How to determine polarity of components in TLC Estimate how a TLC-analysis would look like for the following reaction when you take a TLC at: a) the start of the reaction, b) after 65% conversion and c) after full conversion of A. (You may assume that A-C separate on the TLC-plate)

I am having a very difficult time understand how you can determine which compound is more polar by just looking at the structure. I don't know what to look at or what to prioritize.
When attempting to answer this question I assumed that B would be more polar (why: because it contains a nitrogen which is more electronegative than oxygen). So B should be closest to the baseline and A should be further away. What I then thought was that C should be in between A and B because the reaction reacts something polar and something less polar. This was not correct. Our professor drew the TLC-analysis like this:

And I don't really understand how I can determine that the product is more polar than both of the reactants. Is it because it contains both an oxygen and a nitrogen?
I would truly appreciate it if someone could give me some general rules or steps to follow to easily determine the polarity of different compounds.

I am assuming this is TLC run on silica. Silica is a mildly acidic surface so basic compounds such as amines do not come off the baseline unless you use a very polar eluent with added base (e.g. ammonia/triethylamine). Starting material B is a basic amine. Starting material A is an epoxide and fairly non-polar (kinda similar to an ether) and will run high in most systems, certainly any system where B comes off the baseline. Your product C is more basic (tertiary v secondary), this is the strongest effect. The second non-polar aromatic group raises the likelihood that it will run above B, but this is a weak effect. The alcohol is a polar group and will also pull it down thus the sum of these effects is that it runs lower

This missing information is the eluent. From what what is given in the question it is clear that the eluent is something like DCM/MeOH/NH3 (that would be my first attempt to separate such basic components). As B is in excess what you are really looking for is loss of A. When you can no longer see any A by TLC, the reaction can be worked up.

The groups that make for low running are those that interact strongly with the silica surface: basic amines, acids, alcohols. This is a general rule of thumb, but in my experience trying to guess where a multi-functioned molecule is going to run is quite a lottery. I do not believe that it is a safe assumption that a product similar to C would always run below a SM similar to B.


How do I perform thin-layer chromatography to identify alkaloid content in fruits? Fruits and as well as other plant parts usually contain alkaloids, which are nitrogenous compounds. I'd like to know how the  Thin-Layer Chromatography method can be used to determine the alkaloid content in a fruit.

TLC in alkaloid research

Both analytical and preparative TLC have been used to separate and identify alkaloids from plant material.

Pretreatment of TLC plates and adsorbents

If you are using silica plates, it is often advantageous to wave the plate above an open bottle with aqueous ammonia or pretreat the plates with an organic solvent containing some triethylamine. Otherwise, the $\mathrm{R_f}$ values of the alkaloids might be too small.

Alkaloid detection on a TLC

As pointed out by Zhe in a comment, this is crucial for preparative TLC. Usually, you run an analytical TLC (same material and mobile phase) in parallel and stain the analytical TLC using Dragendorff's reagent.

General hints on the extraction

Process different parts of a plant separately

Different parts of a plants have different functions. The distribution of particular compounds or compounds classes, such as

chlorophylls and carotenoids
quinones
polyphenols
carbohydrates
terpenoids
alkaloids

will vary in flowers, fruits, seeds, leaves, stems (and bark), roots, and tubers.

It is therefore reasonable to process the different parts of a plant separately.

Use dry plant material

The water that you remove in the beginning doesn't disturb the extraction in a later stage. Dry the plant material. If you assume that the compounds you are looking for are air-sensitive, dry the material in vacuum.

Use powdered plant material

Extraction is more efficient on solid materials with a larger surface.

Use the right solvent

Two different strategies are conceivable:

Extract all soluble compounds with a polar protic solvent (methanol, ethanol) at once and partition between solvents of different polarity later
Leach the dry plant material stepwise with solvents of increasing polarity:

cyclohexane
ethyl acetate
chloroform
methanol

Use the right technique

You can

stir/shake the dried plant material with a solvent at room temperature
boil plant the material in a solvent
use a Soxhlet extractor

If necessary, perform the extraction under a nitrogen or argon atmosphere

Specific advice on extraction

Rambutan (Nephelium lappaceum)

In Antioxidants, 2014, 3, 371-386 (Open Access article), the authors write

  The fruit samples were washed and air-dried, followed by drying in an oven at 40 °C for 24 h. All the dried samples were ground to fine powder with a grinder, after which they were vacuum packed into a nylon-linear low density polyethylene film before analysis. An extraction with 80%
  ethanol was carried out, and the mixture was shaken at 50 °C for 120 min using an orbital shaking incubator as optimized in a previous study. The residue was subjected to another round of extraction under similar conditions to obtain maximal yield of antioxidant activities. The extracts were combined and centrifuged, followed by filtration and concentration of the supernatant using a rotary evaporator at 45 °C. The concentrated extract was freeze-dried, wrapped with aluminum foil and kept at −20 °C. The crude extract as then partially fractionated using silica packed open column chromatography and using increasing polarity of solvent, ethyl acetate, chloroform and methanol (100:0:0 to 0:0:100 v/v).

In the article cited above, alkaloids were not mentioned.

Ambarella (Spondias dulcis)

In Asian Pac J Trop Biomed., 2013, 3, 682-691, the authors give the following procedure

  The pulverized plant materials were soaked in 3.5 L of methanol in separate containers and kept for 1 week with occasional shaking. The whole mixtures were then filtered through sterilized cotton followed by Whatman No. 1 filter paper and concentrated using a rotary evaporator (IKA , Germany). The weight of crude extracts of fruits and leaves were 135 g and 101 g, respectively. A total of 40 g crude extract of each plant materials was then partitioned with chloroform and dichloromethane. The percentage
  yields of fruit and leaves were methanol ( 21.3 % , 15.8 %), chloroform ( 19.5 % , 23.1 %) and dichloromethane (20.9 % , 25.4 %).

The authors mention that extracts both from the fruits and the leaves of S. sulcis gave a positive test with Dragendorff's reagent, but do not provide further information on the amount and/or nature of the alkaloids.


Interpretation of the following IR? I have the following IR Spectrum that I am trying to interpret:

So far I am thinking this compound have:

a primary amine due to two peaks at 3624 and 3533.
The sharp peak 3416 could indicate an alkyne(sp C-H).
And the other sharp peak at 1716 is an aldehyde or ketone.

Is this analysis correct ? And what have I missed ? Thanks for any help !

The sharp, intense band at 1716 $\mathrm{cm^{-1}}$ is indeed characteristic for a carbonyl group ($\ce{C=O}$ stretch, in this case aldehyde or ketone). The medium and broad bands between 2732-2957 $\mathrm{cm^{-1}}$ suggest that it is more likely an aldehyde. In this case, they would be assigned to the $\ce{=C-H}$ stretch, and the intensity of the broad band fits better to a saturated aldehyde, i.e. there is no double or triple bond in $\alpha, \beta$-position to the $\ce{CHO}$ group.

Your assignment of the peaks at 3624 and 3533 $\mathrm{cm^{-1}}$ to a primary amine is reasonable, since primary amines have two bands, while a secondary amine would have only one, weaker band at lower wavenumbers. One of the bands in the area of 700-940 $\mathrm{cm^{-1}}$ can then be assigned to the $\ce{N-H}$ wagging vibration mode.

The sharp band at 3416 $\mathrm{cm^{-1}}$ could indicate a terminal alkyne ($\ce{C \equiv C-H}$ stretch), the $\ce{C-H}$ stretch band of this group at approximately 2900 $\mathrm{cm^{-1}}$ would then be included in the broad band at 2957 $\mathrm{cm^{-1}}$. The weak band in the area between 2000-2250 $\mathrm{cm^{-1}}$ can then be assigned to the $\ce{C \equiv C}$ stretch.

Sources: 1, 2, 3, 4


Finding half-life of a nuclide based on gas release Full Question: The hydride of an unstable nuclide of a Group IIA metal, $\ce{MH2 (s)}$, decays by alpha emission. A $\pu{0.025 mol}$ sample of the hydride is placed in an evacuated $\pu{2.0 L}$ container at $\pu{298 K}$. After 82 minutes, the pressure in the container is $\pu{0.55 atm}$. Find the half-life of the nuclide.
Here's what I have so far:

Find nuclear equation MH2(s) ---> He(g) + Noble gas + H2 (g) (not sure if this is right)

Find total moles of gas in container at 80 mins
$$n = \frac{PV}{RT} = \frac{(\pu{0.55 atm})(\pu{2 L})}{(\pu{0.0821 L atm mol-1 K-1})(\pu{298 K})} = \pu{0.045 mol}$$

Use integrated first rate law to calculate k and half-life.
Use $\ln N_t = -kt + \ln N_i$, where $N_i$ is the initial amount of moles of $\ce{MH2}$ ($\pu{0.025 mol}$) and $N_t$ is the moles of hydride remaining after 80 minutes.

Then find half life using k.
$$t_{1/2} = \ln 2 / k$$

Here's what I'm having trouble on: How to find $N_t$, i.e. the moles of metal hydride remaining. This part has been stumping me for quite some time, so help is appreciated! Thank you!

For this we can first write the equation for the reaction again:
$$MH_2 => He_{(g)} + Noble gas + H_{2(g)}$$
Next we can find the moles of gas at 80 mins (I am assuming it is 80 minutes and not 82). You already did this and you got 0.045 moles. Since each mole of $MH_2$ forms 3 moles of gas, $\frac{0.045}{3}=0.015$ moles of $MH_2$ used in 80 minutes. Then we can use the first order integrated rate law now:
$$ln(N_f) = -kt + ln(N_i)$$
$$ln(0.025-0.015) = -k(80 mins) + ln(0.025)$$
$$k = 0.0115 min^{-1}$$
Now that we have k, we can solve for the half-life.
$$t_{1/2}=\frac{ln(2)}{k}$$
$$t_{1/2}=\frac{ln(2)}{0.0115min^{-1}} \approx 61 mins$$


Extract acid from a solution I have a solution containing glucose and some sulfuric acid. I want to distill out the sulfuric acid to then reuse it for other purposes. The problem is the presence of glucose in the solution. I'm afraid that, when increasing the temperature, the sulfuric acid will dehydrate glucose to carbon and other compounds - something that I don't want to happen. Would distillation cause such a reaction?

And in case it would - are there any other methods to extract the sulfuric acid?

As above, your labor all by itself (even if you are grad student) is worth more than the sulfuric acid.  As a pedagogic exercise, run the solution down a strong anion exchange resin washed OH-.  Only glucose comes out.  Recover the sulfuric acid by displacing it with phosphoric acid.


Is it possible to melt HDPE to a point where it has similar viscosity to engine oil? I have a few pieces of HDPE that I want to melt and cast into a cube of around  $\pu{5 cm}$ each side, similar to what one would do with cast resin.

I tried putting it into an oven and heating it to $\pu{250°C}$, where it melted, but never enough to pour it somewhere.

Is there a viable way to melt HDPE scraps to a point where it becomes fluid, with a viscosity similar to engine oil, and retains this state long enough to pour it?

You may see a small reduction in viscosity with increasing temperature, but at elevated temperature the polymer will start to oxidize and/or undergo thermolysis. If I were trying to reduce viscosity I'd try blending in some low molecular weight polyethylene (LDPE). Not sure if you have the capability of doing this or if you could tolerate the resulting changes in property (loss of tensile strength, etc.). This reference:

https://doi.org/10.1080/03602550600609663

suggests you may need to go as high as 40%+ to get a significant change in viscosity.


Electrophilic Aromatic Substitution of phenols What happens if 2,4,6-tribromophenol is forced to undergo further bromination?

In phenol, the benzene ring is activated towards electrophilic substitution. Therefore, phenol on bromination gives 2,4,6-tribromophenol. If now, 2, 4, 6-tribromophenol is forced to undergo bromination, the group attached to the benzene nucleus makes phenol not suitable to undergo Friedel–Crafts reaction due to adduct formation with Lewis acid.

Way overthinking this.  Kohn reports that 2,4,6-tribromophenol can be further brominated to pentabromophenol by adding more bromine and an iron catalyst.  Subsequently the pentabromophenol is oxidized by fuming nitric acid to give a fully brominated quinone.


Allylic Radicals Bromination Mechanism For (a) (iii), how do you get the second product listed in the answers manual? I just can't visualize it.

What about (c) (i), why are there three resonance structures? If a hydrogen is taken away from a specific carbon of the initial molecule to have a radical, there is no way you can have three resonance structures.

The barrier to rotation in the allyl radical is low (~15-16 kcal/m) and it can undergo cis-trans isomerization.

As to question (c)(i), you're right, there are only two resonance structures.

What you show above with three resonance structures is the trimethylenemethane diradical, isobutene with 2 allylic hydrogens removed.


Why does radiocarbon dating only work in nonliving creatures? I understand how carbon dating works, though I do not understand why it doesn't  happen while a creature is living. Because while we are alive we still have carbon 14 in us, so shouldn't it work?

There are plenty of good sources online explaining the principle behind radiocarbon dating. For instance, the wikipedia explains:

  During its life, a plant or animal is in equilibrium with its surroundings by exchanging carbon either with the atmosphere, or through its diet. It will therefore have the same proportion of $\ce{^14C}$ as the atmosphere, or in the case of marine animals or plants, with the ocean. Once it dies, it ceases to acquire $\ce{^14C}$, but the $\ce{^14C}$ within its biological material at that time will continue to decay, and so the ratio of $\ce{^14C}$ to $\ce{^12C}$ in its remains will gradually decrease. Because $\ce{^14C}$ decays at a known rate, the proportion of radiocarbon can be used to determine how long it has been since a given sample stopped exchanging carbon – the older the sample, the less $\ce{^14C}$ will be left. [16]

Your question is somewhat subtle: why doesn't decay begin immediately once the isotopes have been incorporated into an organism? Can't measurements of part of an organism tell us when that particular organ or body part was formed?

The key limitation is explained in a document from a company (Beta Analytic) that provides biobased content / renewable carbon measurements:

  Although carbon-14 is radioactively decaying away in the body, it is constantly being replaced by new photosynthesis or the ingestion of food, leaving the amount relatively constant.

But if a body part no longer cycles carbon (is in effect dead), then radiocarbon dating should be useful. One application, for instance, is in dendrochronology, the analysis of tree growth rings to assess age and location:

  Dendrochronology is useful for determining the precise age of samples, especially those that are too recent for radiocarbon dating, which always produces a range rather than an exact date, to be very accurate. However, for a precise date of the death of the tree a full sample to the edge is needed, which most trimmed timber will not provide. It also gives data on the timing of events and rates of change in the environment (most prominently climate) and also in wood found in archaeology or works of art and architecture, such as old panel paintings. It is also used as a check in radiocarbon dating to calibrate radiocarbon ages.1

Combining tree samples of known age (thanks to dendrochronology) and other sources with radiocarbon ratios provides a calibration method.


Stereochemistry in Sn2 reaction I'm confused why the absolute configuration always has to change after an $S_N2$ reaction:

For example, in primary alkyl halide above there are 4 planes Nu can attack; and absolute configuration is changed after an $S_N2$ reaction because the nucleophile (Nu) only attacks the 'black plane'.
I can understand that the nucleophile wouldn't prefer red or green plane because of steric hindrance, but here's my question:

Why doesn't the nucleophile attack the 'blue plane'?

The attacking nucleophile does not just shove its electron pair to the carbon from some random direction. No, it must donate that pair to the antibonding orbital ($\sigma^*$) of the $\ce{C-X}$ bond, thus weakening and eventually breaking that bond. And it just so happens that the antibonding orbital is mostly located on the side opposite to X.

Upd. Thought I'd better look up a nice picture. Here's one:

(from this page)


E1 elimination and alkyne formation Are E1 elimination reactions possible from a trans dihalide alkane (like 1,2-dibromoethane) all the way to the alkyne (like acetylene)? Or is only E2 possible from the alkene to alkyne? What is the rate of reaction?

My thoughts are that E1 is a possible reaction to form the alkyne, with E2 taking place if there are any anti peri-planar hydrogens and a strong base. So, another question might be that which of these has the faster rate of reaction - an E1 reaction or an E2 reaction with anti periplanar hydrogens that are only available through conformational energies and torsional strain?

Yes, in your example, it is possible to form acetylene. E1 mechanism does not depend on whether the leaving group is anti to $\ce{H}$ or not.

All vicinal halides will not give alkynes, as E1 mechanism involves a carbocation intermediate which can rearrange to the most stable position. Example: In 4-methyl-2,3-dibromopentane, the product will be a conjugated alkene not alkyne.

Rate of E1 elimination depends on the stability of the carbocation whereas rate of E2 mechanism does not. Since in your example, the carbocation formed is a 1 degree, it is not very stable so the rate will be higher for E2 mechanism.


How does temperature affect equilibrium, according to Le Chatelier's principle?? I want to understand this concept in detail. I am getting confused with increase in temperature backward reaction takes place or forward?? And how?? Proper explanation is needed.

In a simple equilibrium we have either $$\ce{A + heat <=> B}$$ where the forward reaction is endothermic and the back reaction exothermic, or $$\ce{A <=> B + heat}$$ where the forward reaction is exothermic and the back reaction endothermic.  So in a simple equilibrium, the reaction in one direction will be exothermic and the reaction in the other direction will be endothermic.

When the temperature of an equilibrium is increased, the rate of both the forward and back reaction is increased.   However the rate of the endothermic reaction (the reaction that absorbs the added energy) is increased more.  So in the first reaction shown above, adding heat will shift the reaction to the right; in the second reaction heat will push it to the left.

The reason behind this behavior is that both the forward and back reaction can be described by the Arrhenius equation $$\ce{rate=Ae^{-E_{act}/RT}}$$  and the rate with the larger activation energy (the endothermic reaction) will always be more sensitive to changes in temperature. Therefor, the endothermic reaction will increase faster with increasing temperature than the exothermic reaction will.  Put some numbers into the equation and convince yourself.


Is there a way to make long delay ink? I would like to make some invisible ink that apears by itself after some time, preferably at least a couple of days. I want to write a dedication in a book with it, so it would probably be fine if the colour apears after half an hour of air exposure. Is this even possible using easy to get ingredients?

You might experiment with fresh-ground apples, which contain polyphenol oxidase and catechol oxidase, enzymes that turn brown in air (a tiny amount of copper or ferrous salt is also present, acting as a catalyst).

For that matter, dilute ferrous salts are often almost colorless (pale green) but darken to brown on exposure to air. (Digging in a podzol, soil of a pine forest, shows that color change.) So, you could extract a bit of liquid from greenish podzool and load your pen or marker with that.


How to calculate the rate constant of a zero order reaction? The reactant concentration in a zero-order reaction was $7.00×10^{−2}~\mathrm{M}$ after $160~\mathrm{s}$ and $2.50×10^{−2}~\mathrm{M}$ after $345~\mathrm{s}$. What is the rate constant for this reaction?
  What was the initial reactant concentration for the reaction described in part one?
  The reactant concentration in a first-order reaction was $7.70×10^{−2}~\mathrm{M}$ after $50.0~\mathrm{s}$ and $1.60×10^{−3}~\mathrm{M}$ after $85.0~\mathrm{s}$. What is the rate constant for this reaction?

Background info: The integrated rate laws for zero and first order reactions may be arranged such that they resemble the equation for a straight line,
$y=mx+b$.

\begin{array}{lccc}
\text{Order}& \text{Integrated Rate Law}& \text{Graph}& \text{Slope} \\\hline
  0 &  [A]=−kt+[A]_0      & [A]\text{ vs. }t    & −k\\
  1 & \ln[A]=−kt+\ln[A]_0 & \ln[A]\text{ vs. }t & −k\\\hline
\end{array}

According to your table, for a zero order reaction, you make a graph of the measured concentrations vs the corresponding times.  It will be a straight line, and the slope and intercept will give you the rate constant and the concentration at time zero.

For a first order reaction, you make a graph of the natural log of the concentrations vs the corresponding times.  It will be a straight line, and the slope and intercept will give you the rate constant and the concentration at time zero.


Proper explanation for depression of freezing point Why does depression of freezing point takes place if a non-volatile solute is added to the solution?

Depression of freezing point

A simple "intuitive" oversimplification: intrusive foreign molecules get caught between solvent molecules, getting in the way of crystallization. That, however, would contradict the similar boiling point elevation.

Another oversimplification: the solute molecules pull the solvent away from the incipient crystals.

Yet another simplistic answer: osmosis exerts a force on the solvent, pulling it back into solution, away from the pure crystal of solvent.

Perhaps the last explanation is closest to a statistical analysis of freezing point depression.

Hopefully, chemistry will not be a depressing subject.


How many valence are in each molecule? I have Benzene, Dewar Benzene, Prismane, and 1,3,5 cyclohexatriene (which looks just like benzene but with different lengths between the $\ce{C-C}$ bonds).

They are all $\ce{C_6H_6}$ chemical formula and I need to find out how many valence are in each molecule. I looked up what valence is but I'm confused if valence is the same thing as valence electrons or if valence is the amount of bonds the molecules can have. I think the amount of valence $\ce{H}$ can have is 1 but and $\ce{C}$ can have 4 so is the amount of valence 30?

i'm pretty sure the amount of valence electrons in all of them are 6.

I assume that the answer is that they are all C6H6 compounds (sorry for the terrible image!):

As others have pointed out in the comments, it doesn't really make sense to talk about the valency of a molecule. You could say they all have the same elemental formula, but then so do around 200 other molecules (in theory).


How to interpret BRICS fragments I have a molecule with a SMILES string as given below.
smi='c1ccccc1OCCOC(=O)CC'

I want to obtain BRICS based fragments which I did as follows.
frag=BRICS.BRICSDecompose(Chem.MolFromSmiles('c1ccccc1OCCOC(=O)CC'))

I obtain the following fragments.
{'[1*]C(=O)CC', '[16*]c1ccccc1', '[3*]O[3*]', '[4*]CC[4*]'}

How to interpret [1*], [16*], [3*] ,[4*] ? What does these number with brackets and stars means with each fragment?

Cutting a molecule (expressed with SMILES strings) into fragments yields sub-structures.  Typically these fragments can't be considered as molecules for their open, non-bonding valencies.  Derived from SMILES, it is then the SMARTS notation which is used to represent them, or to search with this notation of sub-structures in databases, e.g. in bioinformatics.
Open Babel supports the display of SMARTS-encoded fragments with branching points marked by an asterisk. The phenyl group, [16*]c1ccccc1 thus is shown as:

Consequently, bidental fragments are represented with two asterisks, e.g., [4*]CC[4*]:


Is it possible for a bicyclo compound to have three bridges of equal length? My teacher told me that the value of $x$ is always greater than or equal to $y$ and that y is always greater than $z$ in a bicyclo compound, where $(x,y,z)$ are the lengths of the bridges.

But in this compound $x = y = z$.
Can someone clarify?

Why not? The simplest bicyclic bridged compound, bicyclo[1.1.1]pentane, is a well-known example of such a molecule for its use in synthesizing the even more strained propellane. I believe your teacher was trying to prescribe the rules of bicyclic nomenclature, i.e.

  In between the prefix and the suffix, a pair of brackets with numerals denotes the number of carbon atoms between each of the bridgehead atoms. These numbers are arranged in descending order and are separated by periods.


Rate of substitution and elimination reactions at high temperatures From High School, I've been reading that elimination reactions are favorable at high temperatures because their entropy change is positive, so at higher temperature the Gibbs free energy will be more negative so the equilibrium will be more towards product's side.

But I've encountered a different statement in Clayden's Organic Chemistry:

  This explanation is simplified because what matters is the rate of the reaction, not the stability of the products. A detailed discussion is beyond the scope of the book, but the general argument still holds.

So what is the reason behind this? Why are substitution reactions not that much faster at high temperature than elimination reactions?

With temperature, the rates of substitution reactions and elimination reactions both increase.

For the purpose of answering this question, let us consider $S_N2$ and $E_2$ type reactions.

For $S_N2$ type reactions, the HOMO of the nucleophile has to be oriented with the LUMO of the electrophile. Here, a linear orientation is required. So, the entropic requirement is that the nucleophile should approach the electrophile in a certain way.

For $E2$ type reactions, the base attacks the $\alpha$-proton, which finally leads to the elimination. So, the HOMO of the base interacts with the $\sigma^*$ orbital of the $C-H$ bond, and then the $\sigma$ orbital interacts with the $\sigma^*$ orbital of the leaving group. So here the entropic requirement is the linear orientation of the base, the proton and the carbon it is attached to, and the leaving group should be antiperiplanar with respect to the proton.

With an increase in temperature, the "rate of bond rotation" increases and hence, the probability that a collision with a proper orientation will occur in a given time frame increases. This increase is more prominent in the case of the elimination reaction because it is quite dependent on the orientation of bonds (2 constraints vs. 1 constraint) compared to the substitution reaction.

So, a rise in the temperature will not only affect the thermodynamics of the equation, but it will also directly affect the rate of reaction.


Quantitative expression for steric hindrance Steric hinderance is a major component in determining the feasibility and the rate of a chemical reaction. Wouldn't it be useful to measure it quantitatively then? This would make it easier to compare the property of two molecules. Are there currently ways to measure steric hindrance, or is it not possible for some reason?

There are two ways I know of to measure steric hindrance. Once such quantitative measurement are A values, which give the energetic preference of having the substituent on a substituted cyclohexane ring in the equatorial position versus the axial position. This preference is a result of 1,3-diaxial strain.

Additionally, the Taft equation is a linear free-energy relationship used to measure the steric effects of a substituent on the rate of reaction. The equation is given by:

$${\displaystyle \log \left({\frac {k_{s}}{k_{\ce {CH3}}}}\right)=\rho ^{*}\sigma ^{*}+\delta E_{s}}$$

where $\displaystyle \frac {k_{s}}{k_{\ce {CH3}}}$ is the ratio of the substituents rate of reaction relative to a methyl group, $\sigma^*$ is the polar substituent constant that describes the field and inductive effects of the substituent, $E_s$ is the steric substituent constant, $\rho^*$ is the sensitivity factor for the reaction to polar effects, and $\delta$ is the sensitivity factor for the reaction to steric effects.$^{[1]}$

Though A values are useful as a quick reference in deducing the major products of an unfamiliar reaction, the Taft equation would be used preferentially when doing an in depth mechanistic study of a reaction for publication.

$^{[1]}$ Wikipedia, Taft equation


Utility of proton vs carbon NMR I've just learnt about both types of NMR, and I haven't ever used NMR in real life, so anything I say may be wrong.

The crux of this question is that, in my own opinion, proton NMR gives you more information than carbon NMR, which I'd imagine to be much more helpful when analysing compounds. Because of this, can anyone tell me whether carbon NMR is ever used instead of proton NMR (which seems to be more 'powerful')? If it is, when?

TL;DR Yes, sometimes. However, in general 13C data is much less useful than 1H (and 2D) data and the truth is we can often do without it.

I will limit my answer to the context of organic chemistry. Firstly, I directly answer your question: is 13C NMR ever used instead of 1H NMR? The answer is, yes.

However, I can only think of one case where one would do so. This is when there is severe peak overlap in the 1H NMR, such that one cannot reliably use it to ascertain whether one has (for example) a mixture of compound, or just a single compound. In such a case the 13C NMR (which has much more dispersed peaks and hence nearly no overlap) is more informative than the 1H NMR. This is generally not very common.

I want to go beyond a direct answer, though, and talk about the usefulness of 13C data in general. You are correct about the 1H NMR being more helpful than the 13C. This is generally true, and the main reasons for this are twofold:

13C NMR simply does not give as much information as 1H NMR. The only useful thing that one can obtain from the 13C NMR is the list of chemical shifts – but in turn, these are virtually impossible to interpret for all but the simplest molecules. On the other hand, 1H NMR gives you information about proton environments (shifts), number of protons (integrals), and limited information about molecular connectivity (couplings).
As porphyrin wrote, 13C NMR takes significantly longer than 1H NMR, because (a) the natural abundance of 13C is much lower (b) the gyromagnetic ratio of 13C is smaller.

As such, out of all the NMR data that organic chemists collect, the 13C NMR is by far the least useful and most time-consuming spectrum. On its own, its only real use is to provide a "fingerprint" of the molecule, and perhaps some information about functional groups present. So, in most cases, the 13C NMR is indeed less helpful than the 1H NMR.

The question then becomes, "why do people still take 13C NMRs?" Before going on, though, I should make it clear that people do not take 13C NMRs preferentially over the 1H, but rather in conjunction with it.

It primarily comes back to the fingerprint idea – when everybody tabulates 13C shifts of molecules they make, this gives chemist A the information they need to figure out whether they were successful in making what chemist B previously made. You might argue that the 1H is good enough as a fingerprint, but this is not always true, and anyway the consensus is that two is better than one: all major journals require submission of 13C NMR data for novel synthesised compounds.

The only other real case where the absolute 13C shifts are useful are when one needs to resolve very close peaks in 2D spectra (e.g. HSQC, HMBC), or in compounds with very few protons (thus rendering H–H or H–C correlations in 2D spectra less useful). Even then, it is not true to say that the 13C data is preferred over the 1H data.

All in all, your question is perfectly justified. In most cases, the 13C NMR is less useful as the 1H NMR, and only in some limited cases is it as useful as the 1H NMR.

In fact, there is a pretty recent perspective article by Liermann and Schlörer[1] arguing that the 1D 13C NMR should not be recorded, except for in the special cases described in the previous paragraph. The authors also make the point that the 2D spectra serve as much more useful fingerprints and are also much better at validating a proposed structure for a molecule. Even though in crowded spectra the 13C fingerprint may be more useful than the 1H fingerprint, it's still less useful than the 2D data, where peaks are dispersed in two dimensions.

Reference

Liermann, J. C.; Schlörer, N. E. Data handling in NMR facilities and assignment of NMR spectra in synthetic chemistry labs: Why electronic structure validation should become part of the routine. Magn. Reson. Chem. 2017, 1–7. DOI: 10.1002/mrc.4675.


Oxidation of naphthalene with KMnO4 What is the product when naphthalene is oxidized by alkaline (or acidic) solutions of $\ce{KMnO4}$?

Some possible reactions show up in a google search, but they have no references:

Naphthalene to phthalic acid by alkaline $\ce{KMnO4}$:

Another product on oxidation of napthalene by alkaline $\ce{KMnO4}$:

If possible, please include the reaction mechanisms in the answer as well.

I already made a comment about some of what I am about to say but I will provide a partial answer. I say partially because I could not find any mechanism for the second product. However, from literature, what I found was that in acidic conditions, $\ce{KMnO4}$ will oxidize naphthalene into the first product, phthalic acid1. In basic/alkaline conditions, $\ce{KMnO4}$ will oxidize naphthalene into Phthalonic Acid2. The results also agree from other papers. The proposed mechanism for the acidic solution reaction is here (taken from the paper in the first link):

It is essentially the same mechanism for the cleavage of a double bond but it happens in two positions in naphthalene. Note this is the mechanism for $\ce{KMnO4}$ oxidative cleavage:

Also note that the mechanism in the first image was based on a 2017 paper studying the kinetics of the reaction and then formulating a mechanism. So this reaction is still being studied. Along the same lines, I could not find the mechanism for the reaction of $\ce{KMnO4}$ and naphthalene in alkaline conditions. I would assume that its mechanism is still being researched.
References

https://www.jchps.com/issues/Volume%2010_Issue%201/40-0581116.pdf
The Oxidation of Naphthalene to Phthalonic Acid by Alkaline Solutions of Permanganate, R. Arthur Daly, The Journal of Physical Chemistry 1907 11 (2), 93-106, DOI: 10.1021/j150083a001
http://www.orgsyn.org/demo.aspx?prep=cv2p0523
https://www.acros.com/_rainbow/pdf/oxidation_brochure_manga.pdf
http://studymaterial.unipune.ac.in:8080/jspui/bitstream/123456789/8059/1/POLYNUCLEAR%20HYDROCARBONS.pdf


Metal vs. Styrofoam cup calorimeter? For a metal cup as opposed to a styrofoam cup calorimeter, do you expect the enthalpy of the reaction to be higher, lower, or the same?

I think the enthalpy of the reaction would be the same in the sytrofoam cup as the metal cup because enthalpy is a state function? So if the same products are yielded at the end along with the same reactants being used in the beginning, they'll have the same enthalpy because it doesn't matter HOW you got that enthalpy, just the final result? That's my reasoning but I'm not sure.

So does it matter that metal cup will conduct heat better because if they're using the same reactants it'll give you the same change as a styrofoam cup?

You're right that IF the metal calorimeter and the styrofoam calorimeter were both perfect and behaved in the way that we usually pretend calorimeters do (they don't absorb or lose heat) then they should produce identical results, for the enthalpy reason you provide.

However, since we're asking about specific materials, metal vs styrofoam, we're considering real calorimeters instead of ideal.

There are a few things that calorimeters can do that will affect how large a temperature change you measure.

Heat loss - As you note, metal conducts heat much more effectively than styrofoam.  This means that the metal calorimeter will carry the heat of the reaction away from the reaction vessel to the outside world faster than styrofoam will.  If that is the case, which reaction vessel will get hotter, styrofoam or metal?
Heat absorption - The calorimeter (and thermometer) gets heated up along with the reaction mixture.  A calorimeter that absorbs large amounts of energy will have a relatively lower final temperature, since more of the reaction's heat went into heating up the calorimeter than went into heating up the reaction mixture.  This quantity is known as the calorimeter constant and varies depending on the materials and mass of the calorimeter.  Styrofoam absorbs more heat per gram than steel, but the steel calorimeter is probably heavier.  With a fast reaction there is the additional fact that the styrofoam calorimeter probably never comes up to the full temperature of the reaction before you quit recording temperatures.  This winds up leading to a small source of lost heat for the metal calorimeter as compared to the styrofoam.
The quality of the equipment -  Metal calorimeters are often made specifically for the purpose of calorimetry.  Styrofoam calorimeters are often a couple of coffee cups jammed together.  If the calorimeter allows a significant amount of heat loss to the atmosphere due to its construction, then it will result in a cooler reaction mixture and a lower calculated enthalpy for the reaction.

You've got to take all three of these into consideration in order to determine the particulars for your experiment.


what causes the smell of a room ionizer? If ozone, what are we actually smelling when ozone interacts with nasal mucous? This touches on how smell works, but my intent is ask about the mechanism for why do ionised smell like they do?  I would like to get one step past ozone if that is the primary mechanism.

Regarding our sense of smell, from the Wikipedia:

  The perception of odors, or sense of smell, is mediated by the olfactory nerve. The olfactory receptor (OR) cells are neurons present in the olfactory epithelium, which is a small patch of tissue at the back of the nasal cavity. There are millions of olfactory receptor neurons that act as sensory signaling cells. Each neuron has cilia in direct contact with the air. Odorous molecules bind to receptor proteins extending from cilia and act as a chemical stimulus, initiating electric signals that travel along the olfactory nerve's axons to the brain.2

Regarding the odor of ozone, from the Wikipedia:

  Ozone (/ˈoʊzoʊn/), or trioxygen, is an inorganic molecule with the chemical formula O
  3. It is a pale blue gas with a distinctively pungent smell.

From this site:

  While oxygen is not detectable by the human nose, ozone has an extremely low odor threshold.  Humans have the ability to detect ozone [concentrations of ~10 ppb] before many expensive measuring devices.  Also, ozone in air at high levels (greater than 0.1 ppm) is hazardous to human health.

  Ozone is a powerful oxidant (far more so than dioxygen) and has many industrial and consumer applications related to oxidation. This same high oxidizing potential, however, causes ozone to damage mucous and respiratory tissues in animals, and also tissues in plants, above concentrations of about 0.1 ppm. While this makes ozone a potent respiratory hazard and pollutant near ground level [...]

There are quite a few articles available online that discuss the "odor of ozone". It is described by some as "metallic" or as "similar to chlorine" (although note that "chlorine smell" is often "pool smell" caused by chloramines).

Given the reactivity of ozone (it oxidizes just about anything containing carbon to $\ce{CO2}$, which explains why it can cause inflammation and is generally regarded as deleterious to health), it is an interesting question whether there exists a distinct odor receptor that senses ozone or whether the response is nonspecific or mediated by pain receptors and registered as the "pungent smell" of ozone. Or perhaps it is other molecules altogether rather than the interaction of ozone with nasal receptors that causes the smell.

All of this is made even more complicated by the apparent loss of olfactory perception caused by exposure to ozone:

  An acute exposure to 0.2 ppm ozone impairs odor detection threshold.
  The cause is not clear, different modes of action might play at role.
  Olfactory threshold seems sensitive to photochemical air pollutants’ effects.

The "The Curious Dr. Z" blog hypothesizes on the source of "ozone smell" (see also reference 1 cited in the blog):

  It turns out that our olfactory receptors might be able to bind free radicals, and, since free radicals and ozone frequently occur together, that’s what we’re actually smelling.

  In 1991, a group of patients undergoing irradiation for cancer all complained about a strong, pungent odor they later identified as ozone. This lead to a study where researchers discovered that humans can actually detect the irradiation of tissue by detecting free radicals trapped in mucus.

  Another, older theory, is that ozone acts as the signal for irradiation. Ozone binds to and reacts with olfactory receptors the just as other odorants do, and effectively acts as a warning of more damaging chemicals nearby.

  We don’t know too much beyond this, and as far as I can tell, ozone olfaction is not a very active area of research. Bummer.

  Since we are so sensitive to the smell of ozone, and we don’t smell much else when we do smell ozone, I suspect that ozone/free radicals have many potential binding sites (a lot of those “hands” can grab ozone). When the ozone/free radicals does bind, it does so with high affinity, meaning it easily binds to receptors. This persistent and overwhelming smell experience is similar to other smells that indicate something noxious, including smoke and rotting tissue.

References

Lloyd M. Beidler. Olfaction, Handbook of Sensory Physiology, Chemical Senses. Springer Science & Business Media, 2012, ISBN 3642651267, 9783642651267


Order of the subshells in an electron configuration According to what my chemistry teacher said, the energy of a subshell is given by the sum of its principal quantum number (n) and its second quantum number (l).

e.g.:

1s = n+l = 1+0 = 1
2s = n+l = 2+0 = 2
2p = n+l = 2+1 = 3
3s = n+l = 3+0 = 3

However, when it comes to make a graphical representation of an electron configuration (which follows an order of increasing energy), she doesn't put the subshells having the same energy in the same level, like in the following picture:

Shouldn't the subshells 2p and 3s (which both have an energy equals to 3), as well as 4p and 3d, be put in the same level, since they have the same energy?

Whilst the said concept is correct, as orbitals increase in number the less they are attracted to the centre of nuclei, consequently the further they move away the higher the energy they possess.

The 2s electrons spent more time closer to the nucleus (when compared to 2p electrons) and this reduces the energy of electrons in s orbitals, the nearer they get, the lower the energy. This is why the s orbital always has a slightly lower energy than p orbital at the same energy level.

The 2s, 3s .... orbitals get progressively away from the nucleus.


Oxymercuration and demercuration mechanism There are two proposed mechanisms for Oxymercuration and demercuration.

The first mechanism is concerted (i.e. single transition state connecting reactants and products) characterized by an attack of the $\pi$-bond on the $\ce{Hg(OAc)2}$.

Why does mercuric acetate act as an electrophile in the first mechanism? It is in its most common 2+ oxidation state. Does this happen because mercury has empty $6d$ orbitals.

The second mechanism is a stepwise mechanism (i.e. contains at least one intermediate species along the reaction pathway between reactants and products) that is characterized by incorporating the first ionization of $\ce{Hg(OAc)2}$ followed by an attack of the nucleophilic $\pi$-bond.

$\ce{Hg(OAc)2}$ dissociates in water as follows:
$$\ce{Hg(OAc)2 + H2O<=>Hg+OAc + OAc-}$$

Which one is the actual mechanism?

The mechanism you are looking for is probably not known and might well be never known. According to Brown et. al. there is a rapid and reversible equilibrium between associated and dissociated reactants.[1] They tested this for cyclohexene, norbonene and apobornylene, but I believe it is fair to generalise in this matter.
$$\ce{R2C=CR2 + Hg(OOCCF3)2 <=> (F3CCOO)R2C=CR2Hg(OOCCF3)}$$

Due to this, even the formation of a mercurinium ion seems to be sometimes not observed. For more information please have a look at the mechanistic studies of Pasto and Gontarz,[2] where they focussed on the stereo selectivity of substituted cyclohexenes, and references within.

Whatever molecule you might assume acting as the electrophile, it is quite obvious, that the bond mercury oxygen bond is highly ionic, with most of the electron density located in the acetyl moieties, leaving a rather substantial positive charge at the mercury itself. The following structures were optimised at DF-BP86/def2-SVP and the charges displayed are natural charges from the Natural Bond Orbital analysis to illustrate the above point.

The lower structure of $\ce{Hg(OAc)2}$ is slightly more stable.

Herbert Charles Brown, Min-Hon Rei, Kwang Ting Liu, J. Am. Chem. Soc., 1970, 92 (6), 1760–1761.
Daniel J. Pasto, John A. Gontarz, J. Am. Chem. Soc., 1971, 93 (25), 6902–6908.


Why does the rusting process start from the bent/squeezed portion of iron? I have noticed that the rusting of iron starts from the squeezed or bent parts of iron piece.
 I want to know what is the peculiar reason for the observation that rusting starts from bent or irregular surface than smooth part of a given iron piece and it spreads to everywhere.

I think this question is same as wondering as to why a certain cancer patient has a cancer in the kidney but not on the skin? Nobody knows. In the same way, rusting is a random process as well. It can start anywhere. This iron sheet is not ultrapure iron sheet of homogeneous nature. One can see a lot of spots in troughs and crests of this sheet. Secondly you do not know the history of this sheet. Is this galvanized or not galvanized? This rusted spot may have a local impurity which can initiate a local electrochemical cell.

In formal atmospheric corrosion studies, my colleague uses flat sheets of test material and rusting pattern is totally random. They also do a scratch test by drawing a cross X on the flat coated sheet, in case they wish to test a coating. The purpose is to expose the "fresh" metallic surface. The rusting then starts from the X pattern. This image is an example of flat sheet used in ASTM coating. The rusting pattern is nothing but random design on the uncoated sheet (left most).

https://hardide.com/technical/astm-b117-corrosion-resistance/


Is there any chemical that can catalyze "reverse combustion?" I am trying to build a machine that converts carbon dioxide and water into gasoline (i.e. octane). Because the energy needs are so large, is there any catalyst chemical that can cause "reverse combustion" at room temperature?

By principle, all catalysts accelerate both forward and backward reactions
in the same extent, so the equilibrium with and without the catalyst is the same.

So the catalyst catalysing the "reversed combustion" would catalyze the normal combustion as well. The equilibrium would be  far at the combustion products side, as if the catalyst was not used.

If it had not been so, one could have constructed a chemical perpetuum mobile. It could generate energy by shifting equilibrium forward and backward just by adding and removing the catalyst.


Boiling Point lower than Melting Point? Under the Wikipedia entry for Metaldehyde is listed a Melting Point of $\pu{246^\circ C}$ and a Boiling point of $\pu{110^\circ C}$ (actually, it says it sublimes).

How can a Boiling point be lower than the Melting Point? Even if $\pu{110^\circ C}$ is actually a sublimation temperature how can it then subsequently melt at a higher temperature of $\pu{246^\circ C}$?

Sublimation is not boiling but there is a reason for listing it instead. Sublimation is direct vaporisation from the solid. The most familiar is solid dry ice producing $\ce{CO2}$ gas. So if metaldehyde melts at $\pu{246^\circ C}$ then one might suppose that the boiling point is higher. However, look at the generic $P$ vs. $T$ diagram below. If we assume that at $P = 1000$ is normal atmospheric pressure then a compound with this $P$ vs $T$ sublimes. As we define boiling to occur only at normal pressure and be the liquid to vapour (gas) transition then this compound can never boil at normal pressure. To become a liquid high temperature and pressure is needed. So you can see why the table lists sublimation at normal pressure rather than boiling.


Calculating the molar iodine concentration of a CT contrast agent I want to calculate the molar concentration of Iodine in a CT contrast agent.

Name: Isovue 300 (Iopamidol - $\ce{C17H22I3N3O8}$)

Molecular weight: $777.09\ \mathrm{g \ mol^{-1}}$

The part that confuses me is this:
"Each $\mathrm{mL}$ of Isovue-300 (Iopamidol Injection $61%$) provides $612 \ \mathrm{mg}$ iopamidol with $1 \ \mathrm{mg}$ tromethamine and $0.39\ \mathrm{mg}$ edetate calcium disodium. The solution contains approximately $0.043\  \mathrm{mg}$ ($0.002\ \mathrm{mEq}$) sodium and $300\ \mathrm{mg}$ organically bound iodine per $\mathrm{mL}$." (source: https://www.drugs.com/pro/isovue.html)

Each molecule of Isovue contain 3 $\ce{I}$ atoms. So if I calculate the molar concentration (300 mg/ml / 777.09 g/mol * 1000 = 386.1 mM), do I need to correct for this?

I am sorry for this basic question, but I cannot find the answer on the manufacturer's website or elsewhere.

Thank you all in advance

I highly recommend not looking for literature from non-authoritative sources such as drugs.com: that's a paid-advertising site and is full of partial/misinformation.

For your compound, see the paperwork from the people who make it - Bracco Imaging.

Each $\mathrm{mL}$ of Isovue-300 contains more than what you list (namely, water, sodium hydroxide, and hydrochloric acid). That means your approach is close.

The wording about the 300 $\mathrm{mg}$ of organically-bound $\ce{I}$ per $\mathrm{mL}$ of solution makes sense when you consider the atomic weight of $\ce{I}$ (about 127 $\mathrm{mg}\,\cdot\mathrm{mmol}^{-1}$) and that 3 atoms of iodine comes out to 381 $\mathrm{mg}$. Thus, for that slightly dilute solution, the 300 $\mathrm{mg}$ makes sense. It's not an additional 300 - it's what's in the $\mathrm{mL}$ as-is.

So, the molarity of $\ce{I}$ is (I abbreviate iopamidol as "iop" below to save space):

$$\left({612\,\mathrm{mg\,iop}\over 1\,\mathrm{mL}}\right)\cdot
\left({3\,\mathrm{mmol}\,\ce{I}\over 777.09\,\mathrm{mg\,iop}}\right)\cdot
\left({1\,\mathrm{mol}\,\ce{I}\over 1000\,\mathrm{mmol}\,\ce{I}}\right) \cdot
\left({1000\,\mathrm{mL}\over 1\,\mathrm{L}}\right)
= 2.36\,\mathrm{M}$$


empty 30 Gal. barrel of Potassium Hydroxide and Sodium Hypochlorite - safe for collecting Maple syrup? I bought an empty 30 Gal. barrel of Potassium Hydroxide and Sodium Hypochlorite - CIP 150 - Wondering if it is SAFE to use for the collection of Maple Tree Syrup?  (Label say the white plastic barrel had been rinsed 3 times - however it does say that the previous contents (above) were corrosive. - AND, I plan on rinsing it a few more times) Answers? Suggestions?

While I also consider the question somewhat off-topic, I would still like to answer here. In a time in which people eat tide pods for fun it seems kind of necessary...

So, for general information, regardless of what chemical substances were inside the container or what you plan to put into, you should never (re)use containers for chemicals to store food!

Here are some reasons why:

Standard food containers are readily available, do not cost much, and are much better suited to store and conserve food
you are sending a disastrous message to children if you are taking food from a container with chemical warning symbols on it
there is a chance that you mix up your food container with one that contains what is actually on the label
you might not be able to remove all the previous content from the container by rinsing it with water, e.g. if the previous substance is non-soluble in water or if it is greasy. Minuscule quantities of the remaining substance may be toxic or hazardous
the substance may have changed over time if it was stored for a prolonged time or under inadequate conditions, like exposure to heat, moisture or sunlight. Products may precipitate or form films on the inside of the container.
you do not know the history of the container. Maybe the previous owner used up the original content and then used the empty container to store chemical waste. This is also not a good practice, but still more common than to store food!


How can I find the ratio of molecular masses using the ideal gas equation? A person evacuated a cylinder and filled in $4~\mathrm{g}$ of gas A at $25~\mathrm{^\circ C}$. Pressure was found to be $1~\mathrm{atm}$. Another person filled another $8~\mathrm{g}$ of gas B in the cylinder at the same temperature. The final pressure was found to be $1.5~\mathrm{atm}$. What is the ratio of molecular masses of A and B (assuming ideal gas behaviour)?

I tried to do it this way:

Taking the molecular mass of A as $x\ \mathrm{g}$ and that of B as $y\ \mathrm{g}$, I formed the respective ideal gas equations but the required terms ($x$ and $y$) just cancelled out.

$$N(a) = 4/x ;\quad N (b) = 8/y$$
So$$V (a) = \frac{4 \times  22.4}{x};\quad     V (b) = \frac{8 \times  22.4}{y}$$

So, ideal gas equation for A:
$$1 X \frac{4 \times 22.4}{x} = \frac {4 \times  R  \times  25} x$$
Here $x$ cancels out. Similarly $y$ cancels out in its equation.
This is the problem I am facing.

Another thing I want to know is that the total pressure (i.e $1.5~\mathrm{atm}$), will that be the sum of the individual pressures of gases A and B?

The ideal gas equation is $PV = nRT$
But for this problem V, R, and T are constants. So we can collect terms as:
$\dfrac{P}{n} = \dfrac{RT}{V}$
So $\dfrac{P_A}{n_A} = \dfrac{P_B}{n_B}$
Now you correctly determined the number of moles of A and B.
$n_A = \dfrac{4}{\text{MW}_A}$
$n_B = \dfrac{8}{\text{MW}_B}$
But you need to consider the partial pressures of A and B.

The cylinder was empty before A was added, and after A was added the pressure was 1 atmosphere. So the partial pressure of A is 1 atmosphere.
Before B is added the pressure is 1 atmosphere and after B is added the pressure is 1.5 atmospheres.  So the partial pressure of B is 0.5 atmospheres.

$P_A = 1$
$P_B = 0.5$
Now substituting into our simplified equation
$\dfrac{1}{\frac{4}{\text{MW}_A}} = \dfrac{0.5}{\frac{8}{\text{MW}_B}}$
$\dfrac{8}{\text{MW}_B} = \dfrac{2}{\text{MW}_A}$
$\dfrac{\text{MW}_A}{\text{MW}_B} = \dfrac{2}{8} = \dfrac{1}{4}$


Why does a chemical have more than one oxidation number? Like for group I metals, it is understood that all the elements in the group would have an oxidation number of +1.

As far as I know, oxidation number is how much electron the element loses/gains.

Hence my question is, why does an element like Iron, Manganese or even oxygen have more than one oxidation state unlike elements like Magnesium or Potassium.

Edit: If I have stated any terms incorrectly, please let me know because I'd want to know where I went wrong.

Because a lot of what you have been taught in school is a lie.

In this particular case, you've probably been taught about the octet rule, a rule which makes it seem as if any given element should have exactly two oxidation states; 'n' and '8-n', where n is a simple function of it's location in the periodic table.

The truth, as always, is a little more complicated:

Compared to the (somewhat) well-behaved non-metals in the upper right corner of the periodic table, most metals have a lot for electrons. At 26, iron is a lightweight, but already there you have 4 (common!) oxidation states, the two most common ones being iron(II) and iron(III).

One reason for this is that while iron atoms have a "the lowest energy state", it's difficult to get at without first coming across several "a (locally) lowest energy state".

Its four oxidation states represent local minima, and moving between them requires an activation energy, making each of the states relatively stable, but possible to dislodge, i.e. metastable.


Change in Gibbs free energy vs. change in enthalpy When learning about other thermodynamic quantities, like $\Delta H$, I was taught that this represents the energy absorbed or released by a system after the reaction has gone 100% to completion. For example, if a reaction has $\Delta H = -100 \text{kJ}$, it means that the final state of the system after the reaction is complete has $100 \text{kJ}$ less energy than it did before the reaction began. As such, $\Delta H$ is a statement about the overall reaction, and is not dependent on where the reaction is at the present moment. (Is this understanding correct? Or does $\Delta H$ change throughout a reaction like $\Delta G$?)

However, when learning about Gibbs Free Energy, I have read that $\Delta G$ changes throughout a reaction, as can be seen from $\Delta G = \Delta G^\circ + RT\ln Q$, because Q varies through the course of a chemical reaction. If $\Delta G$ gives information about a specific point in the reaction, then why is there a "$\Delta$" at all? In other words, $\Delta G$ is the difference between what two values?

Thanks for the help.

To start off, $\Delta G$ is concentration-dependent and $\Delta H$ is not. You already showed the concentration-dependence of $\Delta G$. There is no equivalent relationship for $\Delta H$ because it does not depend on concentration. $\Delta H$ reflects how the bond energies change from reactants to products. Bond energies don't depend on concentration. (It also reflects other more subtle things like solvation and intermolecular interactions, but these don't change in an ideal solution).  In contrast, $\Delta G$ has an entropy term, which has to do with the entropy of mixing, resulting in a significant concentration-dependence even in ideal solutions.

  Is this understanding correct? Or does ΔH change throughout a reaction like ΔG?

The enthalpy H of the system changes throughout the reaction. Because the enthalpy is not concentration-dependent, it changes in a linear way with the amount that reacts. There are two versions of $\Delta H$ with different dimensions and units. One does what you say: compare H between two states. It has dimensions of energy (typical units: kJ). The other asks what the change of enthalpy is per amount reacted. It has dimensions of energy per amount (typical units: kJ/mol) and is called molar change in enthalpy.

An example is the molar enthalpy of formation. If you want to know how big the enthalpy change is to make a certain amount of a compound, you multiply the amount by the molar enthalpy of formation. The dimensions of the result will be energy again (units: kJ / mol * mol = kJ). Introductory textbooks sometimes gloss over the difference between molar (intensive) and extensive enthalpies.

  If ΔG gives information about a specific point in the reaction, then why is there a "Δ" at all?

The Gibbs energy G of the system changes throughout the reaction. The change per amount of product formed decreases as the reaction approaches equilibrium. Again, ΔG can mean two things, either the difference in Gibbs energy between an initial and a final state, or the molar change in Gibbs energy with given concentrations.

You might say concentrations change while the reaction is going on, so it is impossible to make a statement about specific concentrations. You solve this problem by either looking at a very small amount of change (infinitesimal, if you want) or by imagining very large volumes of reaction mix, where concentrations change very little even if - say - one mole of reactants turn into one mole of products. If you want to communicate that this is what you are talking about, you can either use $\Delta_r G$ or $\frac{dG}{d\xi}$. Both have dimensions of energy per amount of substance (typical units kJ/mol).

Again, introductory textbooks tend to gloss over this difference, not giving $\Delta G$ and $\Delta_r G$ different symbols even though they have different dimensions (one is extensive and the other intensive) and mean different things.


Can you steam clean your range hood, by adding dishwashing liquid to boiling distilled water? My grandparents physically can't remove their Sakura R-727II range hood that lacks any filter. Because of COVID-19, no handyperson can visit their house. But their range hood is now clogged and failing to suck out fumes, so they must de-grease their range hood's interior that hasn't been cleaned in 4 years.
Please see the pictures below. They wonder if they can de-oil their range hood by adding dishwashing liquid to boiling distilled water, then boil this mixture to evaporate the dishwashing liquid that will rise into and de-grease their range hood.

Does the dishwashing liquid vaporize as they think? Or does it remain in the saucier?

Can the resultant evaporated dishwashing liquid + steam combination de-grease their range hood? I wonder if this dishwashing liquid + steam will have enough pressure.

The two pictures above are like their range hood. Theirs isn't a Samsung range hood pictured below, but I used it to devise my collage for it shows steam.

Clogged? Failing to suck out fumes? Quantitatively, how bad is it? And in what manner did this situation arise? Slowly, over time? Or suddenly? Can you hear a fan running?
Grease has a high boiling point, and while it is theoretically possible to clog some fine filter, it is doubtful that 4 years of home cooking could stop a hood fan from operating. In commercial practice, hoods over grills are cleaned frequently from spattered grease, not so much for better air flow, but more for sanitation and prevention of possible fires from leaping flames.
If no filter is accessible from the stove end, perhaps the exit end is the cause for blockage. Exit openings frequently have flaps to prevent rain entry. Two possibilities: 1) the flap is stuck down and hinders airflow, or 2) the flap is stuck up and the tube is blocked by a bird nest or a pile of acorns. (Of course, this exit end will be so high that it is not accessible for cleaning, but that's another problem.)
There is a third situation. In my former house, the builders omitted an outdoor exit pipe for the microwave and fume hood. There was simply a fan which sucked the steam/vapor up off the stove, thru the filter and back at the cook. We generally turned the fan off. The idea of a fume hood was nice, but not really necessary.


Spectroscopy and photometric error What is meant by the lowest photometric error?
why it is important to work in the $0.2\ \mathrm{to}\ 0.7$ absorbance range for a single beam instrument?
Why $0.4\ \mathrm{to}\  1.4$ for double-beam instrument?

Start with a reading about the types of errors that arise in spectroscopy. Clearly photometric error $\delta I_{pe}$ is a device uncertainty. The suggestion is, its magnitude depends on the number of "beams". What changes as the number of beams increases? Path length? Number of gratings or optical components? What else? How do each of those factors contribute to an analytical expression for photometric uncertainty?

At the end, a reasonable result might suggest that, as you double the number of beams, you double the relative uncertainty in signal $\delta I_{pe} / I$, where $I$ is the signal intensity. To obtain the same absolute confidence, you therefore have to double the measurement intensity $I$.


Which species can't appear in the rate expression for this reaction scheme? Question
  Which species can't appear in the rate expression for this reaction scheme?
  Here is the reaction scheme in steps:
  \begin{align}
\ce{A + B~(slow) &\rightarrow C} \\
\ce{C~(fast) &\rightarrow D + B}\\
\ce{D + E~(fast) &\rightarrow F}\\
\ce{F~(fast) &\rightarrow G}\\
\end{align}

The answers are apparently F and G, but I don't see why D and E couldn't also be valid (considering the first step is the slowest). A and B will inevitably be in the rate equation since they are reactants in the rate determining step. C produces B (alongside D) and so will also be in the rate equation as a result. However, why can't we rule out D and E of being in the rate equation, considering they don't seem to have any effect on the concentrations of A and  B.

Answering this question requires assuming that all steps are "elementary", e.g. that $\ce{A + B -> C}$ is an irreversible bimolecular reaction that is first-order in $[\ce{A}]$ and first-order in $[\ce{B}]$.  (Obviously we are also assuming that activities can approximated by concentrations.)

With that assumption, the fact that $\ce{C}$ conversion is "fast" means that we can replace the existence of $\ce{C}$ in the scheme entirely:

\begin{align}
\ce{A + B~(slow) &\rightarrow D + B} \\
\ce{D + E~(fast) &\rightarrow F}\\
\ce{F~(fast) &\rightarrow G}\\
\end{align}

Likewise if $\ce{F}$ conversion is "fast", then it can be replaced:

\begin{align}
\ce{A + B~(slow) &\rightarrow D + B} \\
\ce{D + E~(fast) &\rightarrow G}\\
\end{align}

Hopefully this makes it clear why $\ce{E}$ must appear in any rate law for formation of $\ce{G}$: there is no source of it in the two reactions we are left with.

For $\ce{D}$, I agree that the situation is a bit more complex.  It might appear in the rate equation, or it might not.  Necessary reactants at the beginning of the reaction if we are ever to see product $\ce{G}$ are $\ce{A}$, $\ce{B}$, and $\ce{E}$, so it would make sense to eliminate $\ce{D}$ from the rate law if possible.  However, that could be algebraically difficult, or perhaps $\ce{D}$ is easier to measure, etc.  There is no reason $\ce{D}$ can't appear in the rate law.

Huge caveat: "fast" and "slow" are relative terms and as such without further context they are ambiguous.  The problem as constructed sounds a bit sloppy.  For example, is the "fastness" of $\ce{F -> G}$ faster or less fast than the fastness of $\ce{D + E -> F}$?  The problem doesn't say but the simplifications I did above actually depend on these sorts of details.


Why do Grignard reagents or Organolithium compounds not displace halides as easily as acetylide ions do? This is a snapshot from LG Wade Organic Chemistry. I have a doubt with a particular statement in the book. It says that Grignard reagents do not displace halides as easily as acetylide ions do.

But shouldn’t it be the other way round? Acetylide ions, having an sp-hybridised carbon, are much more electronegative compared to the sp³-hybridised carbon in Grignard reagents. So acetylide ions should have less tendency to displace halide ions by attacking with their lone pair compared to the alkyl group in Grignard reagent. Shouldn’t it?

Please help me sort this out!

An acetylide ion proceeds in a nucleophilic substitution pathway (SN2 in the picture due to primary halide). The substitution involves the attack of a highly nucleophilic sp-carbon on a highly electrophilic primary carbon. The enthalpy of the C-C bond formation is relatively low due the pKa of the nucleophile being ~ 25. The ion is treated as a "soft" nucleophile that prefers to substitute alkyl halides instead of acting as a base and conducting an elimination reaction.

A grignard reagent is essentially a highly polarized bond between Carbon and the Bromide(Halide)/Magnesium complex. The electronegative difference between Carbon and Magnesium is 1.3 (http://www.thecatalyst.org/electabl.html <-- Source of values). The Carbon atom in a Gringnard is treated as a carbonanion. This carbonanion characterization is highly unstable in nature and treated as a "hard" nucleophile with pKa of ~ 50. The highly basic character of a Grignard reagent often results in an elimination reaction or no reaction at all. The transition state to substitute the alkyl halide is less stable than the Magnesium/Bromide(Halide) complex. This is due to a ligation formation between the solvent and the Magnesium atom. The solvent (ex. diethyl ether) shares non-bonding electrons with the cation and creates a stable electronic atmosphere with a lower energy than a possible C-C bond with the alkyl halide.

Substitution isn't preferred with a hard nucleophile and therefore, the acetylide ion is more likely to preform a nucleophilic substitution reaction with an alkyl halide. Note that the electronegative difference between Carbon and Magnesium supersedes the hybridization of the sp or "triple" bond complex. The bond is so high in energy that the nucleophilic character of the grignard far surpasses the electronegative character of the sp-hybridization of the acetylide ion.


Physical meaning of tm The definition of dead time ($t_\mathrm m$) is mentioned as the time taken for elution of an unretained species. It is also mentioned that the retention time ($t_\mathrm r$) is sum of the time analyte spends in mobile ($t_\mathrm m$) and stationary phases ($t_\mathrm s$). But how can time spent by analyte in mobile phase be equal to the time required for an unretained species to pass through (dead time, $t_\mathrm m$)?

Both seems independent of each other. An analyte's time in the different phases is dependent on its affinity while the unretained species takes the same amount of time no matter what.

Please explain how they are same.

The analyte spends the same time in the mobile phase as it tooks for the mobile phase to pass through the column.

  An analyte's time in the different phases is dependent on it's affinity.

The time $t_\mathrm{m}$ is given.

The ratio of an analyte's time in the different phases is dependent on it's affinity and is equal:

$$\frac{t_\mathrm{s}}{t_\mathrm{m}}=K_\mathrm{D} \cdot \frac{V_\mathrm{s}}{ V_\mathrm{m}} $$.

where

$t_\mathrm{s}, t_\mathrm{m}$ are time is stationary and mobile phase
$K_\mathrm{D}$ is the phase distribution constant
$V_\mathrm{s}, V_\mathrm{m}$ are volumes of stationary and mobile phase.


How to make pyramid-shaped NaCl salt crystals I'm attempting to make sea salt pyramids like the one below.

I understand this needs to be evaporated at a reasonable speed to stop the crystals forming a cube, as the when formed quickly, the edges grow faster than the faces, creating the inverted pyramid shape.

When evaporating the saturated water at around 75 degrees celsius, I start with what look like great pyramids forming (like the below image). However, the crystals then begin to join together forming a large flat 'crust', with the individual small pyramids losing their unique shape.

I've attempted reducing the temperature of evaporation to a point where this mass of crust does not form, however at that point, the crystals which now form are cubic instead of pyramids.

Any help would be great!

TL;DR: Use 10-20% urea solution for octahedral crystals and a water-ethanol mix for pyramidal crystals.

There is no universal method to predict and to control crystal habits. Very often the experimental results are explained with molecular dynamics; there are models (Wulff condition, Donnay–Harker correlation, Hartman–Perdok theory, Hartman–Bennema attachment energy) that can only be used to represent vapor-grown crystals. Crystal morphology is vastly dependent on many factors:

phase it's growing from (gas, liquid, melt);
growing rate (temperature);
concentration of the mother solution;
ionic strength of the solution;
presence of impurities (habit modifiers);
solvent etc.

Rock salt can take one of the three morphologies dictated by its FCC lattice with a strong tendency to prefer cubic crystals:

Figure 1. Morphologies of $\ce{NaCl}$ crystal denoted with Miller planes: cube $\{100\}$; octahedron $\{111\}$; rhomb-dodecahedron $\{110\}$.

There are numerous literature sources providing experimental information on crystal growth. For sodium chloride there is a recent review by Aquilano et al. [1], not only providing a brief description of common habit modifiers, but also giving an insight on to why one crystal direction is preferred:

  The morphology of natural halite is largely dominated by the cube $\{100\}$; the $\{110\}$ and $\{111\}$ forms are seldom present. When crystals grow by evaporation from
  aqueous solution the perfection of the cube faces is usually lost; owing to the high values of the supersaturation, hopper shaped $\{100\}$ faces appear along with dendritic
  branches developing in the $❬111❭$ directions. This is the main reason of the halite tendency to cake. Hence, the best way to face this problem is to search for additives
  able to favor the appearance of the $\{111\}$ form, thus avoiding the morphological instability such as the “hazardous” dendritic growth.

Here is a brief summary for the non-standard crystal shapes of rock salt and corresponding modifiers:

Octahedra. Formamide; urea; sodium tetraphosphate $\ce{Na6P4O13}$; cadmium chloride $\ce{CdCl2}$; polymer additive containing an amide functional group.

Dendrites and needles. Hexacyanoferrate(II) $\ce{Fe(CN)6^4-}$; polyvinyl alcohol.

Pyramides. Water-ethanol solution; growth from a supersaturated solution [2].

Presence of heavy metals ($\ce{Pb^2+}$, $\ce{Cd^2+}$) generally promotes formation of large single crystals instead of many small crystals. For more practical details, photos of crystals and manuals you can refer to Crystal Growing Wiki (originally in Russian, but now most articles are translated to English).

References

Aquilano, D.; Otálora, F.; Pastero, L.; García-Ruiz, J. M. Three Study Cases of Growth Morphology in Minerals: Halite, Calcite and Gypsum. Progress in Crystal Growth and Characterization of Materials 2016, 62 (2), 227–251. https://doi.org/10.1016/j.pcrysgrow.2016.04.012.
Desarnaud, J.; Derluyn, H.; Carmeliet, J.; Bonn, D.; Shahidzadeh, N. Hopper Growth of Salt Crystals. J Phys Chem Lett 2018, 9 (11), 2961–2966. https://doi.org/10.1021/acs.jpclett.8b01082.


Feasibility and Entropy Suppose you've got some reaction that has a positive $\Delta H$, and also has positive $\Delta S_\mathrm{sys}$. Due to the equation
$$\Delta S_\mathrm{tot} = \Delta S_\mathrm{sys} - \frac{\Delta H}{T}$$
a low enough temperature will cause $\Delta S_\mathrm{tot}$ to be negative. At this point the reaction won't happen anymore because second law and whatnot.

What exactly is happening at the molecular level that prevents molecules from reacting with each other? The orientation of the molecules is more or less random, and the Maxwell–Boltzmann distribution means there ought to be some particles with sufficient kinetic energy, so what's happening? If the reactants are left long enough, 2 molecules with acceptable orientations & kinetic energies should whack into each other.

A while ago I found the "micro-states per macro-state" explanation of entropy. I get that entropy isn't a property of a single particle, or even of a specific arrangement (microstate) of particles (please correct if wrong). Honestly this just makes me even more clueless as to my above question.

I hope I haven't made this too unclear.

To start, it's important to differentiate kinetics (which describe the rate at which reactions happen) and thermodynamics (which describe what net reaction can happen, even if it doesn't happen on the time scale of the universe's existence).

Your equations are all thermodynamic equations, so let's start there. I'm going to assume that you meant standard enthalpy and entropy. Then if we multiply through by $-T$, we end up with the definition of standard Gibbs free energy:

$\Delta G^\circ = -T\Delta S_{tot}=\Delta H^\circ_{sys} - T\Delta S^\circ_{sys}$.

If $\Delta G^\circ<0$, we say the reaction is spontaneous, but only in the thermodynamic sense that it can happen given infinite time. An important point is that this statement refers to the net reaction involved. The reverse reaction (which necessarily has a positive $\Delta G^\circ$) can happen, but always at a rate less than that of the forward reaction, such that the net reaction observed is the forward reaction. Because we're talking about thermodynamics and not kinetics, we can't make any conclusion about what those rates will be in absolute terms, only that the forward will be relatively faster than the reverse.

A different way of saying that is that if we set up a situation in which all reactants and products are at standard states (generally defined as 1 M solutes, 1 bar gases, pure solids, pure liquids), the concentration changes over time will be increases in products and decreases in reactants until the reaction reaches equilibrium. For a reaction that reaches equilibrium quickly, both the forward and reverse reaction happen quickly, but the forward slightly more quickly. At equilibrium, both reactions are still occurring, but now at exactly the same rate.

For a concrete example, consider the dissolution of a soluble compound like NaCl. If we have a solution of 1 M NaCl in water and we add excess solid NaCl, more NaCl will dissolve, but even when the solution becomes saturated (equilibrium), ions are constantly leaving the solid and going into solution while other ions are depositing from solution to the solid at the same rate.

In your hypothetical case where a change in T results in $\Delta G^\circ$ becoming positive, the reactions again are still happening. The difference is just that the reverse is faster than the forward when things are at standard states. It turns out that dissolution of NaCl is one of those cases you described, where $\Delta H^\circ >0$ and $\Delta S^\circ>0$. That means that if you made a 1 M solution and then cooled it, at a certain point solid would precipitate out, meaning that you've gotten to the point where at the standard state, the reverse reaction is faster than the forward reaction. At this temp, $\Delta G^\circ$ for dissolution is positive. If you stay at that temp, you can wait infinite time, and the solid will never fully redissolve.

All of that said, it sounds like your question might really be about kinetics, that is, how much can you slow down a reaction by cooling it? That question has little to do with entropy and instead relates to activation energies.


Calculate the pH of Na2CO3 Given information:

$K_\mathrm{a}(\ce{H2CO3}) = 4.3 \times 10^{-7}$,
$K_\mathrm{a}(\ce{HCO3-}) = 4.8 \times 10^{-11}$,
${{\ce{[Na2CO3]}=0.20\ \mathrm{M}}}$

This was a question in an exam and I'm struggling to understand how to solve it. I understand that $\ce{Na}$ is soluble in water, so $\ce{[CO3^{2-}]}=0.20\ \mathrm{M}$. I also know how $\ce{H2CO3}$ dissociates:

$$\ce{H2CO3 <=> H+ +HCO3- <=> 2H+ +CO3^{2-}}$$

When I try solve it using $K_\mathrm{a}=\frac{[P]}{[R]}$ I get:

$$\frac{\ce{[H+]}^2}{\ce{[H2CO3]}}=1.032 \times10^{-16}$$

Past here I'm not sure. I've been told to find $K_\mathrm{b}$ because it's a base but I'm not sure why that would make a difference. Please could someone make a worked solution or give me some hints?

It suffices to consider only the reaction

$$\ce{CO3^{2-} + H2O <=> HCO3- + OH-}$$

with $k_{B_1} = 2.1 \cdot 10^{-4}$, since the reaction

$$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$

with $k_{B_2} = 2.3 \cdot 10^{-8}$ will only contribute a small amount of $\ce{OH-}$.

$$k_{B_1} = \frac{[\mathrm{{HCO_{3}}^{-}}] [\mathrm{OH^-}]}{[\mathrm{{CO_{3}}^{2-}}]} = \frac{x \cdot x}{c - x} \approx \frac{x \cdot x}{c}$$

$$x = [\mathrm{OH^-}] = \sqrt{k_{B_1} \cdot c} = 6.455 \cdot10^{-3}$$

which gives a pH of 11.8.


Equilibrium Position So if you had the equilibrium $\ce{A <--> B}$ and you increase the concentration of A. I understand that equilibrium will shift to the right. To say this (I believe) is to say that the $\ce{A -> B}$ reaction is favoured. But when at equilibrium again will the position of equilibrium be back in its original position, as I believe the position of equilibrium to be defined as the amount of A relative to B.

If this is true, does it follow that a shift in equilibrium does not necessarily mean a permanent change in the position?

If it was a $\ce{A + B <--> C}$ equilibrium and you increase the concentration of A the equilibrium will shift to the right. How would you determine if the position had changed at equilibrium?
Thanks very much.

Not knowing what the chemicals in your equation are or what kind of properties they have, I can't tell you how you would measure their concentrations.

Nonetheless, given:

$$\ce{A + B <--> C}$$

If you add an additional, preferably known, quantity of A, then the concentration of B should decrease and the concentration of C should increase.

This is actually just the real meaning behind your statement that "the equilibrium will shift to the right".  In this case, with two reactants, obviously more of the second one (B) must be consumed by the addition of more A in order to produce more C (shift to the right).

Again, how you would measure the concentrations of B and C depends on which compounds B and C actually represent.


Why does cationic charge density affect metallic bonding? Group II metals have a smaller cationic size but one more valence electron than Group I metals. Why would the higher charge density of a metal cation affect the strength of the bonding of it in its elemental state?

A good way to conceptualize metallic bonding is that the valence electrons on metal ions become delocalized, leaving positive ions to be electrostatically attracted to the electrons.

Group II metals contribute 2 valence electrons to the delocalized "sea of electrons", and thus have a +2 charge on the remaining cation. These combined create a stronger attractive force.

It is true that the radii of Group II metal cations are smaller than those of Group I cations (within the same row). This also affects bonding, as the delocalized electrons are closer to the nucleus in the Group II case.

In summary: the increased coulombic force between a higher number of delocalized electrons and higher charges on metallic cations create a stronger metallic bond. Transition metals tend to have even stronger metallic bonding due to the participation of electrons from d orbitals.


Balancing Oxidation-Reduction equations Anode: $\ce{2 H2O_{(l)} -> O2 + 4 H+_{(aq)} + 4e^-}$

Cathode: $\ce{ 2 H2O_{(l)} + 2 e- -> H2_{(g)} + 2 OH-_{(aq)}}$

Can someone show me the steps for adding the two. For the overall reaction, it is $\ce{2 H2O_{(l)} -> 2 H2_{(g)} + O2_{(g)}}$

It is not entirely inconceivable that a Galvanic cell could be created with an acidic anode and basic cathode.  Having both half-reactions occurring in the same beaker, however, is unlikely.  That has implications when determining the thermodynamics (cell potential) of the reaction but not in balancing the half reactions.  Given:
$$\begin{align}
\ce{2 H2O_{(l)} & -> O2 + 4 H+_{(aq)} + 4e^-} \tag{1}\\
\ce{ 2 H2O_{(l)} + 2 e- & -> H2_{(g)} + 2 OH-_{(aq)}} \tag{2}
\end{align}
$$

Multiply equation 2 by 2 and add:

$$\begin{align}
\ce{2 H2O_{(l)} & -> O2 + 4 H+_{(aq)} + 4e^-} \tag{3}\\
\ce{4 H2O_{(l)} + 4 e- & -> 2 H2_{(g)} + 4 OH-_{(aq)}} \tag{4} \\
\ce{6 H2O_{(l)} + 4 e- & -> O2 + 2 H2 + 4 H+ + 4 OH- + 4 e^-} \tag{5}
\end{align}
$$

The key here is that the  $\ce{4H+ + 4OH-}$ showing up on the same side equation 5 will yield $\ce{4H2O}$.  Combining acid and base (to make $\ce{4 H2O}$ on the product side) and canceling the $\ce{4e-}$ and $\ce{4 H2O}$ that occur on both sides of the chemical equation gives the desired equation.


how is the relation between Gibbs free energy and cell potential? First of all what is exactly Gibbs free energy in regards to an electrochemical process? How can we visualise it? How does Gibbs free energy change in an electrochemical reaction? How is it related to the cell potential?

Isn't potential the amount of work done per unit charge to bring it from infinite to that point?

Now the ionisation in cell can be thought of as occurring in multiple steps, for example:

$$\begin{aligned}
\ce{Zn(s)} & \ce{->Zn(g)} && (1) \\
\ce{Zn(g)} & \ce{->Zn+(g) + e-} &&(2) \end{aligned}$$

Then the work done to remove the charge should be the change in Gibbs free energy of second reaction and that energy should be equal to potential energy, but we take the combined change in Gibbs free energy as change in electric potential energy of the electron. Why is that so?

The Gibbs energy is just a way to measure the thermodynamically favoured direction of a reaction. It doesn't tell you anything about the kinetics, so it might well be that the activation energy is too high for a reaction to happen, although it is thermodynamically favored.

The Gibbs free energy is related to the cell potential $\Delta E$ via
$$ \Delta G = -nF\Delta E\; ,$$
where $n$ is the number of Electrons transferred per mole product and $F$ is the Faraday constant. It follows that the larger the cell potential is, the more negative the Gibbs free energy of the system.

The cell potential can be calculated using the Nernst equation:
$$ \Delta E = \Delta E^0 - \frac{RT}{nF}\ln\left( \frac{[\text{Red}]}{[\text{Ox}]}\right) $$

$[\text{Red}]$ and $[\text{Ox}]$ refer to the concentrations of the reduced and oxidized species respectively.

Using this information allows you to determine in which direction a given redox reaction will develop.


How does electrolytic conductance depend on solvation? My textbook says: If the ions are strongly solvated their effective size will increase and hence their conductance will decrease

P.s after reading my textbook, I don't think I'm sure what solvation means. If you could explain that, my understanding would be better.

Suppose you are in a field filled with a crowd of children and some of them are holding your hand, some are holding your shirt but you have an urge to reach from point A ( one electrode) to point B (another electrode)? In that case you effective size has increased. How easy would it be to travel from A to B, if you were all alone?

Change in conductance due to solvation is somewhat similar. When an ion wishes to travel under the influence of an electric field between two electrodes, solvent molecules like to travel with the ion, making it "bulkier", hence the ion moves with a retarded velocity.

P.S. Solvation comes from the word solvate which means "A more or less loosely bonded complex formed between a dissolved species and the solvent." It is a general term used in physical chemistry.

Even more formally, http://goldbook.iupac.org/terms/view/S05747
"Any stabilizing interaction of a solute (or solute moiety) and the solvent or a similar interaction of solvent with groups of an insoluble material (i.e. the ionic groups of an ion-exchange resin). Such interactions generally involve electrostatic forces and van der Waals forces, as well as chemically more specific effects such as hydrogen bond formation."


Can someone tell the me the data bank which has all the STO-3G basis sets for many electron atoms? I am trying to express the slater orbitals of Helium atom in terms of 3 gaussian functions (STO-3G basis set). Is there any data bank or reference table in which I can find the exponents and coefficients in it ?
Also in primitive cartesian gaussian , it seems like both $2s$ and $2p$ has the same functional form. For $2p$ it is $x\exp(-\alpha x^{2})$ , $y\exp(-\alpha y^{2})$ and $z\exp(-\alpha z^{2})$. What is for  $2s$ orbital in primitive gaussian function ?

A wide range of Gaussian basis sets for all elements can be found at the Basis Set Exchange. Alternatively, you can try to extract that data from the basis set library of a quantum chemistry software package.
The functional form of a primitive cartesian gaussian with $l=0$ is $\exp(-\alpha r^2)$, independent of the $n$ quantum number.


Safe and responsible disposal of aluminum chloride? This answer affirms that reacting aluminum with HCl is a (seemingly convenient) way to produce hydrogen for use in other experiments, but the byproduct, AlCl3, is rather nasty. What's the best way to dispose of it? My naive guess is that reaction with baking soda would produce carbon dioxide and aluminum hydroxide, which would be fairly harmless. Is this correct?

Sure, you can react the $\ce{AlCl3}$ with $\ce{NaHCO3}$ solution, and you'd also get $\ce{NaCl}$, in addition to $\ce{CO2}$.

Though $\ce{AlCl3}$ is toxic, it reacts readily with water, forming the less toxic chlorohydrate (used in antiperspirants!), and the baking soda should leave an innocuous sludge, safe to dispose of in a sink.

N.B.: The reaction is exothermic and bubbly, so use care when mixing in the $\ce{AlCl3}$ to avoids spatters.


Basicity of Group 15 hydrides Why does the basicity of Group 15 hydrides ($\ce{NH_3,~ PH_3,~ etc}$) decreases down the group ?

I know that's because of its Lewis Base nature due to the lone pair but wouldn't the more EN element be less reluctant to give the lone pair ?
 And nitrogen is most EN in that group .

It can be explained by the very helpful Bent's rule. As nitrogen is more electronegative than phosphorus, there is a bigger s character of the sp3 hybrid orbitals that are used to form the bond to hydrogen than the ones of phosphorus. As a result, the sp3 hybrid orbital that holds the electron pair, has a bigger p character for nitrogen than for phosphorus, meaning that the pair is more distant from the nucleus and more susceptible to donation.


Carbon-based life and helium Help me please with a question about carbon-based life development. I'm almost sure I know answer, could anybody with more solid knowledge in the area confirm it. Could carbon-based life develop on a planet without helium? Did helium help the process on any stage?

As I understand, yes could, because helium is noble gas which does not react with other elements. Or maybe I am wrong?

Helium in stars is an initial product in the fusion reactions that go on to make "more interesting" elements.  In that sense we do need helium to get life going.

Once the more complex atoms are made, however, life appears to have no further use for helium.  As Nicolau mentioned, it seems too hard to see life use helium. The neutral atom shares a spot with neon for the least reactive substance in all of chemistry, whereas the other commonly-found form of helium, the alpha particle (He2+) from radioactive decay is arguably the most reactive substance in all of chemistry. Life can only work with supremely delicate balancing of reactivity in the middle of the scale, not such crazy extremes.


How do I neutralize copper sulfate blue tint on a sofa? I accidentally poured some copper sulfate solution on my sofa.
So there is a blue halo on it.

Is there something I can put on it to neutralize that blue tint?

I was thinking of bleach, but 1- I don't know if it will actually do anything, and 2- It will surely damage the black dye.

There are two forms of copper sulfate you may encounter, the waterfree form ($\ce{CuSO4}$) which is colourless; and the other, the pentahydrate ($\ce{CuSO4 * 5 H2O}$) which is blue/greenish (ref).  You need to get rid of both forms.

In contrast to application on organic matter, the oxidative action by bleach, regardless if based on hypochlorite or the one that is chlorine free, won't remove the stain, yet may -- depending on the dye (mixture) deployed on your carpet attack the colour. On the other hand, though, both copper salts are fairly soluble in water.

Hence, in several cyles, you might attenuate the problem:
1) Dampen the area with some of hand warm water (e.g., with a spray bottle used to dampen your clothing prior to ironing) and subsequently 2)  rapidly dry it by pressing a fresh, clean cotton tissue against the tainted area. The white tissue is meant to act like a blotting paper, to catch both the water as well as the salt dissolved in it, too.  Hence you have to move on, to expose each time a fresh, clean, and dry section of the clean tissue towards the dampened area.

As for every cleaning of a fabric where you do not know yet how well "it works", try first with a small sector of your stained area.  If the fabric is cotton based, the polar groups in cotton actually may catch the copper salt fairly strong.

PS:  As you witnessed the accidental spill, you may be more sensibilized to see the stain, than a guest not familiar with it would be (especially as the dark fabric was not dyed uniformly).


How can I derive whether the methyl cation or anion is planar? According to VSEPR theory, in which species do all the
  atoms lie in the same plane?

  $\ce{CH3+}$
  $\ce{CH3-}$

  (A) 1 only (B) 2 only (C) both 1 and 2 (D) neither 1 nor 2

My try:
Because $\ce{CH3}$ in both forms only are bonded with three hydrogens, they both have one lone pair, which contributes to their tetrahedral structure. So I got D. However, the answer key says that B is the correct answer. How do I get this answer?

The VSEPR model is a rather simple model that actually helps in this case. It assumes that anything ‘on’ a central atom — atoms that it is bound to and lone pairs — requests almost equal amounts of space, and therefore the compound’s final geometry is dictated by the number (and type; lone pairs are deemed ‘invisible’) of substituents only.

It is important to recognise that only lone pairs (and radicals) count towards the VSEPR model; empty orbitals do not. The reasoning is that something cannot take up space if it is empty.

Applying the theory to the methyl cation should give you three substituents: the three hydrogens. These three will, according to the theory, align themselves in a trigonal planar fashion around the central carbon, giving a perfect bond angle of $120^\circ$ between them. Thus, the methyl cation is planar.

Applying it to the methyl anion should give you four substituents: the three hydrogens we had previously plus the lone pair. The lone pair takes up space, so including it into our visualisation would give us a tetrahedral geometry. However, it is also invisible (we can only ‘see’ full atoms) hence the final structure is predicted to be trigonal pyramidal with the four atoms not occupying the same plane.

Therefore, assuming you copied accurately, the answer key should have given A as the correct answer.


Is there any non-toxic and non-solvable-in-air liquid? I want to heat the air with non-aqueous, non-toxic and non-solvable-in-air liquid (with very low relative pressure in air). Is there any such liquid exists?

It would be great if the boiling point of the liquid would be 110 °C.

You may look at silicon Si and its derivatives. For example, the ethoxytriethylsilicane, or triethylsilicol ethyl ether is boiling at 153 °C. Or the tetraethylsilicane is also boiling at 153 °C. This last substance is used as a reference in NMR spectroscopy.


Does a red hot metal reflect light in the same way as a cold metal? I know that a metal at room temperature looks shiny because of the free electrons on the surface, but what about a hot red metal? Would it also reflect light in the same way as it does at a colder temperature? Even if it emits some red light due to it being hot?

The light that you see from hot metals is incandescence.
Incandescence is caused by high temperature. The high temperature promotes electrons to higher energy levels. These electrons then relax and emit light.

On the other hand, the "light" (shiny) that you see from room temperature metals  is caused by re-emission/reflection of other light. Light from another source is absorbed by the metal and re-emitted, or the light from another source is reflected from the surface of a metal.

The reason as to why metals look shiny have to with plasmons. If I recall correctly, most materials will look "shiny" at a certain wavelength range. It just happens that metals are shiny in the visible wavelength range.

What determines what type of light a material will reflect is determined by its plasma/plasmon frequency. Temperature does seem affect the plasmon/plasma frequency. Several studies confirm this.

In general, it appears that the plasmon/plasma frequency decreases with increasing temperature. In other words, the material is able to reflect a narrower range of light as the temperature is increased. This would mean that if you had a material that is shiny in the visible light range, you could make it become dull by heating it. Of course, this is a simplification that doesn't take into account other factors.

However, I wasn't able to find an equation that directly relates temperature and plasma frequency. How the plasma frequency changes with temperature seems to depend on the material and the size of the material (nanoscale, bulk, etc). Furthermore, there are ab initio studies that suggest that the opposite must happen: the plasmon frequency increases with temperature.


What are some uses for an O2/N2 mixture that is more N2-rich than air? What are some uses for an $\ce{O2}$/$\ce{N2}$ mixture that is more $\ce{N2}$-rich than air?
I need help with this, my professor did not do a good job explaining it, and I can't find a source to read about it.

Some laboratories do use, and possibly some processes use "synthetic air", which contains 5% oxygen. This is to carry out oxidation reactions or processes but reduces the amount of excess oxygen to prevent unwanted combustion e.g. if flammable solvents are also being used. Many university research labs will use this if carrying out pressurised reactions.