Why is the periodate anion not found? Chlorine can be found as the following oxoanions $\ce{ClO–} , \ce{ClO2-}, \ce{ClO3-}, \ce{ClO4-}$ in many chemical stockrooms. However, iodine is never found as $\ce{IO4-}$. Explain this observation.

I didn't really find any reason as to why periodate cannot exist, unlike anions such as flourate where it is limited by flourine's small size. After researching this further, I found that the periodate anion actually exists in compounds such as sodium periodate. Even though it has another form $\ce{IO6^{5-}},$ the periodate anion actually dominates its equilibrium. Thus, I was wondering what the intended/likely answer was?

Yes, $\ce{IO4-}$ does exist and it is called metaperiodate anion. It can combine with a number of counter ions to form periodates, which may also be regarded as the salts of periodic acid.

  Metaperiodates are typically prepared by the dehydration of sodium
  hydrogen periodate with nitric acid.

  $$\ce{Na3H2IO6 + 2 HNO3 → NaIO4 + 2 NaNO3 + 2 H2O}$$

  They can also be generated directly from iodates by treatment with
  other strong oxidizing agents such as hypochlorites:

  $$\ce{NaIO3 + NaOCl → NaIO4 + NaCl}$$

  Metaperiodates adopt a distorted tetrahedral geometry with an average
  $\ce{I–O}$ distance of 1.78 Å

  The ortho- and metaperiodate forms exist in equilibrium.

  $\ce{H4IO6− ⇌ IO4- + 2H2O,      K = 29}$

  For this reason orthoperiodate is sometimes referred to as the
  dihydrate of metaperiodate, written $\ce{IO4^-·2H2O}$

One example of salt having metaperiodate ion is sodium periodate(@Gert). Its preparation is mentioned above. In fact, all other alkali metals and alkaline earth metals forms metaperiodate salts. Some examples include potassium periodate, rubidium periodate, calcium periodate etc.


How get a dilution at 150g/l of H2SO4 at 96%? I've never done any chemistry before and I would like to get 5 liters of H2SO4 at a concentration of $150 [g/l]$.

One liter of $H_2SO_4$ at $96\% $W/W has a density of $1.84 [g/ml]$.

It means my bottle should weight $1840~[g]$ where I have $96/100\cdot1840=1766~[g]$ of pure acid.

5 liters of acid at $150~[g/l]$ means $5~[l]\cdot 150~[g/l] = 750~[g]$

The amount of acid at 96% is:

$$\frac{1840~[g]}{1766~[g]}\cdot 750~[g] = 781 [g] $$

Converted in liters I must have

$$\frac{781 [g]}{1840 [g/l]} = 424~[ml]$$

Eventually I need to poor 424 [ml] of $H_2SO_4$ at 96% into $ 5 [l] - 474 [ml] = 4.51~[l]$ of distilled water. Is that right?

I am wondering of the amount of heat will I get during this dilution. Is that ok to gently poor it (about 1 minute) into distilled water contained into a plastic jerrycan?

First of all: Never add water to sulphuric acid. The heat of the reaction may lead to an explosion. Use safety glasses, safety gloves, and wear protective clothing.

The densities at 20°C are

Water: $\rho_{water} = 0.9982~\mathrm{g~cm^{-3}}$
15% $\ce{H2SO4}$: $\rho_1 = 1.10~\mathrm{g~cm^{-3}}$
96% $\ce{H2SO4}$: $\rho_2 = 1.84~\mathrm{g~cm^{-3}}$

To get $5000~\mathrm{cm^3}$ 15% $\ce{H2SO4}$ with a mass of

$m = \rho_1 \cdot V = 1.10~\mathrm{g~cm^{-3}} \cdot 5000~\mathrm{cm^3} = 5500~\mathrm{g}$

you need

$m = 0.15 \cdot 5500~\mathrm{g} = 825~\mathrm{g}$ pure $\ce{H2SO4}$

These are contained in

$m = \frac{825~\mathrm{g}}{0.96} = 859.4~\mathrm{g}$ of 96% $\ce{H2SO4}$.

i.e.

$V = \frac{859.4~\mathrm{g}}{1.84~\mathrm{g~cm^{-3}}} = 467~\mathrm{cm^3}$ of 96% $\ce{H2SO4}$.

Finally, the amount of water you need to pour the acid in is

$m = 5500~\mathrm{g} - 859.4~\mathrm{g} = 4640.6~\mathrm{g}$ water, i.e. $4649.0~\mathrm{cm^3}$

In short: gently pour, while stirring, $467~\mathrm{cm^3}$ of 96% $\ce{H2SO4}$ into $4649~\mathrm{cm^3}$ distilled water to get $5000~\mathrm{cm^3}$ 15% $\ce{H2SO4}$. Break the procedure if the temperature gets too high.

Disclaimer: I do not take any responsibility for the correctness of the above calculation.


How do you make sense of acid base reactions in organic chemistry? My question is in regards to acid base reaction that take place in larger chemical reaction in orgo. Like, I understand acid base reaction, how to qualitatively check which is stronger acid or base or where equilibrium lies, very well. But i feel that when I am trying to study the mechanism of series of other reaction like grignard mechanism, oxidation of alcohol, reaction with aldehydes etc… I see many intermediate acid base reaction in these mechanism that don’t make sense in terms of what I learned earlier when I was exclusively studying acid base reaction.

For instance, I have seen many intermediate acid base reaction in larger reaction where an oxygen atom of one molecule attack H atom attached to oxygen atom in another molecule. And then I am thinking what the point of this reaction because equilibrium is gonna be in middle and not favor products since the conjugate base create will be acid right away anyways.

Other times I have seen H20 act as an acid while molecule with sulfur atom act as a base. And again I am thinking how this is possible because isn’t sulfur a stronger acid then water. When I  asked two of my teacher about this. One of them said she actually made a mistake. But the main professor actually said that she is correct and to explain this, she said that water would even be hydronium ion here and when there is lot of water around, this water can serve as acid. However I couldn't fully understand what she meant by that? Can anyone please explain what she meant by this or how these acid base reaction possible. I have attached picture of what I am mean below to explain it my problem more clearly.

The answer to both questions lies in the fact that a reaction can be directed to a certain extent by controlling the concentration of the reagents.  The first reaction is perfectly fine since an alkoxide ion is more basic than a hydroxide ion. As for the second reaction, a carbon attached to an electronegative element is quite acidic. Hence is capable of possessing a negative charge. This along with the excess amount of water we supply for this reaction and the fact that water is neutral in nature adds to the reaction progressing in the forward direction.

Hope this helps


Find the concentration of the HCl solution when mixed with 1.265g of ACES-K+ $\pu{1.265 g}$ of N-(2-acetamido)-2-aminoethanesulfonic acid potassium salt ($\ce{ACES^-K+}$, $M=\pu{220.29 g/mol}$) is dissolved in $\pu{88.42 mL}$ of water. $\pu{27.59 mL}$ of $\ce{HCl}$ is added to the solution, resulting in $\mathrm{pH}$ of $6.54$. Calculate the concentration of the $\ce{HCl}$ solution. The $\mathrm pK_\mathrm a$ of $\ce{ACES}$ is $6.85$.

I tried setting up the equation as follows:
$$\ce{ACES + H3O+ <=> HACES +H2O}$$

The moles of ACES initial is $\pu{0.0057 mol}$ $(1.265/22.290)$ and we are trying to find $x$ the amount of moles of $\ce{HCl}$ $\ce{(H3O+)}$ we have. Therefore, using an ice table $\ce{ACES}=0.00574-x~\pu{mol}$ and $\ce{HCl}= x~\pu{mol}$.
When you plus into the equation
$$\mathrm{pH}=\mathrm pK_\mathrm a+\log(\text{base}/\text{acid})$$ I did $6.54=6.85+\log(0.00574-x/x)$ and got $\pu{0.0038 mol}$ of acid divided by total volume = $0.03321\ \pu M$. Is that correct?

Let us call the potassium salt "A-" and its corresponding acid for "HA".
The total volume of the final solution is 88.42 + 27.59 = 116.01 ml.
The final concentration of [HA] + [A-] = 1.265/(220.29 *0.11601) = 0.049499 (M).
[K+] = [HA] + [A-] = 0.049499 (M).

The equilibrium equation gives:
6.54 = 6.85 + log ([A-]/[HA]) => [A-]/[HA] = 0.48977 => [A-] = 0.48977 * [HA].
=> [HA] + 0.48977 * [HA] = 0.049499 => [HA] = 0.03322 (M) and [A-] = 0.016279 (M).

The charge balance is:
[H3O+] + [K+] = [OH-] + [Cl-] + [A-]
At pH = 6,54 [H3O+] = 10^-6.54 and [OH-] can be neglected in comparison with [A-].
=> [Cl-] = [H3O+] + [K+] - [A-] = 0.03322 (M).

The amount of chloride ions in the final solution will be 0.03322 * 0.11601 = 0.003853 mole. Thus, the concentration of the hydrochloric acid used will be:
(0.003853 * 1000) /27.59 = 0.1396 or about 0.140 M.


Chemical equation that can be balanced in infinite number of ways? When I was studying for my chemistry test, I found an equation that illustrates how to get water gas (carbon monoxide + hydrogen) out of methane. It is stated like this:
$$\begin{align}\ce{2CH4 + CO2 + H2O &->[\space \Delta ] 3CO +5H2}\tag{1}\\
\end{align}$$

First, I tried to balance it by myself in order not to memorize it but I ended with this equation:
$$\begin{align}\ce{3CH4 +CO2 +2H2O &->[\space \Delta] 4CO +8H2}\tag{2}\\
\end{align}$$
Then, I noticed that if we make the coefficient Of CO2 one, we can get infinite number of ways to balance the equation by changing the coefficient of CH4

We can make a third trial:
$$\begin{align}\ce{4CH4 +CO2 +3H2O &->[\space \Delta] 5CO +11H2}\tag{3}\\
\end{align}$$
Have you seen the pattern? I can conclude my observations in this equation:
$$\begin{align}\ce{nCH4 +CO2 +(n -1)H2O &->[\space \Delta] (n +1)CO +(3n -1)H2}\tag{4}\\
\end{align}$$
I am confused. Which one is true? If all of them are true, why my textbook mentions only one of them? What really happens in reality? Is equation $2$ just a simplification of more complicated equations? Does this have to do anything with the reaction mechanism?

I need to get more intuition about what is happening and the causes of this special feature.

We first note that your reaction (4) can be broken down into two reactions by splitting up those chemical species that depend on $n$ and those that don't: $$\begin{align}\ce{CH4 + H2O &-> CO + 3H2}\\
\ce{CO2 + H2 &-> CO + H2O},
\end{align}$$
so your reactions (1) through (3) are just statements about the relative rates of these two reactions.

  Which one is true? If all of them are true, why my textbook mentions only one of them?

I presume none of them are really true, in the sense that the ratio of the reaction rates isn't something as nice as 1:1 or 1:2 or 1:3, etc., and we haven't accounted for side reactions either. But assuming a 1:1 ratio is certainly a useful simplification.

  What really happens in reality?

The ratio of the rates will depend on the temperature and pressure and composition of the feed. There will be side reactions that occur. You can perform experimental measurements of the product composition to determine what happens.

  Is equation 2 just a simplification of more complicated equations?

Yes.

  Does this have to do anything with the reaction mechanism?

No. The overall reaction does not, in general, tell you anything about the reaction mechanism.


Conversion factor for conductivity of an aqueous solution I am not 100% sure if this is the right place to ask, but please point me in the right direction if not.
I am designing a conductivity meter, and when it comes to displaying the results in mS/cm or µS/cm, I also need to convert to EC. The first place I found was THIS website which seems great as it has all the units that I can convert here and there, no problem. This site states that 1EC is equal to 1µS/cm.
I had started to do most of my calculations based on the conversions from this website. However, when getting a reference calibration liquid, I noticed it was set to 2.76EC/2.8mS, which goes against what was stated in the website. I then found THIS website which shows 1EC as equal to 1mS/cm. I have now seen another place online (closed the tab and can't find it!) that also agrees with this.
Another website I have found HERE gives a bit more information on the conversion, and states that EC is measured in deciSiemens per meter (dS/m), and 1EC = 1dS/m, which would be 1mS/cm. Then later, it has a conversion thing at the bottom which has a statement saying:

Conversions:

1 deciSiemens per meter (dS/m) = 1 mmho/cm = 1
milliSiemens per cm (mS/cm) = 1000 microSiemens per cm (uS/cm) 1
mhos/cm = 1 S/cm = 1000 mS/cm

Electrical Conductivity units (EC's) IF
measured in uS/cm: 1 EC= 1uS/cm = 0.001 dS/m = 0.001 mS/cm = 1 mS/m

(Example 1.5 dS/m = 1.5 mS/cm = 1500 uS/cm = 1500 EC)

And surely that contradicts itself where it says 1EC = 1µS/cm, then the example after says that is not the case?
I just wanted to ask and be sure on this.... When converting, is 1EC equal to 1mS/cm, or 1µS/cm?

I think you have badly confused yourself by mixing the concept of a property of a solution with the units used to express this property.
Total dissolved solids (TDS) is a property of a solution. Theoretically, it can have any units kg/ kg of a solution, part per million, percentage, or whatever you like. However, people have standardized what they mean by TDS is usually mg solids/L of solution or even ppm.
In the same way, EC is electrical conductivity. It is a property of a solution. It can be expressed in units of microSiemens/cm or any multiple of it. EC itself is not a unit.


Mechanism of SN1 reaction Let us say that we have water reacting with t-butyl chloride. When the chlorine atom departs we have a t-butyl cation which becomes an attractive substrate to the nucleophilic oxygen in water. My question is in the reaction that follows, does the hydrogen break off from water first allowing oxygen to attack the carbocation or does water immediately form a bond with the carbocation and then deprotinates the extra hydrogen?

Surely there are both $\ce{H2O}$ and $\ce{OH-}$ which attacks the carbocation, as well as, heaven forbid, $\ce{H3O+}$ (and countless other species formed by the auto-ioniziation of water).

Usually textbooks record $\ce{H2O}$ as the attacker because it is much more abundant in the mixture.

After that, usually another $\ce{H2O}$ comes and take away the extra $\ce{H}$ with itself converting to $\ce{H3O+}$, but as I mentioned above, this step could also be done by $\ce{OH-}$ among many more species, but the concentration of those other species are too small that they are not worth mentioning.

PS: the last step can also be done by the product, i.e. t-butyl alcohol.


Can this sp2 lone pair resonate? In methyl isocyanate, $\ce{CH3CNO}$, shown below,

 it seems that the nitrogen lone pair can resonate with the carbon oxygen double bond, generating a resonance structure something like acetonitrile oxide:

In that case, why is the geometry around the nitrogen in the methyl isocyanate compound bent and not linear? Can the lone pair even resonate like that?

I remember in Klein's organic chemistry, it is stated that

  Whenever an atom possesses both a π bond and a lone pair, they will not both participate in resonance

In methyl isocyanate, $\ce{CH3CNO}$, shown above, it seems that the nitrogen atom has both a π bond and a lone pair. So perhaps the sp2 lone pair on nitrogen atom cannot resonate with the carbon oxygen double bond? I am not sure.

Whenever an atom possesses both a π bond and a lone pair, they will not both participate in resonance

The sentence means that only one, either the  bond or the lone pair will participate in resonance if the atom has both.

For example, in a pyridine, the nitrogen has a lone pair and it is also attached to a  bond.

Here, however, only the  bond participates in resonance because it is in the plane perpendicular to the molecule, while the lone pair is not.

Methyl isocyanate is best represented by a group of three resonance structures:

Here is the 3D structure of the most significant resonance structure of methyl isocyanate:


CASTEP Geometry Optimization Movie I am using a calculation software called CASTEP, within Materials Studio, and I want to make a "movie" of my geometry optimization, but don't know how to. CASTEP allows one to do many many different types of calculations, and one of the simpler ones it does is a geometry optimization for a unit cell. I remember hearing that it was possible to show a "movie" of the geometry optimization. This "movie" would show all of the steps from the beginning atom positions in the unit cell to the final converged one. I don't know how to make this "movie" and was wondering if someone else knows how.

The geometry optimization routine does not appear to have a method to output frames to an .xtd file for CASTEP in Materials Studio. Usually a movie makes more sense for a molecular dynamics simulation.

However, if you construct a script that performs a sequence if geometry optimizations with MaxIterations = 3 (must be >2), you can output each result to .std (study table) or .xtd. You would do this by outputting a geometry optimization after three iterations, then repeat the calculation on the result, until convergence. Then you take the frames and make a video

The help file in Materials Studio is rather a good resource, be sure to dig around in there, especially for the scripting API


Sodium Bisulfate vs Sodium Bisulfite I'm wanting to do a follow up to this Q&A on the cooking stack regarding browning avocados. I specifically want to expand my experiment to include the options in Wayfaring Stranger's Answer. Unfortunately, I managed to get the wrong stuff, and shipping rules make exchanging it problematic.

I am the proud owner of $500~\mathrm{g}$ of sodium bisulfate ($\ce{NaHSO4}$), instead of sodium bisulfite ($\ce{NaHSO3}$).

Dammit Jim! I'm a cook not a chemist!

What I have learned is that that both compounds are used as food additives towards the same end, and that my stuff has a slightly higher LD50. That's it.

Can anyone advise as to substituting my compound for the other? I'm not interested in tasty guacamole at this point, I'm really only looking at browning. If I'm trying to more or less duplicate the more successful experiments of the University of Florida, should I start with the same ratios? Or more or less of the additive? I can get sodium metabisulfite (also an anti-oxidant food additive) locally, if that would be a better option.

EDIT: So far, answers have all been consistent in that I have a half-kilo of white powder that is useless to me. I will make a trip to the wine-making shop and acquire some sodium metabisulfite. That still leaves me with the question of how much? The University of Florida got the results I'd like to duplicate with $30~\mathrm{mg}$ of sodium bisulfite per $100~\mathrm{g}$ of avocado. Should I start with the same ratio of sodium metabisulfite?

Sodium bisulphate (NaHSO4): an acid
Sodium bisulphiTe (NaHSO3): an antioxidant
Sodium metabisulphiTe (Na2S2O5): another antioxidant

In short, bisulphate and bisulphite are not interchangeable, but bisulphite and metabisulphite are.

Sodium metabisulphite is prepared from sodium bisulphite via dehydration, as in this equation:
$$\ce{2 NaHSO3 -> Na2S2O5 + H2O }$$
It's reversible in aqueous solution. The assumption that 30 mg of metabisulphite contain the same amount of sulphite as 30 mg of bisulphite is OK, the error is small, about 10 %.  The correct amount would be 27 mg of metabisulphite.

You'd ask yourself if pasteurization (75 deg. C, 30 seconds) would work to deactivate the polyphenoloxidase without affecting flavour too much.

I'd also suggest searching Pubmed (it's a decent alternative to commercial citation databases) for such keywords as "polyphenol oxidase" (that's the enzyme that causes browning), "inhibition" and "sulfite".

If you want hands-on answers I'd suggest going to the library and get (via interlibrary loan) a copy of this paper: Plant Foods Hum Nutr. 1995, 47, 245-56, "The control of polyphenol oxidase activity in fruits and vegetables. A study of the interactions between the chemical compounds used and heat treatment."

  From the abstract: ... Interactions between the use of ascorbic acid,
  citric acid, EDTA, sodium metabisulphite and heat treatment (70
  degrees C for 2 min) in the control of PPO activity were studied in
  avocado ..., banana ..., apple ..., pear ..., peach ..., potato ...,
  eggplant ..., mushroom ... and
  hearts-of-palm ... . The results demonstrated that
  PPO of avocado and eggplant was most resistant to inhibition by the
  methods used. ... The results indicated that,
  with the exception of PPO from avocado, the most adequate alternative
  method to substitute for the use of SO2 in the control of PPO was a
  combination of ascorbic acid, citric acid and heat treatment.

This shows that how to get on top of browning while still having a presentable product is a tough problem and an active area of research.  Also, avocado seems a specially hard case.


Which units do I use in calculating heat capacity? The specific heat capacity of liquid water is $4.18 \, \text{J} \, \text{K}^{-1} \, \text{g}^{-1}$. Calculate the energy required to heat $1.0 \, \text{mol}$ of water from $298 \, \text{K}$ to $363 \, \text{K}$.
My question is: Do I have to change the moles of water into grams or should I sub in $1.0$ into the formula for specific heat capacity?

Yes, you would have to change it to grams.

Units must be consistent. Specific heat is defined as the energy required to raise one gram of water by one kelvin. At atmospheric pressure in a closed container, the specific heat of water is 4.184 joules.

One mole of water is 18.016 grams. So:

$E =mc \Delta T $

Where $m$ is the mass of the water, $c$ is the specific heat capacity of water, $\Delta T $ is the change in temperature, and $E$ is the energy required. Substitute in your numbers, and you shall get the right answer:

$E = (18.016\ \mathrm{g})(\frac{4.184\ \mathrm{J}}{\mathrm{g\ K}})(363\ \mathrm{K} - 293\ \mathrm{K})$

$E = 5280\ \mathrm{J}$

If you wanted to use moles directly instead of converting to grams, you would have had to use the molar heat capacity. Generally, people prefer to use specific heat. Engineers sometimes use the volumetric heat capacity as well.


Unsaturated ketone coupling pattern Can someone explain to me the coupling pattern by proton $5$? Why is it triplet of quintet? Is it long range coupling?

I'm assuming that the proton $5$ couples to two other protons on carbon $4$. But what's next?

You are close. The triplet part is the coupling of the enantiotopic hydrogens at C4. The width of the C5 blowup is ~20 Hz in width which means J = ~7 Hz. The smaller coupling is due to allylic coupling with the methyls (C7 and C8). Assuming J5,7 and J5,8 are the same, then the six hydrogens produce a septet. You just can't see the outer peaks because they are too small. If coupling to the two methyls is different then the septet will appear broadened because of lack of resolution. A larger blowup of C5 would reveal the outer peaks (one and seven).


Relationship between Van der Waals forces and the boiling point Choose the right answers in the following statement:

The boiling point of a liquid substance is higher as the Van der Waals forces (in/between) the molecules of the substance are (lower/higher).

I don't understand why the Van der Waals forces are lower when the boiling point is lower. Second, do you say: the forces in the molecules of a substance or the forces between the molecules of a substance?

We talk about the forces between molecules.  The Van der Waals force represents the summation of all attractive and repulsive forces between molecules, other than those due to electrostatic charges or actual bonds.  This summation most often turns out to be a weak attractive force.  Typically the magnitude of the Van der Waals force is small compared to electrostatic or bonding forces and (again typically) only apply at relatively short range.  So they usually serve to weakly hold molecules together.  The stronger the Van der Waals force, the more strongly molecules will be held together and the more energy will be needed to be put into the system to break them apart from each other.  In the case of boiling a liquid, you are really talking about breaking molecules apart from one another.  So in the case of boiling, the weaker the Van der Waals attraction between molecules, less energy will needed to break them apart from each other and allow them to boil.
Here is a link to a Wikipedia article on Van der Waals forces if you'd like to read more about them.


vapor pressure of one part of binary compound Let's say we have a binary compound AB in an enclosed container, and let's say there's a tendency to form B-vacancies so that B atoms/molecules escape and form a vapor. Let's assume there's enough of AB that the pressure saturates before AB$_{1-x}$ decomposes too much. Would the vapor pressure of B be the same as for the pure element B?

No. The equilibrium vapor (gas) pressure is defined by the corresponding equilibrium. When it's just the element $\ce{B}$ that evaporates, the equilibrium is $$\ce{B(liquid)<=>B(gas),}$$ and the thermodynamics of evaporation (or sublimation if $\ce{B}$ is solid), such as the Clausius–Clapeyron equation, gives you the equilibrium $p-T$ relationship.

When you have nonstoichiometric compound $\ce{AB_{1-x}}$ which loses $\ce{B}$, the process is quite different. Let's imagine purely hypothetical $\ce{AB}$, in the crystal structure of which neutrally charged $\ce{B}$ vacancies, $\ce{V^x_B}$, can be formed. Then, our hypothetical equilibrium between the gas phase and defects in solid can be written in Kröger–Vink notation as $$\ce{B^x_{B}<=>V^x_{B} + B_{gas}}.$$ In this case, until $\ce{AB_{1-x}}$ decomposes too much, the equilibrium constant $K=\frac{p_B\cdot[\ce{V^x_{B}}]}{[\ce{B^x_{B}}]}$ and its temperature dependence define the "partial pressure of $\ce{B}$ - temperature - composition" relationship for $\ce{AB_{1-x}}$.

Please note that the example above is oversimplified for clarity. In the real-world applications of defect chemistry, point defects in nonstoichiometric solids are often charged, and they tend to interact with each other, complicating the equilibria.


What is b in the Gompertz model? I am wondering how to use the Gompertz model in modelling drug diffusion:
$$x(t) = x_\mathrm{max}\exp(-a\exp(b\log t)).$$
$x_\mathrm{max}$ is to be taken as $100,$ $a$ is the undissolved proportion at time $t = 1.$ But what is $b?$ I've seen it variously described as half of the max asymptote, the $m$ of the linear portion of the graph, or the $m$ divided by $\mathrm e.$

I can see at Gompertz function they use the formula below. See the link for Graphs of Gompertz curves, showing the effect of varying one of a,b,c while keeping the others constant

$$f(t)=a\mathrm{e}^{-b\mathrm{e}^{-ct}}$$
where

''a'' is an asymptote, since $ \lim_{t \to \infty} a\mathrm{e}^{-b\mathrm{e}^{-ct }}=a\mathrm{e}^0=a$
''b'' sets the displacement along the ''x''-axis (translates the graph to the left or right). When ''b'' = log(2), f(0) = a/2, also called the halfway point.
''c'' sets the growth rate (''y'' scaling)
''e'' is Euler's Number (''e'' = 2.71828...)

The halfway point is found by solving $f(t) = a/2$ for $t$.
$$t_{hwp} = -\frac{\ln(\ln(2)/b)}{c}$$

Regardless of if the equation $$x(t) = x_\mathrm{max}\exp(-a\exp(b\log t)).$$ in the question is the true one, analyzing the equation as is, being aware ift has different shape than the above.:
It can be rewritten as $$x(t)=x_\mathrm{max} \cdot  \exp{\left(-a \cdot t^b\right)}$$
If $\ce{b=1}$, the equation reduces to the classical equation of the kinetic of the first order and the half-time $t_{1/2}=\frac{\ln 2}{a}$, where an asymptote at any point in time crosses x axis after $t_{1/2}$.
$$x(t)=x_\mathrm{max} \cdot  \exp{\left(-a \cdot t \right)}$$
The parameter $\mathrm{b}$ affects the time evolution of the diffusion kinetic model..
$b = 1$ keeps the relative decrease with time constant.
$b \gt 1$ accelerates the relative decrease with time.
$b \lt 1$ decelerates the relative decrease with time.
Illustrative table about the $b$ value influence.
$\begin{array}{|c|c|c|c|}  \hline
 & b & b & b \\  \hline \hline
t & 0.8 & 1 &1.2 \\
0 & 100 & 100 & 100 \\
1 & 50.0 & 50.0 & 50.0 \\
2 & 29.9 & 25.0 & 20.3 \\
3 & 18.8 & 12.5 & 7.5 \\  \hline
\end{array}$
$t$ was padded by small value to avoid log 0. Results were rounded.
$a$ was set to $\ln{2}$ to have $t_{1/2}=1$ for $b=1$.


Are Daniell cells rechargeable (reversible)? I know that the Daniell cell is a primary cell and the reaction occurs as:

$$\ce{Zn -> Zn^{2+} + 2e^-}$$
$$\ce{Cu^{2+} + 2e^- -> Cu}$$

However, I do not know if we can reverse the reaction by applying a potential greater than $1.1~\mathrm{V}$. Also, please give the reason if it is not

P.S.: Please do not confuse this question: Why can't a primary cell be recharged?

I asked it as neither the above link nor the internet could give me satisfactory answer.

The Daniell cell is not rechargeable. According to this Wikipedia article:

  In the Daniell cell, the porous barrier cannot prevent the flow of
  copper ions into the zinc half-cell. Hence, recharging (reversing the
  current flow by an external source of EMF) is impossible because, if
  the zinc electrode is made to become the cathode, copper ions, rather
  than zinc ions, will be discharged on account of their lower
  potential.


Strength of HClO4 and resonance Why is $\ce{HClO4}$ the strongest oxyacid?

My effort: The structure of $\ce{HClO4}$ should be tetrahedral. So when there is a negative charge on oxygen, it should not be in resonance with other oxygen atoms as a condition for resonance is that the atoms taking part in resonance should lie in a plane.

Assume an acid $\ce{HA}$. An acid is strong if

protons are donated easily, i.e., the $\ce{H-A}$ bond is readily broken;
the conjugate base $\ce{A-}$ is stable, i.e., $\ce{A-}$ is reluctant to combine back to $\ce{HA}$.

When are protons easily donated?
First parameter to consider is the polarity of the $\ce{H-A}$ bond. More polar bonds mean less electron density on the bond. Thus the dissociation equilibrium
$$\ce{{{HA_{(aq)}}+H2O_{(l)}}<=>H3O_{(aq)}+ + A_{(aq)}-}$$
is tilted strongly towards the right. For this reason, $\ce{HF}$ is a stronger acid than $\ce{NH3}$.
In addition, the size of $\ce{A}$ is important. If $\ce{A}$ is very big, essentially it is harder to hold on to the hydrogen. Thus the $\ce{H-A}$ bond is weaker. This is why $\ce{HI}$ is a better acid than $\ce{HF}$.
Stability of the conjugate base
Charge delocalisation is key here. If all atoms bonded to a central atom are equal, then the hydrogen ion does not have a preferred location at which to bond. A large charge in a small volume would also translate into an unstable ion.
Oxoacids

Look at the number of oxygens bonded to the central atoms. More oxygens -> stronger acid, e.g., $\ce{HNO3}$ is stronger than $\ce{HNO2}$.
If the number of oxygens is equal, consider the electronegativity of the central atom. Higher electronegativity -> stronger acid. For instance, $\ce{HBrO3}$ is stronger than $\ce{HIO3}$.
When electronegativity is similar as well ($|\Delta EN|\le 0.1$), it sometimes helps to look at the size of the central atom. For example, $\ce{H2SeO4}$ is stronger than $\ce{H2SO4}$.

The case at hand
Polarity of the $\ce{O-H}$ bond in $\ce{HClO4}$ is guaranteed via

the electronegativity of the $\ce{O}$ atom itself;
the electronegativity of the central atom $\ce{Cl}$;
the electronegativity of the three additional $\ce{O}$ atoms.

Thus, the electron density is pulled from the  $\ce{O-H}$ bond and the proton can be easily donated.
Here is a picture where you can compare the charge delocalisation of various chlorine oxoanions.

Clearly, the perchlorate ion $\ce{ClO_4-}$ is the most stable. Every valence orbital and valence electron of chlorine takes part in the bond formation.
Extra
This reply discussed relative acid strength with respect to Brønsted–Lowry acid–base theory (in water). Do not forget that these are pointers; there will be other parameters to consider and exceptions to take into account.
$\ce{HOF}$ is the single known fluorine oxoaxid. The analog $\ce{HFO4}$ is simply not possible. It would violate the octet rule.
EDIT: Answer to comment

You asked: "When you say charge delocalisation in ClO4- anion is it due to resonance? Is resonance possible?"
From IUPAC Gold Book: "[Resonance] refers to the representation of the electronic structure of a molecular entity in terms of contributing structures. Resonance among contributing structures means that the wavefunction is represented by 'mixing' the wavefunctions of the contributing structures."

Therefore, the answer is yes. The main resonance structures are given in the picture below.

The resulting ion is a resonance hybrid. In classical valence bond theory, charge delocalisation is not due to resonance; rather, resonance is delocalisation.
Furthermore, coplanarity is not a requirement. It is fair to say that there is (almost) always some delocalisation (again, see the definition given). Still, having atoms that lie in the same plane can be advantageous. Qualitative models of aromaticity and conjugation may even give coplanarity as a necessary condition. This is not always the case, however. Quantitative models are used to more rigorously predict aromaticity and antiaromaticity (see NICS, QTAIM models).
P.S. Resonance energy is important. High, positive resonance energy indicates increased stability; slightly positive or even negative values point to unstable particles. For instance,  antiaromatic molecules generally have low resonance energies. And even though planarity is a prerequisite of qualitative antiaromaticity, too, it is not actually a rule, instead a rough indicator. Example: 2,3,4,5-tetraphenylcyclopenta-2,4-dienone here and there.


To add acid or base to water There is a rule stating that we shall add a strong acid to water, and not the other way because of safety; if we would add water to the acid, the reaction could be dangerous (boil).

What about adding strong bases and water? Which is the safer order to mix them?

I would like to know the safer addition orders, and the justification, for water+acid or water+base(?)/base+water(?). Thank you!

The reason for adding strong acids or bases to water rather than the other way around is that the dissolution/reaction of these compounds with water tends to be very exothermic and can result in splattering or even boiling of the strong acid or base, particularly if water is added to the concentrated acid or base.

On the other hand, if you slowly add the acid or base to water, you will never have a concentrated acid or base present to splatter as it will be rapidly diluted by the much larger amount of water. Even if you accidentally add the acid or base too quickly, at least the splattering will primarily be of water or dilute acid/base rather than splattering concentrated strong acid or base. It is still important to do the addition slowly so that you only slowly increase the acidity/basicity of the solution. This way, the rate of reaction (and thus also the heat evolution) will happen at a safe level and greatly reduce the risk of splattering.

As a side note, this concept does not only apply to the addition of strong acids and bases to water. While mixing any chemical reagents (unless otherwise specified of course), you should add the more reactive or concentrated solution to the less reactive or more dilute solution, for the same reasons as described for the addition of strong acids and bases to water.


Why is diboron (B2) paramagnetic? Why is $\ce{B2}$ paramagnetic? It's just 2 borons with a triple bond between them. Aren't all the electrons paired?

For the same reason $\ce{O2}$ is paramagnetic: degeneracy.
The filled orbitals are $\sigma(\mathrm{1s})^2$, $\sigma^*(\mathrm{1s})^2$, $\sigma(\mathrm{2s})^2$, $\sigma^*(\mathrm{2s})^2$.  The half-filled orbitals are $\pi(\mathrm{2p}_x)^1$ and $\pi(\mathrm{2p}_y)^1$.

But you're right to think that this is unexpected:  why are the $\pi(\mathrm{2p})$ orbitals  lower than the $\sigma(\mathrm{2p})$? That's the reverse of what we are told to expect! I've never encountered a rationalization that satisfied me.

See Matt Law's lecture notes for the expected and observed ordering:
MO Diagrams for Diatomic Molecules; Chapter 5; Friday, October 9, 2015; PDF via the Internet Archive.


Voltaic cell: why does the rate of rusting increase when NaCl is added? There is a beaker in which there are two iron electrodes (there is no salt bridge between two beakers-there's just one beaker) and one iron electrode is being bubbled with a supply of oxygen from a pipe (causing it to rust and fall to the bottom of the beaker). There is a flow of current in an external circuit (which I think is because the oxygen prevents one electrode from dissociating into $\ce{Fe^2+}$ and $\ce{2e-}$). The one that is not being bubbled with oxygen is the anode, but I'm not sure why. I think it's because the other one can't dissociate since it's reacting with oxygen. Also, the rate of rusting increases when $\ce{NaCl}$ is added to the water. Why is that?

If your diagram is correct (the notation of the polarity of the electrodes) then the anode should be the one that is not being pumped with oxygen, because the negative end of any electro-chemical cell is the anode.

Let's see how this happens.

First of all let us consider the ions and molecules at two electrodes the system.

We have $\ce{O2}$ at electrode A and $\ce{Fe}$ at B   .

Now since the reduction potential of $\ce{O2}$ is greater than $\ce{Fe}$ a reaction takes place at the $\ce{O2}$ pumping electrode. This is a reduction:

$$\ce{O2 + 2H2O + 4e- -> 4OH-}$$

Now at the other electrode, an oxidising reaction takes place:

$$\ce{Fe -> Fe^3+ + 3e-}$$

So you can see there's an excess of electrons at the $\ce{Fe}$ electrode. Now these electrons travel across to the $\ce{O2}$ containing electrode since the cell reaction there needs electrons. This travel of electrons give rise to an electric current. You might know from your previous lessons that current flows from negative terminal to the positive terminal inside a cell. Hence the $\ce{Fe}$ electrode becomes the negative anode.

As to your second question why does $\ce{NaCl}$ increase corrosion, for all these reactions to take place electrons should be moved about. $\ce{NaCl}$, being a strong electrolyte, increases the conductivity of water. That means electrons are moved rapidly giving a rise in the rate of corrosion.
And you have asked in the comments "How is just water doing the trick?". You should keep in mind that water exists as $\ce{H+}$ and $\ce{OH-}$ in the aqueous liquid state.


How to calculate the oxidation state and the number of equivalents oxidized with YBa₂Cu₃O₇ I've hit some trouble with these calculations and need help sorting out confusion.

In the lab we synthesized $\ce{YBa2Cu3O7}$, per the lab manual $\ce{Y, Ba, O}$ have the usual charges of +3, +2, and -2 respectively. Spectroscopic studies show that no copper (III) centers are present in the material, but rather there are missing electrons from the copper-oxygen bonds. For the purpose of the titration though, the missing electrons are thought of as coming from the copper center and we're assuming there are copper III centers present.

We used idodine to tritrate, and there is the assumption that copper is doing the oxidizing. The text states, "Copper(I) is not an oxidizing equivalent: It cannot oxidize iodide to iodine."

The relevant equations are
$$\begin{align}
\ce{Cu^{3+} + 2I- &-> Cu+ +I2}\\
\ce{Cu^{2+} +I- &-> Cu+ + \frac{1}{2}I2}
\end{align}$$

I've calculated the theoretical OS to be
$$\ce{Y} (3^+ )+\ce{Ba} (2^+ * 2)+\ce{O} (2^-*7)= \frac{7}{3 \ \ce{Cu}}\ce{Cu}^{2.3+}$$

We have to calulate the theoretical number of equivalents of $\ce{I-}$ oxidized by 1 g of the sample.

Would this be correct for the oxidation state? If so would that mean that the total equivalents of Iodine are an addition of $\ce{Cu(II)}$ and $\ce{Cu(III)}$ (the two equations)? Also I'm feeling brain dead, because I can't figure out the fraction of $\ce{Cu(II)}$ and $\ce{Cu(III)}$ that would average to 2.3.

Your fundamental assumption is that all of the oxygen is $\ce{O^2-}$. In oxide ceramics, however, it is possible for oxygen atoms to dissolve into the matrix. zirconia,  titania, and bismuth oxide can all easily accumulate non-stoichiometric oxygen. In fact this ability to "dissolve" oxygen into a solid matrix is how fuel cells are able to operate.

You are correct that the yttrium is $\ce{Y^{3+}}$, barium is $\ce{Ba^{2+}}$, and copper is $\ce{Cu^{2+}}$, but in YBCO the oxygen can form $\ce{O^{(2x-2)-}_x}$ sites as part of $\ce{Cu2O_x}$ ($1\leq x \leq 4$) clusters. Note: for $x = 2\ \ce{O^{2-}2}$ is not inherently a peroxide ion, you need cooper pairs of electrons to have superconductivity (see image below). Thus another way to present the structure of YBCO may be as $\ce{YBa2Cu3O5(O2)}$ or rather $\ce{YBa2Cu3O_{5+x}(O2)_{1-x}}$ or even $\ce{YBa2Cu3O_{3+3x}\nu^0_x(O4)_{1-x}}$ $(0 \leq x \leq 1)$ as the ceramic can be non-stoichiometric and produce oxygen vacancies which affect the superconductivity.


Is it possible to use the school supplied algorithm to build the Lewis diagram of the iodate ion? I am upgrading the grade 12 chemistry credit and have to use the school supplied algorithm to build the Lewis diagram of $\ce{IO3-}$ ion. It seems to me that iodine oxidation state in this ion should be 5, therefore it would have one single and two double bonds with oxygen. When I follow the steps below, I am arriving at a different formula.

I do not need to answer whether the ion is polar or not and what is its shape.

  Step 1: Count all of the valence electrons in the molecule or ion. In the case of an ion, add or subtract electrons to account for the ionic charge.

$$\ce{I = 7e^-, O = 18e^-}; \text{ionic charge} = \ce{1e^-}; \text{total} = \ce{26e^-}$$

  Step 2: Arrange the peripheral atoms symmetrically around the central atom. Use a pair oil electrons to form a bond that links these atoms to the central atom.

6 electrons placed on $\ce{I}$, $\ce{26e^- - 6e^- = 20e^-}$ remain

  Step 3: Add pairs of electrons to complete the octet of the peripheral atoms.

$3\cdot 6$ electrons placed on $\ce{O}$, $\ce{20e^- - 18e^- = 2e^-}$ remain

  Step 4: Place any unassigned electrons on the central atom.

2 electrons placed on $\ce{I}$, $\ce{2e^- - 2e^- = 0e^-}$ remain

  Step 5: If the octet of the central atom is incomplete, move a lone pair of electrons from a peripheral atom to a new position between the central and peripheral atom.

After completing step 4 of the rules above I am arriving at $\ce{I}$ having 8 electron dots. At this point all electrons have been used up, the octet of $\ce{I}$ is complete, and I can only create a structural diagram with two single and one double bonds.

In the course I have taken previously we were taught that $\ce{I}$ was an exception to the octet rule, but when I asked this teacher, I was told that I was not allowed to use the exceptions and had to apply the octets only.

Am I missing anything? Is it possible to comply with the algorithm above and come up with the correct answer?

When I follow your school's algorithm for iodate I get a central iodine atom with three single bonds to oxygen.  I've shown my work below.  This is problematic, as you well know, but I can potentially see where your teacher is coming from.

When I teach formal charge and resonance structures and the like, I often have students use their intro chemistry rules to determine the structure of a polyatomic ion such as this.  If they follow the rules they get a pretty structure that looks proper - I then introduce the idea of formal charge by pointing out the wild imbalance of charges in the ion and how energetically unfavorable that would be, then we work out a way to rearrange the electrons to minimize that problem.

I also have a lot of students who immediately go to google to find Lewis structures rather than drawing them themselves, so when a student tells me that they think the structure should be something different, it takes an actual argument from them before I believe that they know why that should be.  I don't know that that is the case here; it also seems entirely possible that your instructor is just wrong.


Why are protein hydrophobic cores denatured by heat when entropic forces should be favoured at high temperature The hydrophobic interaction is thought to be driven by an entropic force. If it is, shouldn't hydrophobic interactions be stronger at higher temperatures, where states of higher entropy are favoured? Why, then, are protein hydrophobic cores denatured by heat?

You are correct that the hydrophobic interactions are stabilized at higher temperatures. However, I believe the addition of heat mainly disrupts protein structure by breaking hydrogen bonds through increased thermal motion.

Once the secondary structure starts to go, any tertiary interactions will get scrambled, likely resulting in the separation of hydrophobic residues. The refolding of the polypeptide backbone must be stronger than the enhanced hydrophobic interactions.


Why and how does hydrogen peroxide decompose in the presence of light? When exposed to UV light, hydrogen peroxide decomposes into $\ce{H2O}$ and $\ce{O}$. Why does this happen and more importantly how? Is the energy from light absorbed by the bonds which are specific to UV frequency and thus leading to decomposition?

Chemical reactions in which, a single substance splits up into two or more simpler substances are called decomposition reactions. These reactions are carried out by energy, supplied by different sources. The required energy can be supplied by heat (thermolysis), electricity (electrolysis), or light (photolysis).

Let’s talk about photolysis reactions (not photosynthesis):

  Photolysis (also called photodissociation and photodecomposition) is a chemical reaction, in which a chemical (an inorganic or an organic) is broken down by photons and is the interaction of one or more photons with one target molecule. The photolysis reaction is not limited to the effects of visible light, but any photon with sufficient energy (higher than dissociation energy of the targeted bond) can cause the chemical transformation of the said (inorganic or organic) bond(s) of a chemical. Since the energy of a photon is inversely proportional to the wavelength, electromagnetic waves with the energy of visible light or higher, such as ultraviolet light, X-rays, and $\gamma$-rays, can also initiate photolysis reactions.

Like all other peroxides, hydrogen peroxide ($\ce{H2O2}$) also consists of a relatively weaker $\ce{O-O}$ bond, which is susceptible for light or heat. In the presence of light (the UV light from the sun catalyzes the reaction), $\ce{H2O2}$ spontaneously decomposes into water and oxygen.

The net equation for the reaction is:
$$\ce{ 2H2O2 -> 2H2O + O2}$$

The step-wise reaction mechanism is suggested as follows (Ref.1):

$$\ce{ H2O2 + h\nu -> 2 HO^.}$$
$$\ce{ HO^. + H2O2 -> HO-O^. + H2O}$$
$$\ce{ HO-O^. + H2O2 -> 2 HO^. + H2O + O2}$$

Using isotope studies ($\ce{^{18}O}$ labelled $\ce{H2O2}$), early work has been confirmed that the $\ce{O2}$ formed is cleanly derived from $\ce{H2O2}$ (Ref.2).

Notes: The rate increases rapidly in the presence of catalysts such as $\ce{MnO2}$ and $\ce{KI}$ (Ref.2). The rate of decomposition is slow at room temperature, but it increases with temperature. It is believed to be due to thermal decomposition of $\ce{H2O2}$, which seemingly accelerates the photolysis (Ref.3).

References:

J. P. Hunt, H. Taube, “The Photochemical Decomposition of Hydrogen Peroxide. Quantum Yields, Tracer and Fractionation Effects,” J. Am. Chem. Soc. 1952, 74(23), 5999–6002 (https://doi.org/10.1021/ja01143a052).
A. E. Cahill, H. Taube, “The Use of Heavy Oxygen in the Study of Reactions of Hydrogen Peroxide,” J. Am. Chem. Soc. 1952, 74(9), 2312–2318 (https://doi.org/10.1021/ja01129a042).
F. O. Rice, M. L. Kilpatrick, “The Photochemical Decomposition of Hydrogen Peroxide Solutions,” J. Phys. Chem. 1927, 31(10), 1507–1510 (https://doi.org/10.1021/j150280a004).


Can very dilute acids exposed to air become more concentrated as water evaporates? If a very minute quantity spilled acid (e.g. $\ce{H2SO4}$, though the question would apply to other acids as well) is cleaned up with water (e.g. yielding a $0.0001~\mathrm{M}$ solution), and some of the water (dilute acid solution) is left behind, will the resulting solution become more concentrated as the water evaporates?  Would there be any danger that a drop of an initially-harmless concentration might become a visually-unnoticeable quantity of acid that was strong enough to pose a contact hazard?  Would wiping an area down with a moderately-dilute solution of a weak base such as sodium bicarbonate be an appropriate safety precaution, or would it be unnecessary if any acid droplets that might exist on a surface have been thoroughly diluted?

Sulfuric acid (and other acids) can be concentrated through evaporation of water.  It would be a good idea to wash the surface down with dilute, basic solution.  That said, if I'm reading your post correctly, your droplets are already 0.1M.  That is still a strong acid and care should be taken in its clean up.  I was once working with a 0.01M solution of sulfuric acid and when cleaning the flask a few drops of the dilute solution splattered onto my foot, I didn't even notice it at the time.  An hour or so later my foot started to burn.  When I looked, my sock was gone in the affected area.  If you're dealing with anything more than drop or two use a spill kit and let someone know what you are doing.


Why does a protein crystallize with a specific salt but not with another? So I am conducting an experiment and I do not understand this super fundamental part.
I know salting out works via solubility, and I know crystallization works via ordered structures. But could someone explain to my why one salt allows a crystal to form, but another salt doesn't. A component of one of my salts (Mg) has been found to be important to my protein in optimization and activity, but why would another cation with the same anion (I know the anion is not the problem) discourage the formation of crystals. If this is a clearly stupid question I apologize.

EDIT:looking for theory. I have already crystallized everything, and have CRYSTALS. Will be performing x-ray crystallography.This post was more to get me on the right track in thinking.

It would be a lie to postulate we knew much about the formation of protein crystals. Yes, we know the basic theory — that we need a supersaturated solution in which the protein is not denatured and which would then slowly evaporate until hopefully a single crystal starts growing — but we don’t have the slightest of clues of what exactly helps and what doesn’t help.

This is why protein crystallography is still trial and error (and nothing else). You try thousands of different conditions in an automated manner, select those with promising results for further examination and finally — if and only if you are lucky — end up with crystals suitable for X-ray diffraction.

Therefore, we don’t know and nobody knows why magnesium chloride might help the crystallisation of your protein but sodium chloride, calcium chloride, aluminium chloride or magnesium bromide don’t.
We don’t know and nobody knows why a $\pu{1.3M}$ solution of magnesium chloride might help the crystallisation but a $\pu{1.4M}$ or a $\pu{1.5M}$ solution don’t.
We don’t know and nobody knows why $\pu{23^\circ C}$ might be a good temperature for crystallisation but $\pu{21^\circ C}$ and $\pu{25^\circ C}$ aren’t.
And so on and so forth.


Ultrasound absoprtion by polypropylene beaker for HF acid In my Ph.D. research, I need to carry out an etching of glass samples in HF acid. I use an ultrasonic bath for the experiment, but to avoid corrosion of steel I put a small plastic (polypropylene) jar with 5% HF with glass samples into the ultrasound bath full of water. Recently, my colleague told me that the plastic jar absorbs most of the ultrasonic energy and my etching process goes the way it would be without the bath at all. So, first of all, is it true that a polypropylene beaker absorbs ultrasound? What kind of beaker should I use instead which wouldn't be affected by HF acid?

Polymers, like other materials as well, attenuate / absorb acoustic waves.  Beside reflection (echo, similar to active sonar), this renders ultrasound suitable for non-destructive inspections.  The extent of absorbing ultrasound depends on several factors (e.g., temperature, density/crystallinity of the sample, deployed frequency) and may be expressed as attenuation and by acoustic impedance.

The following table is part of a lecture about medical diagnostic applications of ultrasound:

(source, slide 5.)

A more detailed listing, specifically addressing polymers, may be found here. From data like

|---------------------------+----------+---------+-----------+-------------|
| Plastic material          | Velocity | Density | Impedance | Attenuation |
|                           |    [m/s] | [g/qcm] |           |     [dB/cm] |
|---------------------------+----------+---------+-----------+-------------|
| Polyethylene, low density |     1950 |    0.92 |      1.79 |         2.4 |
| Polypropylene, Profax     |     2740 |    0.88 |      2.40 |         5.1 |
| Polypropylene, white      |     2660 |    0.89 |      2.36 |        18.2 |
|---------------------------+----------+---------+-----------+-------------|

some differences in polymers of same name, but different brand (and degree of polymerisation, etc.) become apparent. (Interestingly, Teflon/PTFE is not mentioned in their listing, yet according to this study possess a much higher acoustical impedance than PP.)

Hence, beside chemical resistance against HF, you likely have to test the efficiency of the ultrasound waves passing through the walls of your beaker anyway.


Determine compound acidity based on structure In an exercise I have the following structures

Moclobemide:

Paracetamol:

Hexobarbital:

I know for example by heart that paracetamol is an acid. What is the best approach in determination of acidity of a compound based on it's structure?

I'd like to know for example if these compounds are:

Basic
Acidic
Amphoteric

I know for example that many metals (Zn, Pb, etc) can be amphoteric (act as $H^+$ donors or acceptors). But I don't know how to determine what a complex structure is.

Please offer a detailed explanation, with links if possible.

Best Regards,

Panagiotis, in order to solve the exercise, one needs to identify the relevant functional groups and explore their environment a bit.

Supposed that we're talking about "standard conditions", i.e., what is likely to happen if the dissolve/suspend the compounds in water, the situation is as follows:

In the structure of Moclobemide, we spot an amide, and - in the morpholin moiety - an ether and a tertiary amine. Amides do have a lone pair on the nitrogen atom, but due to the conjugation with the carbonyl group, the electron density on the N atom is strongly reduced, as compared to an amine. Ethers are rather lousy bases, in fact, the $pK_a$ of protonated tetrahydrofuran is as low as -2.05.

The remaining tertiary amine thus is the functional group that influences the pH: tertiary amines show a $pK_b$ around 3.3.

In Paracetamol we identify an amide (that we've already excluded as irrelevant) and a phenol group. The latter, once deprotonated, will stabilize the negative charge by resonance with the $\pi$-system (G M has already mentioned this). Therefore, phenols are far more acidic than alkanols. Consequently, the compound definitely acts as an acid.

I'm sure that you've already noticed two imide units in the case of Hexobarbital and in fact, it is possible to abstract the proton from the nitrogen atom. However, you'd need a base such as a carbonate under anhydrous conditions to deprotonate here. As a result, Hexobarbital will hardly do anything at all in water.


Reason for shorter bond between sulphur and oxygen? I recently came across this sentence in my textbook:

  the bonds between sulphur and oxygen in oxides of sulphur ($\ce{SO2}$ and $\ce{SO3}$) are much shorter than might be expected for a single bond.

I feel it could be due to partial double bond character due to resonance.

And the same textbook gives another explanation:

  In these molecules, in addition to normal π bond, a π bond is also
  formed by the sidewise overlap of a filled 2p orbital of oxygen with a
  vacant 3d orbital on the sulphur). This is called pπ - dπ bond and
  results in bringing the two atoms closer and thus accounts for shorter
  bond length of $\ce{S-O}$ bond.

The reason provided in the textbook (as mentioned above) is definitely untrue and incorrect

Because of reasons mentioned in these posts

Why is the bond order in the SO₃ molecule 1.33 and not 2?
.
Hybridization of sulfur in sulfur dioxide

Please provide a explaination for the same (the title question).

If you would consider it more carefully, you would see that the bond order in SO2 and SO3 are, respectively, 1.5 and 4/3. This means that they are shorter than a single bond of sulphur and oxygen.


How to calculate and denote small uncertainties? I am writing a report and I'm stuck on determining how to write the uncertainty.

I have this:

$$1.0 \times \left( 0.1000 \pm 0.0005 \right).$$

If I multiply the values, I need to keep two significant figures, but then my uncertainty would be beyond the correct amount of decimal points. What would I do in this case?

When a number is given without any further information, it is generally interpreted so that the last digit is rounded.

Therefore, the first given value $a=1.0$ is generally assumed to represent a value between $0.95$ and $1.05$, or $a=\left(1.00\pm0.05\right)$. Thus, the assumed uncertainty of $a$ is $u(a)=0.05$.

Conversely, the second given value $b=\left(0.1000\pm0.0005\right)$, where the assumed uncertainty of $b$ is $u(b)=0.0005$, might simply be written as $b=0.100$.

When multiplying numerical values, according to the simplified rules of significance arithmetic, the result is rounded to the number of significant digits in the factor with the least significant figures. Therefore, when you multiply $a=1.0$ (two significant digits) by $b=0.100$ (three significant digits), the result is $\boldsymbol{y=a\cdot b=0.10}$ (rounded to two significant digits).
This result may be estimated to represent a value between $0.095$ and $0.105$, or $y=\left(0.100\pm0.005\right)$; i.e. the assumed uncertainty of $y$ is $u(y)=0.005$.

By way of comparison, a generally better result for the uncertainty $u(y)$ may be obtained by using the rules of propagation of uncertainty (or propagation of error) as follows.

When calculating the product of $a$ and $b$

$$y=a\cdot b$$

and if we assume that the uncertainties of $a$ and $b$ are not correlated, the relative uncertainty of the result $y$ may be estimated as

$$\begin{align}
\left(\frac{u(y)}{y}\right)^2 &= \left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2 \\[6pt]
\frac{u(y)}{y} &= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2}
\end{align}$$

Thus, the absolute uncertainty of $y$ is

$$\begin{align}
u(y) &= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2} \cdot y \\[6pt]
&= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2} \cdot a \cdot b\\[6pt]
&= \sqrt{\left(\frac{0.05}{1.00}\right)^2 + \left(\frac{0.0005}{0.1000}\right)^2} \times 1.00 \times 0.1000\\[6pt]
&= 0.005024938\\[6pt]
&\approx 0.005
\end{align}$$

The same result may be obtained using the general formula

$$\begin{align}
u(y) &= \sqrt{\left(\frac{\partial y}{\partial a}\right)^2 u^2(a)+\left(\frac{\partial y}{\partial b}\right)^2 u^2(b)} \\[6pt]
&= \sqrt{b^2\cdot u^2(a)+ a^2\cdot u^2(b)} \\[6pt]
&= \sqrt{(0.1000)^2\times (0.05)^2+ (1.00)^2\times (0.0005)^2} \\[6pt]
&= 0.005024938\\[6pt]
&\approx 0.005
\end{align}$$

Thus, in this case, the result that is obtained by using the simplified rules of significance arithmetic is a good approximation of the result that is obtained by using the rules of propagation of uncertainty.

(Note that the numerical values of the estimate $y$ and its uncertainty $u(y)$ should not be given with an excessive number of digits, although in some cases it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.)


How is the hydrazone formed in the Wolff-Kishner reduction? In the Wolff-Kishner reduction the first step is to react a carbonyl with hydrazine to form a hydrazone. The later steps of the reaction requires a strong base to cause loss of nitrogen.
However, the mechanism for the formation of the imine/hydrazone is acid catalyzed:

Wikipedia mentions that the formation of the hydrazone is typically done in situ.  So how does the formation of the hydrazone occur in basic conditions?
References:

https://www.organic-chemistry.org/namedreactions/wolff-kishner-reduction.shtm
https://en.wikipedia.org/wiki/Wolff–Kishner_reduction

Wolff–Kishner reduction is the method for the reduction of aldehydes and ketones to corresponding alkanes. First, the carbonyl compound condenses with hydrazine to form the corresponding hydrazone. The resulting hydrazone is treated with base to induce the reduction of the $\mathrm{sp^2}$-carbon to $\mathrm{sp^3}$-$\ce{CH2}$, and the oxidation of the hydrazine to gaseous nitrogen in the process to yield the corresponding alkane:

Source of the image
The following is the accepted mechanism of the reduction:

Source of the image

How does the formation of the hydrazone occur in basic conditions?

The formation of the hydrazone can be done in either acidic or basic conditions (recall imine formation of aldehydes and ketones with primary amine and 2,4-dinitrophenylhydrazone formation of  them with 2,4-dinitrophenylhydrazine). And, yes, as Wikipedia mentioned, sometimes it is advantageous to use a pre-formed hydrazone as substrate for the reduction. Yet, remember, Wolff–Kishner reduction is better suited for substrates with acid sensitive groups. Therefore, use of acid catalyst to pre-form hydrazone for the reduction is meaningless in that situation.


Why does Finkelstein reaction only work with R-Cl and NaI? Why does Finkelstein reaction necessarily involve: R-Cl and NaI (in Acetone)? Why can it not be done with KI? How does potassium instead of Sodium affect the reaction? Why can't R-I react with KCl in the same way?

(I've not included R-I (Iodine) or Fluorine (in the ionic compound) since they are extremes and are very slow or violent in many reactions respectively.)

The classic Finkelstein reaction entails the conversion of an alkyl chloride or an alkyl bromide to an alkyl iodide by treatment with a solution of sodium iodide in anhydrouss acetone (Wikipedia). You can find here why this reaction works in this manner.

The reaction is essentially driven toward products by mass action due to the precipitation of the poorly soluble $\ce{NaCl}$ or $\ce{NaBr}$. Solubility of one specific salt (only $\ce{NaI}$) is important to the reaction. I emphasize above that your acetone has to be very dry, otherwise $\ce{NaCl}$ and $\ce{NaBr}$ are also going to be soluble large scale in acetone and reaction won't go to completion, and thus, won't be useful. For example, the solubility of $\ce{NaI}$ in acetone is found to be 29.2% by weight at $\pu{25 ^\circ C}$ but it is increased to 40.5% when 6.1% (by weight) of water added to the solution (Ref.1). Similar effect should be expected for the solubility of
$\ce{NaCl}$ and $\ce{NaBr}$ (Ref.2).

To answer your particular question about $\ce{KI}$, you need to look at the solubility of it in acetone: Unlike $\ce{NaI}$, which has maximum solubility in acetone at $\pu{25.7 ^\circ C}$, maximum solubility of $\ce{KI}$ in acetone is at $\pu{-57.5 ^\circ C}$ (Ref.3):

  [....]. In this paper are listed the solubilities of potassium iodide in acetone from $+54.5$ to $\pu{-78.5 ^\circ C}$. While, in general, the solubility curves of sodium and potassium iodides are similar in form, the maximum solubility of potassium iodide is about threefold less than that of sodium iodide, and it occurs at $\pu{-57.5 ^\circ C}$, instead of $\pu{+25.7 ^\circ C}$ as was reported for the sodium salt. The solid phase which is in equilibrium with the solution below $\pu{-57.5 ^\circ C}$ is a solvated form containing five (or six?) molecules of acetone per molecule of potassium iodide.

As stated in above abstract, $\ce{KI}$ molecule is surrounded by 5 to 6 acetone molecule, instead of 3 in $\ce{NaI}$ case. That makes it crowded (steric effect) as well. Its solubility is reported as 1.330% by weight at $\pu{+25 ^\circ C}$ (Ref.3), compared 29.2% of $\ce{NaI}$ under the same conditions. These facts make $\ce{KI}$ unsuitable for used in Finkelstein reaction. Yet, there are several instants of using $\ce{KI}$ in this type of reactions, e.g., see Ref.4.

There are also instances of using substrates other than $\ce{R-Cl}$ or $\ce{R-Br}$, e.g., see Ref. 5.

References:

R. Macy, E. W. Thomas, “The System: Sodium Iodide-Acetone-Water,” J. Am. Chem. Soc. 1926, 48(6), 1547–1550 (https://doi.org/10.1021/ja01417a015).
M. Li, D. Constantinescu, L. Wang, A. Mohs, J. Gmehling, “Solubilities of $\ce{NaCl, KCl, LiCl}$, and $\ce{LiBr}$ in Methanol, Ethanol, Acetone, and Mixed Solvents and Correlation Using the LIQUAC Model,” Ind. Eng. Chem. Res. 2010, 49(10), 4981–4988 (https://doi.org/10.1021/ie100027c).
R. Livingston, R. R. Halverson, “Solubility of Potassium Iodide in Acetone,” J. Phys. Chem. 1946, 50(1), 1–6 (https://doi.org/10.1021/j150445a001).
F. G. Bordwell, W. T. Brannen, “The Effect of the Carbonyl and Related Groups on the Reactivity of Halides in $\mathrm{S_N2}$ Reactions,” J. Am. Chem. Soc. 1964, 86(21), 4645–4650 (https://doi.org/10.1021/ja01075a025).
D. J. Maloney, S. M. Hecht, “A Stereocontrolled Synthesis of $\delta$-trans-Tocotrienoloic Acid,” Org. Lett. 2005, 7(19), 4297–4300 (https://doi.org/10.1021/ol051849t).


Calculating enthalpy change for a real gas I know this question would be closed citing it to be a homework question. I posted it earlier also. But believe me I have tried it for myself before and that too many times but then couldn't arrive at a solution maybe due to lack of my conceptual understanding. Any help would greatly be appreciated.

  Calculate $∆H$ for $1$ mole of real gas undergoing change of state from $\pu{10^5 Pa}$, $\pu{300 K}$ to $\pu{2e5 Pa}$, $\pu{600 K}$. The values of real gas constants are $a = 0$ and $b = \pu{50 mL/mol}$.

Besides, we have been given $C_p = 20$ (in SI unit) and the relation

$$ \frac{∂H}{∂P} = T\frac{∂V}{∂P} + V$$

(the partial derivative has been taken keeping the temperature constant on both sides of equality.)

My approach has been like this:

First, I wrote the real gas equation. Then I partially differentiated this whole equation w.r.t. $P$ keeping temperature constant so as to find $∂V/∂P$ and substituted the value of $∂V/∂P$ in the equation given in the question. My actual aim was to use the formula

$$\mathrm{d}H = nC_p\mathrm{d}t + \left(\frac{∂H}{∂p}\right)\mathrm{d}p.$$

Next, I substituted $(∂H/∂p)$ from the given equation in the the given formula, but now I am stuck here.

You started out correctly by getting the relationship for $\mathrm{d}H$ for your van der Waals gas algebraically. The usual approach to evaluating a change like this for a real gas is to break the change into three steps (since you will typically know $C_p$ only in the limit of ideal gas behavior at low pressures):

Step 1: Calculate $\Delta H$ for the change from $\pu{10^5 Pa}$ and $\pu{300 K}$ to $\pu{0 Pa}$ at $\pu{300 K}$.

Step 2: Calculate $\Delta H$ for the change from $\pu{0 Pa}$ and $\pu{300 K}$ to $\pu{0 Pa}$ and $\pu{600 K}$ (i.e., using the ideal gas heat capacity).

Step 3: Calculate $\Delta H$ for the change from $\pu{0 Pa}$ and $\pu{600 K}$ to $\pu{2e5 Pa}$ and $\pu{600 K}$.

Based on Hess' law, add the three changes together.

For this particular problem, the 3 step procedure is not necessary because of the following:

$$\left(\frac{\partial H}{\partial P}\right)_T=b$$

So,

$$\mathrm{d}H = C_p\mathrm{d}T + b\mathrm{d}P$$

Since $b$ is independent of temperature, $C_p$ is independent of pressure. So for this system, the equation integrates immediately to

$$\Delta H = C_p\Delta T + b\Delta P$$


Why does nitration of N,N-dimethylaniline occur at the meta position? The nitration of N,N-dimethylaniline with $\ce{H2SO4}$ and $\ce{HNO3}$ gives mainly the meta product, even though $\ce{-NMe2}$ is an ortho,para-directing group. Why is this so?

In the presence of these strong acids the $\ce{-NMe2}$ group is protonated, and the protonated form is electron-withdrawing via the inductive effect. This discourages attack at the electron-poor ortho position.

Under the conditions I know for that experiment, you get a mixture of para- and meta-product, but no ortho-product due to steric hindrance.


What is the electronic configuration of La⁺? Second IP of La is most likely to second IP of which element:

  (1) Be
  (2) Ba
  (3) Ca
  (4) Zn

The given answer is calcium. I don't understand if the 6s electron or the 5d electron leaves on ionization. Any help is appreciated.

During ionization only the electrons in last shell leaves.So its 6s


How to determine whether the enthalpy of solution is positive or negative by calorimetry? How can we determine if the enthalpy of solution of a compound is positive or negative?

In order to make a clear idea about what I mean, take the following problem:

A 3.5 g sample of $\ce{NH4NO3}$ was added to 80. mL of water in a coffee cup calorimeter of negligible heat capacity. As a result, the temperature of solution decreased from $21.6 \ ^\circ\text{C}$ to $18.1\ ^\circ\text{C}$. Calculate the enthalpy change of the solution ($\Delta H_{\text{sol}}$ ). (specific heat of the solution = $4.18 \frac{\text{J}}{\text{g}\cdot ^\circ\text{C}}$; density of water $1.00 \frac{\text{g}}{\text{mL}}$)

Answer: $+28 \frac{\text{kJ}}{\text{mol}}$

Why it is positive despite there is a decrease in temperature?

The sample takes in the heat from the water, so it takes in energy to "form" the solution and the water thereby loses it, becoming a lower temperature. You need to realize the reference point is not the water, but rather the sample in this case. The heat transferred (lost) by the water is to the sample, so it is the negative of the heat lost by water, making it positive.


Are any bonds broken when NO2 becomes N2O4? This reversible reaction occurs in room condition:

$$\ce{2NO2 <=> N2O4}\quad(\Delta H=-57.23\mbox{ kJ/mol)}$$

This is the Lewis structure of $\ce{NO2}$ (courtesy Ben Mills via Wikipedia):

This is the Lewis structure of $\ce{N2O4}$ (courtesy ChemSpider):

To me, it seems that no bonds need to be broken for the forward reaction ($\ce{2NO2 -> N2O4}$) to occur, which is strange indeed.

No bonds are broken, but there are several isomers of N2O4.  The unpaired electron can be on either the N or the O, for each NO2 molecule. The long bond between the monomers involves these two unpaired electrons.

See sections 7-1 to 7-3 of Bonding in Electron-Rich Molecules: Qualitative Valence-Bond Approach via Increased-valence Structures for more information.

According to First-Principles Study of the Role of Interconversion Between NO2, N2O4, cis-ONO-NO2, and trans-ONO-NO2 in Chemical Processes J. Am. Chem. Soc., 2012, 134 (31), pp 12970–12978:

there is no barrier to the bonding of two monomers to form the planar isomer with the N-N bond.


A theoretical framework for the formalization of chemical interaction and configuration based properties Is there an attempt in any area of chemistry to formalize the way a molecule's properties change based on its configuration and the resulting changes in interaction with other molecules. For example, exchanging an ethyl for methyl group and vice versa.

What I am thinking of is analogous to the pursuit of a ToE(theory of everything) in theoretical physics. If any research like this is occurring or even being conjectured on, what is its current state?

Technically, one could argue that the entirety of chemistry is trying to formalize how properties of molecules change under a given perturbation. From a computational perspective, some work is being done in the context of machine learning towards developing a continuous mathematical representation of molecules, essentially mapping out chemical space. In this way, you could determine how properties change based on how close different configurations of a compound are in this chemical space.

For details/an example, see 1.
For more on this topic, search "chemical space" or, for some of the commonly used mathematical tools used, "principal component analysis" and "variational auto encoder".

Overall, I think a chemical Theory of Everything would have to necessarily be much more qualitative than that of Physics, just due to the emergent complexity that comes from the vast number of combinatorial possibilities of different compounds.

Aspuru-Guzik, A. et al. Automatic chemical design using a data-driven continuous representation of molecules. ACS Central Science. 2018, 4.


Difference between continuous and line spectrum of hydrogen The difference between a continuous and a line spectrum is well understood in that a clear cut boundary is present in the case of a line spectrum. When will continuous spectrum be obtained and when a line spectrum for example  in  hydrogen? What method can be used to observe either or both these spectra?

I will quote part of this answer:

  Quantum mechanics dictates that the energy levels of atomic hydrogen are discrete--just from the way the math works out. The reason why hydrogen's peaks are so spaced out is that it has so few degrees of freedom (just the one electron). Xenon, in contrast, has many closely-spaced peaks because it has 54 electrons (so it has many more accessible energy levels than hydrogen).

The implication is that to a certain resolution, line spectra appear to become continuous once the energy level spacing and the excitation energies become close together. If you look closely enough, they are still discrete.

However, there comes a point where the electron that's being excited is no longer bound to the atom; that would be the first ionization potential (IP) or ionization energy. In the case of the hydrogen atom, that is the Rydberg constant. See this answer for how this arises mathematically.

For more details on the hydrogen atom, see here:

  The infinity level represents the highest possible energy an electron can have as a part of a hydrogen atom. If the electron exceeds that energy, it is no longer a part of the atom. The infinity level represents the point at which ionization of the atom occurs to form a positively charged ion.

At this point it is important to distinguish between hydrogen's absorption/excitation spectrum and its emission/deexcitation spectrum (taken from here):

Since energy is inversely proportional to wavelength, the higher-energy excitations are toward the left of the spectra. This is visual confirmation that as energy level spacing decreases, lines in spectra get closer together. Past that point, what do the spectra look like? Certainly there will be peaks at the ionization energy, which is at a wavelength of about 91 nm. Applying more energy will also strip the electron from the proton, but the excitation from $n = 1$ to $n = \infty$ will still occur at the same energy, and likewise for the absorption from $n = \infty$ to $n = 1$. So there is only one peak in the hydrogen emission and exitation spectra directly resulting from the interaction between the electron and the proton (related). Any potential continuous band structure at higher energies due almost entirely to the free electron in the continuum, or you could think of any point in that spectral region being composed of two contributions, one being the fixed electron-proton interaction energy and the other being related to the kinetic energy of the free electron.

Note that this is for a single atom; for a collection of atoms approaching a bulk phase, you may have continuous spectra under other situations. Any time there are free electrons, energy levels may be continuous. There is also above threshold ionization, but without doing any calculations I am not sure how intense these effects are in terms of directly contributing to UV/vis spectra; if the $x$-axis is electron energy in eV and the $y$-axis is related to excitation probability, the effect is most likely negligible in practice.


How to test for halides without ammoniacal silver nitrate solution? If I have an halide e.g magnesium iodide, how can I test what halide the compound contains?

I understand there is a method using silver nitrate and ammonia but what other methods are there?

ChemGuide gives another method:

  If you add concentrated sulphuric acid to a solid sample of one of the halides you get these results:
  $$
\begin{array}{c|c}
\text{Ion present}&\text{Observation}\\\hline
\ce{F-}&\text{steamy acidic fumes}\\
\ce{Cl-}&\text{steamy acidic fumes}\\
\ce{Br-}&\text{steamy acidic fumes contaminated with brown bromine vapour}\\
\ce{I-}&\text{some steamy fumes, and lots of purple iodine vapour}\\
\end{array}
$$

Note that those "fumes" are of the corresponding hydrogen halides. Also, in the case of the iodide ion, "various red colours in the tube" are also observed.

This reaction occurs because concentrated sulphuric acid acts as an oxidizing agent. It isn't able to oxidize fluoride or chloride ions, but definitely oxidizes $\ce{Br-}$ and $\ce{I-}$ to $\ce{Br2}$ and $\ce{I2}$ respectively (details here)

Do note that, to distinguish between the fumes of $\ce{HF}$ and $\ce{HCl}$ above, you'll need to dissolve them in water and perform the silver nitrate test!


Does it change the compound if the element letters in its molecular formula are switched? Does it change the compound if the element letters are switched? ex. $\ce{HO}$ or $\ce{OH}$. Are those two different compounds or are they one in the same?

Apart from affecting organic compounds, it also affects coordination compounds.

$\ce{Hg[Co(SCN)4]}$ is a significantly different compound from $\ce{Co[Hg(SCN)4]}$. The central metal atom and the counter-ions in both cases differ, as does the complex anion.
In fact, in the above compound itself, $\ce{Hg[Co(SCN)4]}$ differs from $\ce{Hg[Co(CNS)4]}$ or $\ce{Hg[Co(NCS)4]}$: this is called linkage isomerism. The order in which ligand atoms are written denotes the atom of attachment to the central metal atom, which happens to differ in this case.
The complex salts $\ce{[Co(NH4)6]^3+[Cr(CN)6]^3-}$ and $\ce{[Co(CN)6]^3+[Cr(NH4)6]^3-}$ - formed by swapping the two radicals - are coordination isomers, and not identical.


In practice, how dangerous is liquid oxygen as an oxidizing agent? The Wikipedia page on liquid oxygen tells us

  Liquid oxygen is also a very powerful oxidizing agent [...], if soaked in liquid oxygen, some materials such as coal briquettes, carbon black, etc., can detonate unpredictably from sources of ignition such as flames, sparks or impact from light blows.

I have heard similar warnings about LOX in texts about amateur and semi-professional rocketry, but I know nobody who has ever handled stuff like this.

In practice, say I spill some LOX and soak paper lying around (that maybe has been warmed by a desk lamp), will that burst into a fireball?

Maybe more to the point, how do I assess such risks short of conducting foolhardy experiments? Is there a way to estimate the maximum effect a given amount of spillage on a given material may have?

I'll have to acknowledge my question might be too vague for a straightforward answer, as can be seen by the various comments. The correct answer, as given by Paul J. Gans, probably is

[V]ery. I know of no way to quantify this.

However, using the comments as a starting point, I read some more around the topic, and I would like to expand on the comments to the question and to summarize what I found out.
Since there's no unit to measure "dangerous" with, we'll first have to get more specific. In the context above, we can offer the following dimensions:

ease with which LOX + X is set on fire or detonated, the easier, the more dangerous
predictability of ignition, the less predictable, the more dangerous
energy output of ignition, the more energy the more dangerous
speed of reaction, the faster the worse

The core problem with all of the above is that the LOX is not a fuel and will, on its own, do nothing at all. So before we can even speculate on possible reactions, we will have to narrow down the fuel options our spilled LOX might encounter. When we also want to account for oxygen-enriched atmosphere by boiled off LOX, a danger in itself, even without spilled, as matt_black correctly points out, the fuel can be almost anything burnable in the room.
That pretty much takes care of 2), predictability, which is low, so keeping the whole setup safe is difficult. -> Danger: high
Ease of ignition depends on the fuel, and the number we're after is the activation energy (Wikipedia). I found various Internet pages mentioning LOX spill accidents that turned tarmac into an explosive that can be set off by walking or driving over it, although I have not found an actual report of such an accident. So apparently, there are common materials that combine with LOX to a mixture requiring a very low activation energy. -> Danger: high.
Energy content and speed of reaction also depend on the fuel, but the fact that it's being used as rocket propellant and that LOX was used to build explosive charges in mining (so called Oxyliquits (Wikipedia), via permeakra's comment) means that we can expect plenty of both for a number of materials. The article also emphasizes the unpredictability, even of things that are meant to explode. -> Danger: high.
Finally, various safety guidelines inspire vivid imaginations of some scenarios:

The whole floor of an office is known to have caught fire when oxygen vapors contacted a lighted cigarette butt. -- Saftey Precautions for Oxygen Plants
If liquid oxygen spills on asphalt or other surfaces contaminated with combustibles, do not walk on or roll equipment over the area of the spill. Liquid Oxygen, Environmental Health and Safety


Rate of reaction and dependence on concentrations Consider the decomposition of ozone as follows

(fast)
$$\ce{O3  <=> O2 + O}$$

(slow)
$$\ce{O + O3  -> O2 }$$

The question comes here if the concentration of $\ce{O2}$ is increased , then comment on the rate of the reaction.

increases
decreases
constant
cannot be predicted

My try: I have a confusion that the rate law is not given so we cannot tell about the actual dependence of the rate on the concentration of $\ce{O2}$, I know that the rate law will depend upon the slow step as it is the rate determining step. Answer given is 2) decreases , can we apply Le Chatelier's principle here? Increase in $\ce{O2}$ , backward reaction so less rate?

If we carefully see them ozone formation is a complex ( not elementary) reaction.
Yes the slowest step would be rate determining .
so for slowest step ;

Rate =$   k [O] [O _3] $

where $k$ would be rate constant.

but  final reaction is :
$$\ce { 2 O_3 -> 3O_2} $$
and so ,
equillibrium constant K(for 1st reaction)

$K= \frac{[O][O_2]}{[O_3]}$

thus substituting $[O] $from above equation to the rate equation .

We get

Rate =$\frac {  k × K [O _3]^2 }{[O_2]}$

And hence if $[O_2]$ is increased it causes rate to decrease.


Will UVC light/ozone affects color on fabrics? I'm planning to put a low power UVC light bulb into my wardrobe to kill bacteria and prevent bad smell on clothes but I'm wondering will it affect color on the clothes? I've done some research but haven't got into a conclusion if UVC light will affect color faster than sunlight or not (because there's no UVC light reaches the earth surface).

I have 2 light bulbs, 1 able to produce ozone and the other doesn't, ozone can help reaching the dead spots which UV rays can't reach but I've read that ozone will cause cracking on some plastic do fabrics also get affected by that?

I'm planning to put a low power UVC light bulb into my wardrobe to kill bacteria and prevent bad smell on clothes but I'm wondering will it affect color on the clothes?

If the UV light is strong enough to break bonds to kill bacteria it is certainly strong enough to bleach the organic dyes and degrade the clothing.

  I've done some research but haven't got into a conclusion if UVC light will affect color faster than sunlight or not (because there's no UVC light reaches the earth surface).

This depends on the brightness of the bulb and the sun where you live

  ...but I've read that ozone will cause cracking on some plastic do fabrics also get affected by that?

Yes fabrics are also degraded by ozone.

Keep in mind that though the UV and ozone may cause degradation of your clothes, they may or may not do so faster than mold or normal use would too many factors to say with certainty which is best.


Counting number of acidic-H in a molecule In a recent exam, following question was asked:

Number of acidic hydrogen present in X are:

I've found 8 of them as shown:

7 of them (marked with numerals) were correct according to solution, but 8th one (pointing with red arrow) was not. Is it because of the fact that the carbon was bridge-headed or something else?

An aldehyde or a ketone with an alpha hydrogen forms a carbanion that resonates to enolate form. This leads to two canonical structures that are in resonance. The example below is a polyphenol compound extracted from turmeric, called curcumin.

source: Biomaterials. 2010 May;31(14):4179-85. doi: 10.1016/j.biomaterials.2010.01.142. Epub 2010 Feb 23. (https://www.ncbi.nlm.nih.gov/pubmed/20181392)

The hydrogens between two keto groups are most acidic. This is due to formation of two equivalent enol structures stabilized by hydrogen bonding. The structural requirement would be planarity of atoms involved in hydrogen bonding.

Below are 3-D structures of the given compound in the question above.

Structure without hydrogens

Bridgehead carbon is 4. Keto groups are at 2 , 5 and 8 (shaded in grey). Keto groups on 5 and 8 are not in the same plane . Hence Hydrogen on 4 , 22 (see figure with hydrogens) does not form stable hydrogen bonded enol structures as discussed in the above example.

Therefore, hydrogen at 8th one (pointing with red arrow) may not be acidic.

Note: 3-D structures have been drawn on Chem3Dpro

Structure with hydrogens

Structure with hydrogens (2nd view)


What affects bond angles? I have three factors in mind:

1) vdW repulsions

2) Bond length

3) Electron-electron repulsions.

I know that vdW repulsions between ligand atoms push the atoms apart. And bond length obviously influences bond angle; if the bonds were very long, then the bond angle can shrink without bringing the ligand atoms too close together.

Are there any other factors or are these it?

Bond angles mainly depend on the following three factors:

Hybridization: Bond angle depends on the state of hybridization of the central atom
Hybridization: $\ce{sp^3}$, Bond angle: $109^\circ$, Example: $\ce{CH4}$
Hybridization: $\ce{sp^2}$, Bond angle: $120^\circ$, Example: $\ce{BCl3}$
Hybridization: $\ce{sp}$,   Bond angle: $180^\circ$, Example: $\ce{BeCl2}$
Generally s- character increase in the hybrid bond, the bond angle increases.
Lone pair repulsion: Bond angle is affected by the presence of lone pair of electrons at the central atom. A lone pair of electrons at the central atom always tries to repel the shared pair (bonded pair) of electrons. Due to this, the bonds are displaced slightly inside resulting in a decrease of bond angle.
Electronegativity: If the electronegativity of the central atom decreases, bond angle decreases.

Good Side note:

Triple bonds repel other bonding-electrons more strongly than double bonds.
Double bonds repel other bonding-electrons more strongly than single bonds.

Heres a nice read into the different components (extra)


Unknown surface coating Forgive me if this isn't the best forum for this question, and guide me elsewhere if so.
I have two identical smartphones, both made by Sony (Xperia z2 to be exact). One I purchased new and one second hand. I want to affix an adhesive-backed mounting system to the back of both of them.
With one phone (that I had from new) there's no trouble, everything (e.g. adhesive-backed Velcro) sticks really well. But with the other phone nothing will stick . I've checked that there's no protective overlay sheet and can only assume it's been treated with some sort of chemical coating. Every solvent I've tried to remove this coating just forms a meniscus and runs straight off. Water, detergent, isopropyl, acetone, WD-40 - nothing makes any difference. Any suggestions?

Very likely, the title of your question already hits the nail.

Since mobile phone tend to be "smart", filled to the rim with computing power and data, users wish to protect their companions with a coating that repells water.  Often, the devices are not only coated on the front side, which would at least hamper accumulation of residues of gliding fingers, but all around, including the back.  As a result, you might get a cell phone that may drop into a bowl of water, and still remain functional.

Among the different technologies available (a short presentation by CNN, back 2013 is found here), there are polymer coatings, or a mixture of anastase / $\ce{TiO2}$ and PFTE / teflon (like the coating of your frying pan) particles with some perfluorinated liquid (example).  For some, you need a larger installation that appears like a scilab, for others, you may prepare it at home with a spray can (example).  Hence the results in terms of performance, maintenance, and longevity may vary, too.  As a result, if applied both on front and back of the phone, not only water, but the components of "normal glue" (referring to what backs a velcro tape) may be repelled more or less efficiently, too.

Your observation that the coating appears little to non-affected both by water as well as iso-propanol (which for some coatings is a no-go, as it would tear the layer off), as well as for WD 40 (which may be summerised as a blend of apolar hydrocarbons (ref)) allows the speculation that the coating might be either polymer-based, or the triple mix of anastase + teflon + perfluorinated binder.  To discern between the two, remotely, is not possible.

If a mechanical, pocket-shaped mount like

(source)

or a pouch, to be bound around your upper arm while riding a bicycle, like

(source)

is not what you may deploy (or alter a little to fit your needs), and if the coating consists of the "polymer-only" variant, then maybe it is possible to establish a connection between the back plate of the phone, and either the velcro tape or something already connected to the velcro tape with the help of cyan-acrylate based gorrilla super glue.  These glues often bind very strong and to polymers, however partially dissole the polymer a little.  This chemical approach however comes with many "ifs" and uncertain outcome.


Why doesn't Epsom salt dissolved in water act like sulfuric acid? People take baths in it so it clearly doesn't. But if epsom salt is the salt of sulfuric acid when it dissolves the chemical ions present in sulfuric acid should also be present in the bath water and in abundance enough to melt you.. right?

I also remember that water autodissociates into hydronium and hydroxide, so my other question is why wouldn't plain water simultaneously be a acid and a base and also just melt you?

If there's a trillion (or whatever) water molecules next to your skin and even a tiny fraction of them turn into an acid and some of them turn into a base, there'd cumulatively be a lot of acid on your skin. I can't imagine adding salt which would also ionize could help things.

Well, as you might have guessed, it's the $\ce{H+}$ in sulfuric that is dangerous, not the $\ce{SO4^2-}$. So, Epsom salts, being $\ce{MgSO4}$, aren't very dangerous.

In the same vein, table salt ($\ce{NaCl}$) is perfectly OK despite hydrochloric acid ($\ce{HCl}$) being rather nasty.

And $\ce{H+}$ is only corrosive in high concentrations, mind you - otherwise we'd be burning ourselves drinking orange juice, which itself has a small concentration of $\ce{H+}$. That also answers why water doesn't kill you: only a tiny fraction of it (roughly 1 in ten million molecules) actually undergoes autodissociation at any one time. It's not about the actual number of $\ce{H+}$ ions, but rather the concentration. If you dilute $\ce{HCl}$ enough you can pretty much just drink it. (Please don't be stupid and take this as a licence to do a home experiment and drink $\ce{HCl}$.)


Fundamental vibrational modes for QM9 molecules QM9 dataset contains about $134000$ small organic molecules with following properties computed using DFT approaches:
A = "rotational_constant_A"
B = "rotational_constant_B"
C = "rotational_constant_C"
mu = "dipole_moment"
alpha = "isotropic_polarizability"
homo = "homo"
lumo = "lumo"
gap = "gap"
r2 = "electronic_spatial_extent"
zpve = "zpve"
U0 = "energy_U0"
U = "energy_U"
H = "enthalpy_H"
G = "free_energy"
Cv = "heat_capacity"

In many papers, fundamental vibrational modes $\omega_1$ is also mentioned along with these properties. I was wondering:

Can we calculate $\omega_1$ given the properties above?
Are rotational constants used for computing $\omega_1$?
Do we need DFT calculations to compute $\omega_1$ and zero point vibrational energy (ZPVE)?

Only for a diatomic molecule and only assuming the vibration is harmonic. In this case, $\mathrm{ZPVE}=\omega_1/2$. If its anharmonic or has rovibrational coupling, even the expression for a diatomic doesn't not allow you to determine $\omega_1$ from just the ZPVE. With polyatomic molecules, the ZPVE has contributions from all the vibrations, so you can't just extract a single vibrational frequency.
Rotational constants are not used in computing $\omega_1$ (unless you are considering forms of rovibrational coupling).
You can determine $\omega_1$ from other electronic structure methods and even classical molecular dynamics, but assuming you don't want to compare against experimental data, you would need to repeat these DFT calculations to get the corresponding $\omega_1$. For this dataset to include ZPVE, the underlying calculations would have had to determine all the vibrational frequencies of the molecule, but they seem not to have included this data in the QM9 set.


Number of stable molecules of artificial molecule XY4 Suppose element X has $5$ stable isotopes, and element Y has $6$ stable isotopes. Find the number of natural molecules, knowing that an X has a charge of $+4$ and Y has a charge of $-1$.

What I tried was this: for element X we have $5$ options, and for each element Y we have $6$ options. Therefore the total number of options would be $5 × 6 × 6 × 6 × 6$.

But this answer is clearly wrong, as it counts the same type of molecule, multiple items. (AAAB) and (AAABA) count twice which we don't want to happen. (A is an isotope of Y and B is some other isotope of B)

What can I do?

There are 5 ways to choose the central atom.

Choosing the possible combinations of the other 4 items from 6 isotopes is complicated as this article explains. But it provides a formula for combinations with repetition that gives the answer of 126.

This implies that there are 630 distinct isotope combinations in the molecule (assuming that we can distinguish combinations of isotopes with the same atomic number which is possible with high resolution mass spectroscopy). If only integer atomic numbers count then the distinguisahble combinations will be lower. But the easiest way to work that out is to do a Monte-Carlo simulation not to look up probability formulae.


Principle of the Biuret Test What is the principle of the biuret test? While reading online, it is unclear to me how a bond is created between the $\ce {Cu^{2+}}$ and the proteins? What are the chemical reactions that lead to the color change?

Edit

The biuret test is a chemical test combining two products, $\ce{CuSO_{4}}$ and $\ce{OH^{-}}$ in my case. While exploring online the possible effects on protein, some sites suggest that the $\ce{OH^{-}}$ is responsible of removing on hydrogen from the amine and that is why there is a color change while other suggest that the $\ce{OH^{-}}$ would rather act as a redactor for the copper and the release of electrons from the copper would transfer the copper ion from a cupric form to cuprous one. (Copper-based assay chemistries)
I have also understood that the copper geometry is changing during the reaction, but the idea of the reduction is unclear to me.

I will ask my question again, what are the chemical reactions that lead to the color change?

The $\ce{Cu^{2+}}$ ion is capable of coordinating to the lone pairs of the nitrogens of the amide bonds.

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The formation of this complex is associated with a certain energy change in the orbitals of the $\ce{Cu^{2+}}$ ion, and hence a color change.


Differences between Urushibara nickel and Raney nickel The Wikipedia article about Urushibara nickel catalyst is rather short and vague. What is the mechanism for its preparation, and furthermore, in what sense does it differ from Raney nickel (i.e. when should you prefer the one over the other)?

You can find information about the preparation of Urushibara nickel (U-Ni) in the original paper by Urushibara[1].

Regarding the differences with Raney nickel (R-Ni), surface characterization by means of XPS[2] had shown that R-Ni catalysts contain metallic aluminium in the activated phase as well as oxidized aluminium, while U-Ni catalysts contain metallic zinc in the activated phase in addition to oxidized zinc. Also, U-Ni catalysts can be described as supported nickel catalysts, while R-Ni catalysts are shown to have the well known skeletal structure.

The activity of U-Ni catalysts is very similar to R-Ni catalysts, both in nature of the reaction catalysed and in the rate of catalysis. One of the advantage of U-Ni is their simplicity of preparation, while R-Ni catalytic synthesis requires a long time. Despite this, U-Ni catalysts are also not very sensitive to impurities, so they can be prepared from commercial chemicals of ordinary grade, and they can be stored for a long time without decreasing the activity. Another advantage is the safer handling, since U-Ni catalysts are not inflammable upon contact with air, in contrast to Raney catalysts which are pyrophoric. This allows the catalyst to be recovered and regenerated. However, it should be noted that U-Ni catalyst durability is not satisfactory during the reaction, but this is also the case of R-Ni catalysts, so there is not really a big difference when it comes to practical use, maybe slightly supporting the use of U-Ni vs R-Ni as the former is less expensive and easy to produce and regenerate. The only real disadvantage of U-Ni catalysts is that for some particular reductions, they show less activity than R-Ni catalysts, favouring one reaction over another, for example U-Ni catalysts particularly favours the partial hydrogenation of acetylenic compounds to ethylenes[3].

     1 Urushibara, Yoshiyuki. "A New Method of Catalytic Hydrogenation". Bulletin of the Chemical Society of Japan 25 (4): 280–280
     2 Y. Okamoto, Y. Nitta, T. Imanaka and S. Teranishi, J. Chem. Soc., Faraday Trans. 1, 1979, 75, 2027
     3 Kazuo Hata, “New Hydrogenation Catalysts: Urushibara Catalysts” (1971), pp 235-237