. Snoeijer, and Bruno Andreott, Why is surface tension a force parallel to the interface?. Physique et Mecanique des Milieux Heterogenes, 2012 - is very nice in case you wanted to do additional study in the case.

References:

R Nave. Surface Tension. URL (accessed April 25th, 2018)
R Nave. Surface Tension and Droplets. URL (accessed April 25th, 2018)

for a given volume sphere is having minimum surface area as compare to circle it means sphere have less surface tension as compare to circle (surface tension have tendency to the liquid to keep it free surface area minimum)
 which in turn means minimum surface area


Gas Law problems with stoichiometry Consider the following reaction : $\ce{H2 + Cl2 -> 2HCl}$

A) If you have 15.00 liters of hydrogen reacting with 10.00 liters of chlorine at STP, how many liters of HCl would you produce?

B) What would be the limiting reagent?

C) How many moles of excess reagent would remain?

Please explain.

Molarity = moles/Liters
to check for $\ce{H2}$: M = 2/15.00 = .13334
to check for $\ce{Cl2}$: M = (35.45 x 2)/10.00 = 7.09

But I do not think that is correct. can somebody show me the correct way to do it.

Avogadro's law says that $V/n = \text{constant}$ at constant temperature and pressure.   That means your volumes can be used "as if" they were moles in the chemical reaction stoichiometry.   So, if the mole ratio is 1:1 for H$_2$ and Cl$_2$ reacting, then the limiting reagent is obviously the smaller volume of 10.0 L for Cl$_2$.

For (b) you can easily calculate the volume of HCl since 1 mole of Cl$_2$ produces 2 moles of HCl.  So, you get 20.0 L of HCl.

For (c) the excess reagent is the H$_2$.  Since it reacts with 10.0 L of Cl$_2$ in a 1:1 mole ratio, then you have 15.0L - 10.0 L = 5.0 L of H$_2$ left over.   Since this all ends up at STP, you can use the ideal gas law to calculate the moles of H$_2$ left over:

$n_{H_2} = \displaystyle \frac{p V}{R T} = \frac{(1 \text{atm})(5.0 \text{L})}{R \times 273.15 \text{K}} = 0.2230749373505515$ mol

Don't forget sig. figs., which in this case is determined by the 2 sig. figs. in the volume of 5.0 L, giving you

$n_{H_2} = 0.22$ mol

For Review check out:
gas laws: http://www.grandinetti.org/Teaching/Chem122/Lectures/Gases

If you have a mac, check out PhySyCalc. It will automatically do all the unit conversions in your calculations.

https://itunes.apple.com/us/app/physycalc/id644105995

You only need to know that at STP $V_m=24.4\:\mathrm{L\:mol^{-1}}$, and that $n$ (number of moles of of a compound or molecule) is equal to $\frac{V_{gas}}{V_m}$.

Find limiting and excess reagents using stoichiometric ratios (in this case it is 1:1), therefore $\ce{H2}$ is limiting the limiting reagent.
Use stoichiometry to find amount of $\ce{Cl2}$ reacted, and then subtract the initial number of moles from the final number of moles.


Polarity index vs. Dipole moment I'm looking to find whether water or methanol is more polar, and I'm getting conflicting answers.

My textbook says "the polarity of a bond is quantified by the size of its dipole moment." According to this table, at 20 °C, the dipole moments of methanol and water are 2.87 D and 1.87 D. This implies to me that methanol is more polar.

But there seems to be another measure called "polarity index," which I can't seem to find a definition for online. This table says the polarity indices of methanol and water are 10.2 and 5.1, which implies to me that actually water is more polar.

So what's the difference between the dipole moment and the polarity index? How is the latter defined? And is methanol or water more polar?

From Roger E. Schirmer "Modern Methods of Pharmaceutical Analysis":

  Solvents are generally ranked by polarity, but polarity is not a uniquely defined physical property of a substance. Hence the relative polarity of a solvent will be somewhat dependent on the method used to measure it....

  Solvent polarity is a complex function of many parameter in addition to adsorption energy. A more recent ranking of solvents by Snyder is based on a combination of parameter such as dipole moment, proton acceptor or donor properties, and dispersion force solvent...

  Snyder's polarity index ranks solvents according to a complex theoretical summation of these properties. As a rule, the higher the polarity index, the more polar the solvent.

Snyder's paper was published in   Journal of Chromatography A, Volume 92, Issue 2, 22 May 1974, Pages 223-230 . Take a look at it to see how the index was calculated.

There are different types of polarity indices. And each of them has different parameters and ways to calculate the polarity.

Unlike polarity, dipole moment is a physical property. Polarity indices often take dipole moments as a parameters when calculating the polarity of solvents.  The reason why there are polarity indices is because dipole moments alone couldn't explain the nature and interactions of solvents.

Part of your confusion stems from the repetition of the word "polar" to describe different phenomena. There is polarity of bonds and there is polarity of solvents. They are different things.

Badyal YS, Saboungi M-L, Price DL, Shastri SD, Haeffner DR, Soper AK. Electron distribution in water. The Journal of Chemical Physics. 2000;112(21):9206-8 gives the dipole moment of water 2.90 +-0.6D, what means that water is more polar then methanol.


Esterification with tertiary alcohol I saw a different mechanism for esterification of tertiary alcohol in a book.

They say that first the $\ce{OH}$ is protonated and it leaves to form the tertiary carbocation. Then the oxygen's lone pair from the carboxylic acid attacks to form the ester.

Is this mechanism actually correct?

Yes, the mechanism is correct and is different from the normal mechanism because here the more stable tertiary carbocation formation.


Does 3,6-di-X-cyclohexa-1,4-diene have stereo centers? The following question has multiple correct answers:

The book out of which the question is taken  shows that a, b and d are correct answers.

According to me c should also be correct, as it shows cis-trans isomerism, both X could go to the front (wedge bond), or one could go to the front (wedge) and one to the back (dash bond).  After all geometrical isomerism is also a sterioismersim.
I came to this conclusion considering the definition given on Wikipedia:

  A stereocenter or stereogenic center is any point in a molecule, though not necessarily an atom, bearing groups, such that an interchanging of any two groups leads to a stereoisomer.

Good point. The (c) compound has two tetrahedral stereogenic carbon atoms, with four “substituents”

$\ce{-H}$,
$\ce{-X}$,
$\ce{-CH=CH-\overset{\text{*}}{C}HX-C=C-\bond{...}}$,
and the same one, $\ce{-CH=CH-\overset{\text{*}}{C}HX-C=C-\bond{...}}$ but from the other side, thus with the opposite configuration, or its mirror image.

(‘*’ denotes (the other) stereogenic carbon atom)

They are called pseudo-asymmetric carbon atoms.

A different relative configuration at these two pseudochiral centers leads to these two ‘cis’ and ‘trans’ isomers.

The definition of stereogenic center (for more complete one, see e.g. this answer) is fulfilled.

(Maybe this topic is too early, for this textbook stage, and the author did not notice this complication.)

Now, the key to understanding stereoisomerism is that we have to look for elements of symmetry of which:

Plane of Symmetry(PoS)
Centre of Symmetry(i)
Alternating Axis of Symmetry(AAoS)
are the prominent ones.
Again, please keep it in mind that chiral centres may or may not enough to decide the chirality of the entire molecule(examples are meso compounds with two or more chiral centres, or "pseudo chiral" molecules which are chiral in one configuration but not in another)
Now, since you are clear on the other three options, I am going to jump straight to the third option.

Now, as you can notice, the molecule does possess a centre of symmetry. What this means is that now, when you will try to make those cis-trans configurations, you will always find a plane to cut through to make identical halves. Also, since they are present on a ring,so there is no scope of "weird" configurations like keeping both the X and H on cis or trans. Hence, it is safe to say that as soon as you have located even one symmetry element, you can call the molecule achiral.

If the molecule turns out to be achiral, so any change at any group of atoms won't give you a steroisomer. So it also won't possess any stereocentres


Why only one peak is observed in NMR spectrum of H2? Assuming a downward magnetic field is applied and the nuclear spin of two equivalent protons in $\ce{H2}$ are all in ground state $\downarrow\downarrow$ initially. Would the required energy of one of the protons excited to spin up state $\uparrow\downarrow$ be different from the energy of proton excited form state $\uparrow\downarrow$ to state $\uparrow\uparrow$ ?

And if it's true why only one peak is observed in NMR spectrum?

For starters, it is incorrect to think of the ground state being substantially more populated in NMR spectroscopy. Because the energy difference between aligned and misaligned nuclear spins in the external magnetic field is very low, we can say that both states are almost equally populated due to thermal excitation alone. For a $500~\mathrm{MHz}$ experiment, the energy difference can be calculated to be $0.21~\mathrm{J \cdot mol^{-1}}$ which corresponds to a population ratio as follows:

$$\frac{n_\mathrm{ex}}{n_\mathrm{g}} = \exp \left(\frac{\Delta E}{RT}\right) = 0.999916$$

Next, consider the extremely high symmetry of the $\ce{H2}$ molecule: It has the point group $D_{\infty\mathrm h}$, one of the most symmetric ones around. Due to this high symmetry, we cannot distinguish between either proton at all — both can be transformed into each other by a rotation as defined by the point group. Thus, the two nuclei are magnetically equivalent — not only must their chemical shift be identical but also the coupling to the other one.

Think of it this way: Assume, that the spin transition of one proton happened at a slightly different frequency than the other’s. That would mean that one proton had to be different from the other in some way (a different transition energy is equivalent to a different chemical shift is equivalent to a different environment). But how would you explain a different environment on one side of the $\ce{H2}$ molecule compared to the other? Correct: You cannot.

Now we could still assume that the energy required to excite a nuclear spin whose neighbour is parallel be different from one whose neighbour is antiparallel. However, no matter how we excite, the transition is always $\ce{parallel <=> antiparallel}$. And since we cannot distinguish between the hydrogens, we also cannot distinguish which one pointed in which direction, so those two energies must be identical. Therefore, we only see one peak and no $^1J_\ce{HH}$ coupling.

If by referring to peak you mean a single chemical shift — in low-resolution H-NMR — the number of peaks available on the spectrum would depend on the type of hydrogen environments there are. For example, ethanol ($\ce{C2H6O}$) has three different H environments: $\ce{C\mathbf{H}_3-C\mathbf{H}_2-O\mathbf{H}}$.

(source: rsc-cdn.org)
The H's in the two methyl groups are of different environments because one methyl group has its electrons pulled on stronger by the oxygen in the hydroxide functional group ($\ce{OH}$).
In that case, $\ce{H2}$ only has one hydrogen environment, with both hydrogen atoms in the same environment, resulting in a single chemical shift signal.
However, you could have meant the spin-spin coupling determined by high-resolution H-NMR — the one where the tinier individual peaks within a certain signal are visible. The number of tinier peaks that show up for a particular chemical shift is dependent on the number of hydrogen atoms on a neighboring carbon atom.

(source: alevelchem.com)
In ethanol, the hydrogen environment of the central carbon atom ($\ce{CH3-C\mathbf{H}_2-OH}$) would have a quartet (4 peaked) signal because the neighboring carbon has 3 hydrogen atoms. The number of peaks due to spin-spin coupling is = number of neighboring hydrogen atoms + 1
With respect to $\ce{H2}$, the hydrogen atoms have no neighboring hydrogen atoms — there is no carbon in $\ce{H2}$ — so a singlet or single peak would apeak above the chemical shift for $\ce{H2}$. Even if a compound contained carbon atoms, like the H-NMR standard tetramethylsilane $\ce{Si(CH3)4}$ only a singlet is produced because all hydrogen atoms are in the same environment of a methyl group attached to a silicon atom.


Does nitrogen inversion affect the basicity of amines? If I were to compare the basic strength of 1-azabicyclo[2.2.1]heptane and triethylamine:

Can I say that 1-azabicyclo[2.2.1]heptane is more basic than triethylamine because the lone pair of electrons is less available in the latter due to rapid nitrogen inversion? Nitrogen inversion is not possible in the bicyclic amine.

There is not much difference in the inductive effect. In both cases, the nitrogen atoms are $\mathrm{sp^3}$ hybridized.

Background

Amine basicity correlates with, among other things, the hybridization of the nitrogen orbital that is holding the lone pair of electrons.  The less s-character in this orbital, the more basic the amine. So, as the following figure indicates, a trialkylamine (the lone pair is in an $\ce{sp^3}$ orbital) is more basic (less acidic or higher $pK_{a}$) than the lone pair in pyridine (the lone pair is in an $\ce{sp^2}$ orbital), which is less acidic then the lone pair in a nitrile (the lone pair is in an $\ce{sp}$ orbital).

(image source)

The rate of nitrogen inversion also correlates with, among other things, hybridzation. But in cases where these "other things", such as ring constraints, limit nitrogen inversion, then the rate of nitrogen inversion may not correlate with hybridization. So the rate of nitrogen inversion does not always correlate with hybridization and is therefore not a good indicator of lone pair "availability".

Answer

Looking at your two compounds, a reasonable first guess might be that the nitrogens in these compounds are both roughly $\ce{sp^3}$ hybridized.  Therefore, even though one compound can undergo nitrogen inversion and the other can't, their hybridzations are similar and we would expect similar basicities.

We could refine this prediction by noting that in the bicyclic compound, due to geometric constraints, the carbon-nitrogen bonds around nitrogen are slightly pulled back a bit from the tetrahedral angle.  This would increase the p-character in these bonds (as a bond angle moves from 109° to 90° the p-character increases with 90° being pure p).  As a result, the p-character in the lone pair orbital decreases and the s-character increases.  As mentioned above, more s-character in the lone pair orbital leads to lower basicity.

Therefore we might expect both compounds to have similar basicity with the bicyclic compound perhaps being a bit less basic.

This analysis is corroborated by the $\mathrm pK_\mathrm a$ values of the conjugate acids (a more positive number means that the amine is more basic). According to J. Org. Chem. 1987, 52 (10), 2091–2094, quinuclidine has a $\mathrm pK_\mathrm a$ of 10.90, and 1-azabicyclo[2.2.1]heptane a $\mathrm pK_\mathrm a$ of 10.53. The decrease in basicity can be attributed to the increased strain and greater s-character of the lone pair. Triethylamine itself has a $\mathrm pK_\mathrm a$ of 10.75 (Evans table), making it marginally more basic than 1-azabicyclo[2.2.1]heptane.

The bridge head is the most unreactive, and thus if we see the conjugate acid of the first compound, it will be extremely unstable. Amine flipping does lower the basicity, but still in this context, the basicity of the second amine will be greater than that of the first.


How does one tell (or conclude) if a substance is carcinogenic? It's common to read statements like:

  Tetrachloroethene ($\ce{CCl2=CCl2}$) was earlier used as a solvent for  dry cleaning, but since it contaminated ground water and is a suspected carcinogen, it has been replaced by ... .
  Various hydrocarbons harms plant tissues, shed leaves, flowers and twigs. Many of them are also carcinogenic.

How do we tell or suspect one compound to be carcinogenic? And are all carcinogens organic compounds?

How do we tell or suspect one compound to be carcinogenic?

As written in the comments to the question, this the result of large studies on the human population, correlating blood or urine levels of some substance to instances of cancer. Alternatively or concurrently, studies on animal models, such as mice, rats or even in-vitro studies may be carried out. As a result, regulatory bodies may include a substance in lists, which may preclude them from use in consumer products, manufacturing procedures etc.

Are all carcinogens organic compounds?

No, this does not have to be the case: plain arsenic and basically all inorganic cadmium compounds have been classified as suspected or definitively carcinogenic.

Now arsenic has been mentioned and interestingly, inorganic arsenic compounds interfere with the basic mechanism of DNA repair, which apparently increases frequency of chromosomal aberrations. So this is chemically predictable.

Cadmium has also been mentioned, but what is little known is "even tumor formation [52] can be suppressed by the simultaneous Zn treatment" cited in this work. So chemically predictably treated.

For organic compounds, I suggest a simple process to determine the POTENTIAL carcinogenic ability of an organic compound (I believe this is what you are, in effect, asking in being able to 'tell').

In the lab or reported in the research literature, is it apparent that the compound readily undergoes decomposition creating (directly or otherwise) radicals or groups in an environment of aqueous saltwater containing transition metals, CO2, oxygen and naturally produced HOCl or other reactive oxygen species (ROS) at pH levels found in the human body?

Some generated radicals, like the hydroxyl radical and the carbonate radical, for example, can attack DNA as it would any large organic compound to extract a hydrogen atom.

Damaged DNA can lead to cancer cells.

While, for example, short-lived hydroxyl radicals are good in breaking down problematic pesticides in natural waters, they are not recommended in large amounts in the human body.

Note, I am not recommending the direct testing of the decomposition of DNA,  but just the expected (and conveniently measured) side products created with the associated breakdown of the organic compound in question. This may be suggestive of potential DNA damage, and at best, a predictive tool. Varying dosing levels may also be quite informative. This is categorically distinct from expensive short-term animal studies (examining tumors) and also investigations of long term low dose exposure and reported incidents of cancer in human populations with inferences drawn from correlation analysis.

If a compound displays a chemical potential to be carcinogenic, per above, you may still want to await the opinion of say Germany, England, or other countries with socialized medicine,.., to verify. But, in my opinion, I would just avoid/limit exposure and consume over the counter anti-oxidants (Vitamin C,...), especially if chemically related suspect compounds are now publicly believed to be carcinogenic.

Thanks Andrew for the reference to the Ames test! It, however, apparently employs "several strains of bacteria (Salmonella, E.coli) that carry mutation" which could limit the accessibility of the test.

DISCLOSURE: I have no financial interest, or am I employed, in the health supplement industry or the health care industry.


What advantages does the mole have over counting large numbers using SI prefixes? I'm trying to understand why chemists use the mole unit instead of just counting and using SI prfixes to simplify the big numbers with units such as zetta- or yotta-molecules (yotta=10^24).

Here's what I've found so far, but it doesn't seem to be enough to make moles particularly important:

Reasons for moles:

You can easily approximate the number of atoms from the mass of a sample because the mass number of an isotope approximately equals the mass of 1 mole of atoms in grams.
It's a historical convention that would be too expensive or uncomfortable to change now.
You can use millimole and micromole instead of having to learn several more obscure SI prefixes like exa and peta. EDIT this reason is added from Jan's answer
Shorter descriptions, eg "mole of carbon" vs "yotta atom of carbon" (from matt_black's answer).

Reasons against moles:

For accurate calculations (beyond about 3 significant figures), the relationship between mass number and mass in the first reason above breaks down. So this is a risky thing to use.
It's a additional concept and set of facts that chemists have to spend effort learning but which describes human convention rather than nature.
We need extra conversion factors such as 1/mol which people often neglect and Faraday's constant used in Q = n(e-) x F [1] which wouldn't exist without moles.

Non-reasons for moles:

You can easily calculate the number of atoms from the mass of a sample because the atomic mass of an element equals the mass of 1 mole of atoms in grams. This relationship only exists because of the special units commonly used for atomic mass. If periodic tables listed atomic mass in grams (or perhaps yoctograms), then we could do the same calculations just as easily without moles.
In practice, we can't measure the number of molecules so we have to measure mass or volume instead and should therefore count them using a unit that's defined in terms of mass. I don't think the precise details of how a unit is defined matter for practical purposes. If you have 1 litre of an ideal gas, you still have to do calculations to find out many moles it contains just as you would to find how many molecules it contains. There is even a proposal [2] for SI to redefine the mole to be independent of the mass of any substance, indicating that keeping the definition isn't very important.
Chemists would make mistakes with those big numbers. They'd use "yotta" in the same way they use "mole", not doing calculations with the actual number it represents and not being at risk of other types of error. (from matt_black's answer) However it would be more complex replacing millimole with zetta because you'd have to remember that a zetta-atom is a milli-yotta-atom.

Here's a similar question but it's mixed in with the idea of measuring number rather than mass or volume -
Why do people still use the mole (unit) in chemistry?

This other similar question mainly addresses the uncommonness of quantities as big as 1 mole in other areas of life -
The mole is used extensively in chemistry, why not elsewhere?

[1] http://www.ausetute.com.au/faradayl.html

[2] https://en.wikipedia.org/wiki/Proposed_redefinition_of_SI_base_units

Of course, it’s convention and has been so for a good century. And there is no real reason why this convention (and not a different one) happened in the first place — it is entirely conceiveable to define an ‘imperial mole’ so that the atomic mass of an element is equivalent to the same mass in ounces. 1 imperial mole of carbon atoms would then be twelve ounces of carbon atoms.

So there are basically two questions in this:

Why replace a large number with a unit just meaning ‘this certain large number’?
Why is the presently defined mole a good choice for this unit?

The thing about large numbers is that they are large. Almost everybody who grew up in a metric country can name at least three SI-prefixes: kilo, centi and milli. Thanks to IT, many people now also know mega and giga (and maybe tera), even if they don’t use metric units at home. But tera only gets you to $10^{12}$. We need some $10^{21}$ for moles.

I often work with milli- or micromoles of substances in my research. In plain numbers, that’s $10^{20}$ or $10^{17}$ — I don’t know those prefixes and thus would have to learn an entirely new subset. With the mole, everything one uses in the lab will nicely fall into something between nano and kilo.

It also helps to have a single unit there. Molar mass is expressed in grams per mole, concentrations in moles per litre and many more. If there were no unit, it would be simple grams per 1 or 1 per litre — precisely the reason why some people prefer to use rad or some other way to show radians rather than just writing the number. If the unit is there, you’re unlikely to forget it, you know if your calculations are good and more. If there was no name for this unit, it would have to be invented.

So if the mole didn’t exist, it should be invented for simplicity.

The good thing about the size definition of the mole is, as noted above, that it brings everything into one general range. Whether it’s mass, volume, concentration or amount, every unit is going to be prefixed by only a small subset of the SI-prefixes: kilo, milli, micro, maybe nano. Thankfully, those are the ones that are most used in everyday life, too (excluding nano and maybe micro).

It doesn’t really matter where one ends up. If the mole had originally been defined imperially,[1] that would be fine, I wouldn’t have memorised $12.01\,\frac{\mathrm{g}}{\mathrm{mol}}$ for carbon but $340.48\,\frac{\mathrm{g}}{\mathrm{mol}}$. That would create significantly larger molar masses, but that shouldn’t be a problem it should only mean that nano is more prevalent.

The thing about the definition of the unit is that it doesn’t bother $99.9\,\%$ of the scientists working with the unit. Seconds are defined (by SI) according to a number of transitions occuring in some weird isotope that I wouldn’t even know how to measure. I would explain a second by saying it’s the 86400th fraction of a day if someone asked me. Same meaning, different exactness. If the moles are soon defined by mere counting rather than weighing atoms then so be it. Nothing will change for me in practice. (Maybe the fourth digit of a molar mass but I don’t really count those.) So as long as a new definition doesn’t break anything, we’re good to carry on.

Notes:

[1]: With imperially, I meant to assume the following definition using imperial units:

  One mole is the number of atoms in exactly $12.00~\mathrm{oz}$ of atoms of the carbon isotope $\ce{^{12}C}$.

I am a chemist and I am still trying to understand why SI prefixes are not used by chemists.

I am familiar with all the SI prefixes, but even if I was not familiar, I would prefer to learn a pair of universal prefixes rather than deal with the mole and all the issues, inconsistencies, and headaches it brings to us:

I have to learn SI prefixes the same. I.e. the mole is redundant.
I have to learn two mole concepts: the chemical mole and the thermodynamic mole [1].
I have to learn three definitions of mole in the SI. The original definition in the SI (1971), the amended definition (1980), and the new definition proposed for the future SI [2]. I have to learn the differences and check what one is being used in a given reference. For instance, the definition of mole in the current SI gives a uncertainty to the Avogadro constant, the proposal for the new SI sets a exact value for this constant and moves the uncertainty to $M(^{12}C)$.
Then I have to deal with "correction factors" to compensate for the
incompatibility between the redefinitions of mole and kilogram [3].
I have to learn an abstruse name for the physical quantity that
supposedly is measured in moles, "amount of substance", like if a
mol of electrons is a substance. And then I have to deal with the
proposals to change that physical quantity to "chemical amount",
"stoichiometric amount", or "enplethy" [2].
I have to deal with the ontological confusion between counting and measuring [4].
I have to deal with redundant equations such as $pV=nRT$ and $pV=Nk_BT$.

REFERENCES

[1] Failures of the global measurement system. Part 1: the case
of chemistry. Gary Price. Accred Qual Assur (2010) 15:421–427.

[2] What is a Mole? Old Concepts and New. Jack Lorimer. In CHEMISTRY International January-February 2010.

[3] Comments on recent proposals for redefining the mole and kilogram. B. P. Leonard. Metrologia 47 (2010) L5–L8.

[4] The Mole is Not an Ordinary Measurement Unit. Ingvar Johansson
Accred Qual Assur (2011) 16:467–470.

EDIT

I want to add the words that a synthetic chemist wrote in a letter to C&EN:

  In theory, we can dispense with the mole (C&EN, Aug. 4, page 32). The mole, molarity, and molality were adopted as convenient ways to express relative amounts and concentrations of substances when we didn’t know the actual mass of atoms and molecules—when the very existence of atoms and molecules was in dispute.

  Now we know, which has enabled us to come up with the number $6.022 × 10^{23}$, give or take, and requires us to explain to students—who question why anyone would specify such a ridiculous number— that it wasn’t chosen but is experimentally derived. Now that we know the actual mass of atoms, ions, and molecules, we can describe the actual number of entities directly—for example, an acid as $1.5 \; \mathrm{YH^{+}}$ per liter ($\mathrm{Y} = \mathrm{yotta} = 10^{24}$), which is about $2.5 \; \mathrm{mol \, H^{+}/L}$. We could rid ourselves of an unnecessary and confusing-to-beginners concept, but of course it will never happen.

I agree with his viewpoint, except maybe with his "it will never happen". Time will say.