text:
Which of the following hydrocarbons react with sodium in liquid ammonia?
Problem

The hydrocarbon which can react with sodium in liquid ammonia is
(1) $\ce{CH3CH2CH2C#CCH2CH2CH3}$
(2) $\ce{CH3CH2C#CH}$
(3) $\ce{CH3CH=CHCH3}$
(4) $\ce{CH3CH2C#CCH2CH3}$

Solution

 Solve it as a case of terminal alkynes, otherwise all alkynes react with sodium in liquid ammonia.

Question
I know sodium in liquid ammonia is used to produce trans-alkenes from alkynes. But from the options, it seems all of them react with sodium in liquid ammonia.
The solution doesn't make any sense to me. If all of the alkynes given react with sodium in liquid ammonia, then why the question? And what difference does it make if the alkyne is terminal?

Could the answer actually be all of them? Sometimes they do that on multiple choice questions. In any case the second choice should have an additional carbon between the triple bond and the methyl group on the right. They apparently mistyped.

The terminal alkyne will not give the birch reduction as they will just form the aklyne anion. Maybe this was something the textbook was referring to.

A Birch reduction uses alcohol as a proton source. No mention of alcohol here. Choice (2) was probably meant to be 1-butyne.

Although this is more of a case of communication rather than actual chemistry, I'll explain exactly what the author meant.
They basically want you to ignore the Birch reduction reaction. They want you to find the compound that reacts with sodium in liquid ammonia, but not via Birch reduction.
I however, do stand by that the question is very poorly framed. It is definitely ambiguous, and relies on strange, unrealistic assumptions.

The terminal alkyne has acidic $\ce{H}$ and therefore will give acid base reaction with $\ce{NaNH3}$ and go $\ce{NH4+}$ and $\ce{RC#C- Na+}.$


text:
What is the reaction mechanism between a basic oxide and an acid?
For instance in $$\ce{ Li2O (s) + 2HCl (aq) -> 2LiCl (aq) + H2O (l)}$$ I am wondering why the $\ce{Cl}$ displaces $\ce{Li}$ .I figure that oxygen is negatively charged as it pulls the electron from Lithium toward it, and that Chlorine is attracted to the lithium since chlorine is negatively charged, as itself has higher electronegativity than Hydrogen. But why dont lithium and oxygen just stick together, and hydrogen and chlorine stick together as well, instead of swapping places? Maybe it is related to the electronegativities of oxygen and lithium?

It's because the oxide ion is unstable in aqueous solutions. You may know that the dissolution of $\ce{H2O}$ into $\ce{H+}$ and $\ce{OH-}$ has a $K_w = 10^{-7}$. Trying to pull one proton off a water molecule is hard enough, pulling another one, well you get the idea.
What basically happens is that once you drop the lithium oxide into $\ce{HCl}$ solution, it dissociates into $\ce{Li+}$ and $\ce{O^{2-}}$,
$$\ce{Li2O -> 2Li+ + O^{2-}}$$
The oxide quickly grabs two protons from the acid solution.
$$\ce{O^2- +2H+ -> H2O}$$
What's left is the $\ce{Li^2+}$ and $\ce{Cl-}$ ions. They simply exist in the aqueous medium as a dissolved ionized salt.

Thank you for your answer. I was wondering why Li2O dissociates in water?

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text:
What's the difference between povidone K29/32 and povidone K30?
I'm specifically looking at the two as inactive ingredients in the drug lamotrigine. How do the povidone K values differ chemically and/or in the context of lamotrigine?
They're obviously designed to be completely inert, but is there any chance they're acting differently in the body?

Firstly, Povidone (polyvinylpyrrolidone, PVP)  is  a synthetic polymer used in the pharmaceutical industry as vehicle for dispersing and suspending drugs.

Polyvinylpyrrolidone
There are quite a number of commercial grades of povidone (and also its highly cross-linked counterpart called crospovidone). The K-value of different types of povidone is characterised by their viscosity in solution  and also the mean molecular weight:

The method of obtaining these values is stated in various pharmacopoeias e.g British Pharmacopoeia
These different chemical characteristics offer a variety of uses in pharmaceutical formulations, including as a binder for tablets and capsules, a film former for ophthalmic solutions, to aid in flavoring liquids and chewable tablets, and as an adhesive for transdermal systems.
These also affect the formulation processes during tablet manufacturing for example, the low viscosity and high adhesive properties of K29/32 make it highly efficient as a binder for wet granulation. On the other hand, K30 has the ability to complexate with the active ingredient providing taste masking and solubilization.

The grade of povidone that is selected depends mainly on its molecular
  weight, as this dictates the viscosity, binding effect, the
  complexation capacity and how readily it is eliminated by the body.

Povidone is relatively inert and won't necessarily affect processes in the body instead, on the formulation providing different release characteristics of drugs for example modifying its release properties etc
References

British Pharmacopeia ed
Povidone K29/32
Kollidon® 30 – The Original from BASF

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text:
Nomenclature and bonding of the complexes ions

Why the nitrogen and not the sulphur in the SCN is bonded to the chromium?

https://en.wikipedia.org/wiki/HSAB_theory

Could you please clarify how your question is related to nomenclature, or correct the title and tags of your question?

The bonding of the $N$ end of the thiocyanato group to the Metal centre $\ce{Cr^3+}$ can be explained by both HSAB Theory and Symbiotic effect.According to HSAB Theory, Hard acids prefer to bind with  hard bases, while soft acids prefer to bind with soft bases. Here the metal centre is $\ce{Cr^3+}$, which has $3d^3$ electronic configuration. So,  $\ce{Cr^3+}$ satisfies the criterions for a Hard acid centre, as it has very less no. of $d $ electrons, high positive charge density on it, smaller radius and greater polarisation. So, the metal centre in this complex acts as Hard acid centre. And, now if you consider the properties of a hard base, you will find that $N$ is the harder basic centre, whereas, $S$ is the softer base centre. $N$ has higher electron charge density on it and also smaller in size than $S$ which makes it harder base. Thus, due to the presence of the harder metal acid centre, the ligand tries to bind with the harder basic centre, which in this case is $N$. The attachment of the $N$ site can also be explained by the Symbiotic effect. Observe that, in the complex, you already have two $\ce{NH_3}$ ligands donating through the $N$ site, which actually compells the thiocyanate ligands also to coordinate through the $N$ site, which is the main idea of this effect. Actually, donation through $N$ sites of ammonia makes the metal centre even more hard, and therefore it prefers also the $N$ site of other ligands.  But a remark which is necessary is that the HSAB theory doesn't always predict the correct coordinating site for an ambidentate ligand. If even stronger effects like Competitive $\pi$-Bonding effect start to interfere due to presence of $\pi$-accepting and $\pi$-donating ligands, HSAB theory may not give a correct answer at all. So, in the absence of those effects, HSAB theory  works well.


text:
When iodine replaces bromine in 2-bromobutane, does this lead to an overall lower energy level (electrons closer to nucleus) as well?
It seems to me that iodine is a nucleophile that replaces bromine because it has greater attraction to the carbon nucleus due to its electron properties. However, I thought that bonds are formed for there to be a lower overall energy level, meaning that electrons move closer to the nucleus. Since bromine has less electrons, it seems that the bond between carbon and bromine would have a lower overall energy level than the iodine carbon bond? I guess im asking if bond energy plays a role here, or if the nucleophilic attack occurs indepedent of this?

In hydroxylic solvents nucleophilicity increases as you go from F>Cl>Br>I. Fluoride is the most basic of the four and highly solvated. Iodide is the least basic, most polarizable and most nucleophilic. So bromide would rather be solvated than iodide and iodide would rather be a nucleophile. Moreover, in such reactions, an excess of nucleophile is generally employed. In polar, aprotic solvents (dimethyl formamide (DMF) or dimethyl sulfoxide (DMSO)), the order of nucleophilicity is reversed. You cannot just consider bond strengths or heats of formation of the alkyl halides without considering the status of the halides (solvated, unsolvated). One must consider all of the reaction components.


text:
Are viscoelastic materials to be considered solids or fluids?
I recently stumbled upon the concept of viscoelasticity. Not being a chemist, but being interested in the topic, I started to read the wikipedia article.
The main thing I extracted from it is:
Solids are elastic, fluids are viscous, viscoelastics materials are both viscous and elastic.
I didn't get it, maybe due to my poor background. Is viscoelasticity a state of matter or just a property of materials? In the latter case, can we talk about viscoelastic solid and viscoelastic fluids?
I know the state of matter isn't an element nor compound or material property but it changes as temeprature and pressure change. The strength of the bonds between atoms/molecules and the "value" of their kinetic/vibrational energy is what determines how we see them.
Anyway, I couldn't really fit viscoelastic material in my chemistry mental scheme, so I'm here in search of some clarifications.

Would you consider a semi-crystalline polymer between its glass transition and its melting point a solid or a liquid?  Above its melting point it would be considered a liquid, but would still exhibit viscoelastic behavior.

There are two basic models that describe the two fundamental branches of viscoelastic behaviour:

The Maxwell model is made up of a spring and a dashpot in series: For fast deformations, the dashpot does not really move, and your spring gives pure elastic behaviour. For slow deformations, the spring stays relaxed, and the damper fluid in your dashpot turns all energy into heat.
The Voigt-Kelvin/Kelvin-Voigt model is a spring and dashpot in parallel: Even at the slowest/smallest deformation, the spring never relaxes until you let go and let it return your sample to the original shape. And the faster you try to deform it (e.g. oscillatory), the more heat your damper produces.

The former, e.g. a molten polymer, is a liquid, the latter, e.g. crosslinked rubber, is a solid.

Yes, you can talk about viscoelastic solids and viscoelastic liquids, and it is possible to distinguish the two by oscillatory mechanical tests (e.g. with a rheometer). You usually apply a sinusoidal deformation over time to your specimen and measure the time-dependent response.
In case of a true elastic solid, the deformation will induce a force which follows the deformation without time delay. It is like if you pull a spring: you will instantly feel the force required to do so. In other words: The phase angle which describes how far the sinusoidal material response is shifted from the sinusoidal deformation is 0°.
In case an ideally viscous liquid, you will only feel a force if you 'deform' with some velocity. Therefore, the force you feel while oscillatory deformation will follow the deformation with some time delay. The maximum force will be measured when the velocity of deformation is at its maximum. This results is a phase angle of 90°. It is like stirring in a pot of water: the faster you stir, the more force is required.
In a viscoelastic case, your phase angle is somewhere between 0° and 90°. If it is between 0° and 45° you are closer to an elastic solid and you would call your material a viscoelastic solid. If it is between 45° and 90° it is a viscoelastic liquid.
In a viscoelastic solid, a bigger fraction of deformation energy is stored elastically than is dissipated. In a viscoelastic liquid, the opposite is true. As mentioned by Karl, if a material behaves like a viscoelastic solid or liquid also depends on the frequency or velocity of deformation as well as on the deformation itself.

+1 But not exactly. A viscoelastic material behaves elastic in some frequency regime(s), and viscous ("lossy") in others. The distinction solid/liquid is usually made in the low frequency + small deformation limit.

Viscoelastic materials are generally considered solids, but often are
noncrystalline (like the vinyl in records).  The 'visco' property
is akin to ductility in metals, that a slowly deformed piece of material
cannot be made to relax to its original shape.
Presumably, some near-neighbor bonds (hydrogen
bonds?) reform in relatively short times when strained.
The vinyl record will hold its shape indefinitely under small forces,
but if you bend it, the large force applied causes internal strain
only for a short time, the strain relaxes and the material permanently
takes the as-bent shape.   Rapid bending (slap the record) will
shatter it.   Silly putty, if you are familiar with that material,
is also viscoelastic, but won't hold its shape  indefinitely, and
should be considered a very viscous liquid, except that rapid bending
will break it (in propogating-crack fashion).

Vinyl (PVC) is partially crystalline (ten percent or so), otherwise the record would still shatter if you hit it, but it would already deform standing upright in your bookshelve, and any music pressed into it would vanish after twenty minutes. Polymer melts otoh. are viscoelastic liquids.

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text:
Does hot water freezes faster than cold water?
When we put cold water in the fridge it takes time to freeze but sometimes hot water is faster in freezing.Why?

https://en.wikipedia.org/wiki/Mpemba_effect

Related: Frequency of Mpemba Effect and What is the reason for mpemba effect?

Here are some relevant videos:  Here, Here, Here

https://physics.stackexchange.com/questions/32989/what-is-the-status-of-mpemba-effect-investigations

https://physics.stackexchange.com/questions/122742/hot-water-freezing-faster-than-cold-water?noredirect=1&lq=1

This phenomenon is called the mpempa effect. There is a research paper which explains this using hydrogen bonding The paper is slightly difficult to understand which can be found here
This is a simpler version which I found here

Now Xi and co say hydrogen bonds also explain the Mpemba effect. Their key idea is that hydrogen bonds bring water molecules into close contact and when this happens the natural repulsion between the molecules causes the covalent $\ce{O-H}$ bonds to stretch and store energy.
But as the liquid warms up, it forces the hydrogen bonds to stretch and the water molecules sit further apart. This allows the covalent molecules to shrink again and give up their energy. The important point is that this process in which the covalent bonds give up energy is equivalent to cooling.

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text:
reaction coordinate, kinetics, equilibrium in example

In this special reaction coordinate diagram with two reaction mechanisms, I tried to analyze it in two other ways, one with kinetics, another with equilibrium.
A. kinetics, I tried to find reaction rate of forward, and reverse. By the diagram, rate determining step of forward reaction is the first one (because activation energy is the highest and let's say it is one and only factor for RDS in this case). so Vf=k1[A][B].
RDS in reverse reaction is second step of the reaction judging by the diagram activation energy (M->A+B). So I used pre-equilibrium approximation for that. That's Vr in the picture.
B. equilibrium, I wrote all the equilibrium constant and simply added the reactions.
now here is the question. Expression of the rate of forward and reverse reaction in dynamic equilibrium is quite different from each case, analyzing in a way of kinetics and equilibrium constant, as you see in the picture below. Why is it different? What's right expression? What am I missing here?

I don't understand your question. What do you mean by "different"? The two expressions at the bottom left and right e.g. are equivalent.

You have chosen an unusual scheme in that A appears to be an intermediate as well as a reactant. If you use the simpler scheme things are easier and so rather than try to follow yours I have examined the scheme below which is very nonetheless very similar
$$RX\underset{k_{-1}} {\stackrel{k_1}{\leftrightharpoons}}R+X$$
$$R+Y\underset{k_{-2}} {\stackrel{k_2}{\leftrightharpoons}}RY$$
where species RX looses X and is replaced by Y. The intermediate species is R. The overall reaction is $RX+Y=RY+X$ and the equilibrium constant for the two steps are
$$K_1=\frac{k_1}{k_{-1}}=\frac{\mathrm{[R]_e[X]_e}}{\mathrm{[RX]_e}}$$
$$K_2=\frac{k_2}{k_{-2}}=\frac{\mathrm{[RY]_e}}{\mathrm{[R]_e[Y]_e}}$$
and the overall equilibrium constant
$$K=K_1K_2=\frac{k_{1}k_{2}}{k_{-1}k_{-2}}=\frac{\mathrm{[RY]_e[X]_e} }{\mathrm{[RX]_e[Y]_e }}$$
and all the concentrations in square brackets with subscript $e$ are equilibrium values. [This type of equation is true if there are many equilibria one after the other, $\displaystyle K=K_1K_2K_3K_4\cdots=\frac{k_1k_2k_3k_4}{k_{-1}k_{-2}k_{-3}k_{-4}}\cdots$].
In a rate equation approach we can apply a steady state approach to the intermediate species R. The steady state assumes that the rate of change of R is zero;
$$\frac{d[R]}{dt}= k_1[RX]-k_{-1}[R][X]-k_2[R][Y]+k_{-2}[RY]=0$$
from which
$$ [R]_{ss}= \frac{k_{-2}[RY]+k_1[RX]}{k_{-1}[X]+k_2[Y]}$$
The rate $r$ can be given by
$$\begin{align}
r=-\frac{d[Y]}{dt}=k_2[R][Y]-k_{-2}[RY] &= \frac{k_2(k_{-2}[RY]+k_1[RX])[Y]-k_{-2}[RY](k_{-1}[X]+k_2[Y])}{k_{-1}[X]+k_2[Y]}\\&=\frac{k_1k_2[RX][Y]-k_{-1}k_{-2}[RY][X]}{k_{-1}[X]+k_2[Y]}\\
\end{align}$$
and the numerator is zero when equilibrium concentrations are used making the rate equal to zero also. So this connects the equilibrium to the rate equations.
Initially just after the reactants are mixed the amount of product is very small and $k_1k_2[RX][Y]>>k_{-1}k_{-2}[RY][X] $ and the the rate is
$$r=\frac{k_1k_2[RX][Y]}{k_{-1}[X]+k_2[Y]}$$
which can be confirmed by experiment. Hope this helps.

It's beautiful. Thank you so much. It helped me a lot. I just need a little more time to organize.

so overall k(forward) is k1 x k2 x .... anyway. But by the aspects of reaction rates, it's not really important. What's important is to find it by suggesting some cases with experiments. If k-1 is small in your last equation, erase it and finally k(forward) is similar to k1. (even though k(forward) is k1xk2 thermodynamically) Am I right? anything missing?

I'm not sure I understand your last question, but the last equation only applies at very short times, i.e. just as reaction starts; its not just  the rate constants but also the low concentration of products that are important in removing the last term in the numerator.

I think you get confused by thinking reaction rate and equilibrium constant are same. However, they are in different concepts: The rate of a reaction is dependent on reaction mechanism, which may have different steps, orders, different activation energies for different steps, etc. Unlike reaction rates, equilibrium constant is a state function, which is a function defined for a system relating several state variables or state quantities that depends only on the current equilibrium state of the system.


text:
Can an organic compound undergo ring contraction to give more stable carbocation?
Throughout my journey in organic chemistry I’ve only seen ring expansions by a compound in order to attain more stability, but yesterday I thought about the following reaction.

I feel that ring contraction resulted in a more stable carbocation.
Can someone tell whether the idea behind this reaction is correct or not?

"I feel that ring contraction resulted in a more stable carbocation" Why do you think so, btw? I can't directly infer your reasoning from the diagram you've supplied. Mind explaining? Thanks

https://chemistry.stackexchange.com/questions/19090/what-is-the-reason-for-the-exceptional-stability-of-the-cyclopropylmethyl-carboc

Ring contraction can certainly occur in many cases where the stereochemistry compels the molecule to undergo that to form a more stable carbocation.
In your described case, theoretically, it looks fine, because of more stability due to non-classical resonance (sometimes called dancing resonance). But, if you want to perform that experimentally, the percentage yield may not be guaranteed. It is possible to form the desired carbocation, as it is thermodynamically more stable. But the formation step is expected to have a large rate constant. So, at normal temperature, it may be difficult to form. However, at higher temperature its yield may be increased.
An excellent example of ring contraction is observed to happen during the pinacol-pinacolone rearrangement of (1R,2R)-1,2-dimethylcyclohexane-1,2-diol.
In pinacol-pinacolone, to form a stable carbocation, generally, a methyl or a hydride shift occurs. But for this shift, the migratory group has to be in anti to the leaving group. In the taken compound, the methyl shift can't occur, as it is not in the anti position w.r.t the leaving group $\ce{H2O}$. But a more stable carbocation is formed adjacent to the $\ce{-OH}$ group (like this, $\ce{-C+H(Me) -OH}$), if the bond migration occurs, and due to this bond migration, a ring contraction occurs, and the product which is expected to form theoretically is also verified to be produced experimentally.
A possible mechanism for this beautiful reaction is shown below.

Shouldn't the bond of the ring also be anti for migration to be possible?

How would you define a bond to be in anti w.r.t a group ??

It's not the bond that is migrating, it's the carbon atom of the cyclic ring. I was just saying that the carbon atom should also be anti to the -OH group for migration to be possible. I think in your particular example, an epoxide will form instead of the product you made.

Your idea of ring contraction to give more stable carbocation is correct and similar contraction has observed before (see example 4 below):

Courtesy of Department of Chemistry, Michigan State University, East Lancing, MI: link to webpage


text:
phase equilibrium
Is partial pressure in vapor phase is the same as vapor pressure?
pure vapor pressure of x and y are given.
and i know we have to find the mole fraction times with the purevapor pressure to find vapor pressure.
then, the question also gives partial pressure of x and y in vapor phase and asked whether the systm exhibit negative or postive deiviation from raoult's law.
so what is the use of the partial pressure?

For an ideal liquid solution, Raoult's law predicts that the partial pressure of a species in the gas phase is equal to the mole fraction of the species in the liquid phase times the equilibrium vapor pressure which that species would exhibit if it were a pure liquid at the solution temperature.  If the liquid is not an ideal solution, the partial pressure in the gas phase will deviate from the prediction of Raoult's law.

A negative deviation: If the vapor pressure of a mixture is lower than that expected from Raoult's law.  A positive deviation:  The vapor pressure of the mixture is greater than that expected from the Raoult's law.


text:
Would stainless steel corrode in this electrolytic cell I designed?

As far as I know, stainless steel is resistant to corrosion due to an inert oxide layer formed by the other alloy metals. I am learning about redox reactions right now so I am no expert; but since stainless is not to be used as a anode, I am wondering why is it allowed as an cathode? Wouldn't the oxide layer be reduced, forming oxygen gas and thus exposing more of the metal and lead to corrosion?
I have also read that stainless forms hexavalent chromium when being used as a anode, so is that because it gains extra oxygen atoms due to the oxidizing process? If so, it should be okay to put it in a high oxygen environment, right?
When using a carbon anode, does it only produce carbon dioxide? And is this gas formed from ions, or is there oxygen gas present first, which later reacts with the carbon?

Here is my design, I do not care at the moment that the gases are being mixed:

I plan on using sodium acetate as a salt in a future project, it produces carbon dioxide and hydroxide ions as some of the products. Similar to question three, is oxygen gas ever produced in this reaction?

In the future I would hope to have proper equipment and materials to build a cell like this one: link
Also if any of these questions are too broad to answer please let me know, I am quite new to stackexchange. Thanks in advanced!


text:
What is the relation between bond energy and stability?
My book has two confusing statements about bond energy:

The most stable bond is the bond with the highest bond energy.

A system with low energy is a stable system. Thus, a bond is formed when electron energy level is the lowest.

So, does high energy means more stability or less stability?

Bond energy is shows how much lower is energy of the compound.

“Bond energy” is the same as bond dissociation energy — the amount of energy to break the bond.
So, if adding all the 'bond energies' of a molecule gives the highest number, it requires the most energy to break, and therefore is the most stable.

Look at it in this way. Consider the energy to break a bond to be arbitrarily assumed to be at $\pu{0 J/mol}$. Now, there are two bonds, $\ce{A-B}$ and $\ce{C-D}$. Let's say they have energies of $\pu{-242 kJ/mol}$ and $\pu{-378 kJ/mol}$. Clearly, the $\ce{C-D}$ bond has the lower energy.
Now, you can say that a stronger bond needs more energy to break it. So, this means that you need to supply $\pu{242 kJ/mol}$ of energy to break $\ce{A-B}$ and $\pu{378 kJ/mol}$ of energy to break $\ce{C-D}$. You can definitely see that $\ce{C-D}$ needs more energy to break it, and to bring the energy to $\pu{ 0 J/mol}$.


text:
Free API to view 2D representation of Molecules
I'm looking for a free API that can accept a SMILE and produce the 2D structure of a molecule. I would like to include this in a web page.
I have collection of SMILES and would like to have the 2D structure available as part of a database. I tried generating images using RDKit but I can't add that these images to any data frame or tabular data structure. Any suggestions on how to display the 2D representation of and any SMILE queried on the web page will be appreciated.

PubChem offers a way to get the 2D structure of a molecule if you input the SMILES.
However, why settle for 2D, if you can get 3D structures just as easily?
As an added addition, you might as well check out https://chemapps.stolaf.edu/jmol/jmol.php?model=[INSERT-SMILES-HERE]. This tool will produce the 3D structure of any molecule you can get the SMILES of. Just write the SMILES after the equal to, and voila! You got yourself a nice interactive molecule viewer which you can rotate around to see how it looks.

Disclaimer: I am not affiliated with any of the sites posted above.

Open Babel can convert all kinds of input into pictures with 2D structures.

While I appreciate the mention of Open Babel, there isn't a web api, which was the original request. If people do use obabel, I'd highly recommend using the SVG export option.

If you want a free web service, as noted above you can use PubChem to convert from a SMILES string to a 2D representation. But that's only "limited" to the entries in PubChem...
A better option, IMHO, is the very easy to use NIH Chemical Resolver
https://cactus.nci.nih.gov/chemical/structure/"structure identifier"/"representation"
For example, you can use:
https://cactus.nci.nih.gov/chemical/structure/aspirin/image:

As indicated above, the "identifier" can be names, escaped SMILES strings, etc.

Note: Triple bonds in SMILES strings represented by '#' have to be URL-escaped as '%23' (e.g. the SMILES string of ethyne has to be specified as 'C%23C' instead of 'C#C' if encoded as part of a URL

And yes, you can also get the 3D structures.. e.g. https://cactus.nci.nih.gov/blog/?p=249

It's not an API, but JSME (Javascript Molecule Editor) can be used to convert MOL, SDF, or SMILES into a 2D structure.

If you are willing to put a little programming effort into this you could have a look at the Chemistry Development Kit (CDK) that is easily capable of converting a SMILES string into a 2D render of the molecule.
An example (taken from here) that will produce the depiction as a PNG image looks like this:
import org.openscience.cdk.CDKConstants;
import org.openscience.cdk.interfaces.*;
import org.openscience.cdk.silent.SilentChemObjectBuilder;
import org.openscience.cdk.smiles.SmilesParser;
import java.awt.Color;

public class Main {
public static void main(String[] args) throws Exception {
    IChemObjectBuilder bldr   = SilentChemObjectBuilder.getInstance();
    SmilesParser       smipar = new SmilesParser(bldr);

    IAtomContainer mol = smipar.parseSmiles("CN1C=NC2=C1C(=O)N(C(=O)N2C)C");
    mol.setProperty(CDKConstants.TITLE, "caffeine");

    DepictionGenerator dptgen = new DepictionGenerator();
    // size in px (raster) or mm (vector)
    // annotations are red by default
    dptgen.withSize(200, 250)
          .withMolTitle()
          .withTitleColor(Color.DARK_GRAY);
    dptgen.depict(mol)
          .writeTo("~/caffeine.png");
}
}

The result can be seen here

Another option for a little programming is the Indigo Toolkit, which has Python, Java, .NET, and C libraries.
The following Python example takes the isomeric SMILES for Albuterol, which I obtained from PubChem, and renders 2D structures for three forms (one with unspecified stereochemistry and the two stereoisomers). The individual structures are labeled with their respective PubChem CIDs.

from indigo import *
from indigo.renderer import *

indigo = Indigo()
renderer = IndigoRenderer(indigo)

mols = {   '2083': 'CC(C)(C)NCC(C1=CC(=C(C=C1)O)CO)O',
         '123600': 'CC(C)(C)NC[C@@H](C1=CC(=C(C=C1)O)CO)O',
         '182176': 'CC(C)(C)NC[C@H](C1=CC(=C(C=C1)O)CO)O' }

array = indigo.createArray()
for key in mols.keys():
    print(key, mols[key])
    mol = indigo.loadMolecule(mols[key])
    s = "CID=" + key
    mol.setProperty("grid-comment", s)
    array.arrayAdd(mol)

indigo.setOption("render-comment", "Albuterol")
indigo.setOption("render-comment-position", "top")
indigo.setOption("render-grid-margins", "40, 10")
indigo.setOption("render-grid-title-offset", "5")
indigo.setOption("render-grid-title-property", "grid-comment")
indigo.setOption("render-background-color", 1.0, 1.0, 1.0)
indigo.setOption("render-atom-color-property", "color")
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text:
Why 2,2-dimethylpropane is not called isopentane?
why 2,2-dimethylpropane is not called isopentane? since 2-methylpropane is called isobutane

... because it is not the only isomer of n-pentane.

It's called neopentane, typically 2-(mono)methylalkanes are called iso.

@Mithoron Yes; except the famous “isooctane” which is used for the determination of the octane rating of fuel. This isomer of octane is actually 2,2,4-trimethylpentane.

@Loong Indeed, calling in isoneo would be complicated ;D

In common nomenclature, the prefix 'iso-' is applied when the chain of carbons is continuous bar one methyl group. The example you provided, 2-methylpropane, is observed to be continuous with exception to one methyl group bound to the central carbon atom.
However, 2,2-dimethylpropane does not feature an analogous structure: because two methyl groups are bound to the central carbon, the same classification is inapplicable. In place, the molecule could be labelled as 'neopentane', in which the prefix specifies a continuous chain excepting two alkyl groups.

The above image is 2-methylbutane, or isopentane.

The above-shown image is 2,2-dimethylpropane, or neopentane
If you desire further information concerning the nomenclature, I would advise the following link, from which these images were extracted: http://www.chem.ucla.edu/~harding/IGOC/C/common_name.html

Which is "isohexane"?  2-methylpentane or 3-methylpentane?  If I am not mistaken "iso-n-ane" is generally applied to 2-methyl-(n-1)-ane.


text:
How do you know if a nucleophile is strong enough to to break a bond?
For instance, iodine can displace bromine in $\ce{CH3Br}$ Is this due to iodine being lower in the periodic table, thus having more electrons in the outer shell and being a stronger nucleophile?

It doesn’t have more electrons on outer shell...

In a simplistic way you can say that, due to the less effective charge of iodine outer electrons respect to bromine, it will be more likely to react, and will replace bromine. But to say it exactly you need Gibbs free energy, if reaction occurs and constant T and P...


text:
Reduction of nitro compound using protection group

I've tried researching things like cyclisation, nitro reduction and all I've found is little to nothing. I know that for the 1st arrow I need to protect the carbonyl and the 2nd arrow is the carboxylic acid but I'm not sure how to go about with the reaction or mechanism. Can you someone guide me along please?

For the first arrow you are correct that you need to protect the ketone before reducing the nitro group. For this I would use a dimethyl ketal. There are many sets of conditions for this; trimethyl orthoformate in refluxing MeOH with catalytic PTSA is a good bet. This will also give the methyl ester of the carboxylic acid but that is of no consequence as it will be reacted in the next step. Your next step is then the nitro group reduction. I see no reason why you should not use hydrogenation, just don't have any acid present as that may remove the ketal and you still need that for step 3. The aniline thus produced will cyclise easily, just heat it in neutral solvent if it does not spontaneously cyclise. Step 3 is reduction of the amide, LiAlH4 should do this. Finally you deprotect the ketone with a mild acid hydrolysis.
For the second arrow reduction of the nitro group by hydrogenation will give indole-2-carboxylic acid by spontaneous cyclisation as formation of the aromatic [5] is favoured over the non-aromatic [6]. This has been reduced to your target indoline-2-carboxylic acid by sodium/liquid ammonia  https://patents.google.com/patent/US4485241


text:
Is heat an extensive or intensive property?

Is heat extensive or intensive property?

I surfed the Internet for the same and found quite different answers. Like, heat is not a property, so it is neither. One said that since heat depends on the amount of substance, it is extensive. I thought about the latter one as well.
So, my question could be primarily divided into:

1. Is heat a property of matter?
2.1. If yes, is it extensive or intensive? Why?
2.2. If no, why?

In thermodynamics, heat is not a function of state but a process function. So for the question

Is heat a property of matter?

the answer is no.

... is it extensive or intensive?

Since the concept of intensive and extensive properties only applies to physical properties that are a function of state, heat is neither intensive nor extensive.
As Chester said in the comments, it is worth noting that heat (as well as work) applied to a system can cause changes in the thermodynamic equilibrium state of the system.

Am I right if I say that heat is transfer of thermal energy?

Yes, you can say so.

This was a wonderful concise answer.  I would add that heat and work cause a change in the thermodynamic equilibrium state of a material.

You seem to be confused between internal energy and heat.
An object does not "possess" heat. The correct term to describe the energy possessed by it due to molecular motions and other microscopic properties is internal energy.
Heat is the energy that flows due to temperature difference between two points.
Thus, at any point of time, you can state that a body "has" internal energy but you cannot declare that it "has" heat rather, the latter is flowing through the body.
This article might be helpful.
Conclusion: Since heat is a path function, it is not a "property of matter" and hence cannot be described by "intensive" or "extensive".

Heat means different things at different times, and to different people.
1) If some particular matter has been injected with an amount of heat when referred to that same matter at some other temperature, it has a quantity of heat energy with units of calories or joules. This quantity of heat will be dependent on the size of the system; therefore, the heat (energy content) is called an extensive property of the system.
2) The particular matter has a temperature and can be subdivided into smaller pieces. They will all have the same temperature. This property of the system is independent of the size of the pieces, so temperature is called an intensive property.
So, heat content (amount) is extensive, but heat level (i.e., temperature) is intensive.


text:
Results of mixing HSbF6 and O2F2
I'm an amateur writer and one of the topics I deal with is what happens when a slightly-mad scientist and engineer is given an opportunity to work with full-on “magic.”  So there are situations where magic is analyzed and other situations where the technology that comes from that analysis seems on par with “sufficiently advanced technology”... and there are plenty of dangerous shenanigans that result; one of which is when an experiment on regenerative healing magic is contaminated with HeLa cells, accidentally creating a magically-fueled, self-sufficient cancerous growth.
One of the solutions I was looking at was Piranha Solution, but I got to wondering if that concept could be ramped up.
Thus, I was wondering if fluoroantimonic acid and dioxygen difluoride mixture would work in a similar, if more volatile and violent, manner, or if there would be other, undesirable reactions?
If it would work, what ratio would be the most effective?

https://en.wikipedia.org/wiki/Dioxygen_difluoride#Reactivity - If SbF5 alone would be used, maybe dioxygenylium salt might be obtained.

Mithoron:  I tried googling Dioxygenylium, didn’t really find anything on my level of understanding.  Could you or someone give me a quick crash course on what that is?

I'd recommend using artistic license. Very potent, potentially volatile, violent, virtually any desired property.


text:
Is a non-compound molecule a chemical substance?
The Wikipedia definition of chemical substance states:

A chemical substance, also known as a pure substance, is a form of matter that has constant chemical composition and characteristic properties. It cannot be separated into components by physical separation methods, i.e., without breaking chemical bonds. Chemical substances can be chemical elements, chemical compounds, ions or alloys.

A molecule made up of the same element (e.g. $\ce{H2}$) is not a compound; however, molecules aren't listed in the final sentence. Is a non-compound molecule a chemical substance?
I believe it is and here's my overall understanding of the terms in this space, and any corrections would be appreciated:

Found that the answer is 'Yes' in the IUPAC Gold Book definition of chemical substance:

chemical substance
Matter of constant composition best characterized by the entities (molecules, formula units, atoms) it is composed of. Physical properties such as density, refractive index, electric conductivity, melting point etc. characterize the chemical substance.

I also made an edit to the Wikipedia page to add molecules.


text:
How do sulphate esters compare to phosphate esters?
In particular, the energy released from hydrolysis of the -S-O-S- bonds and how stable are they in aqueous solution compared to phosphate esters?

I disagree with this being too broad. I'd think a yes with an example or two would suffice.

Yes. How is it "too broad"? I'm asking for comparison of sulphate esters and phosphate esters when it comes to hydrolysis in aqueous solutions.


text:
How does the Wurtz reaction produce alkynes?
Wikipedia mentions that alkynes can be produced from geminal dihalides through the Wurtz reaction. However, I am unsure of the reaction mechanism for this reaction. I have not been able to find any sources which show the mechanism for this reaction. Does anyone have any idea how the reaction is likely to occur?

This is just a guess but I would guess that if you had 1,2-dichloroethene (idk trans or cis) and added sodium metal, you would get sodium chloride and ethyne.

@tox123 But the wikipedia article mentioned "geminal dihalide", not vicinal dihalide.

My reading of the Wikipedia article's reference # 2 does not involve "geminal dihalides". The Organic Synthesis (ref #2) article is the Wiberg/Lampman synthesis of bicyclobutane from 1-bromo-3-chlorocyclobutane with sodium. Base treatment of a geminal dihalide would be expected to give an alkyne. Sodium would lead to reductive elimination (carbene) and an alkene. Wikipedia is not a primary resource.

I have a better guess now (that takes into account geminal dihalides, that I'll post as an answer just do to it's length.

@tox123: You have generated two radicals on the same carbon by a two-electron reductive α-elimination (1 --> 2). So far, so good. In other words, you have created a carbene (2). Carbenes can rearrange to alkenes, NOT alkynes. Think of the two orbitals on the carbenoid carbon as one bearing two-electrons (negative) and the other one vacant (positive) (3). When a vicinal hydride migrates to the vacant orbital, one has a negative charge on one carbon and a positive charge on the vicinal carbon (4). Structures 4 and 5 are identical. If dichloride 1 is treated with base, one gets a double elimination to an alkyne. There is no change in oxidation state. Therefore, dichloride 1 cannot give an alkyne via a 2-electron reduction.

In other words, you are saying that the Wurtz reaction cannot be used to produce alkynes directly?

what if only one chlorine is removed at a time? also are carbon diradicals necessarily carbenes? I was envisioning the carbon being in a triplet state, rather than the singlet state.

A 2-electron reduction is a 2-electron reduction. Singlet or triplet is irrelevant. It is basic oxidation/reduction chemistry.  The conversion of a geminal dichloride to an alkyne, no matter the pathway or mechanism,  is a net zero-electron change process. Look at the diagram 1 --> 6. As to the whether the chlorine atoms are removed one at a time, they probably are. Doesn't really matter.

@user55119 I have learnt that vicinal dihalides can undergo elimination to give alkynes. Can the same be done for geminal dihalide? Could you provide a source?

@Tan Yong Boon:  This reference may be of help.      Bergstrom, F. W. and Fernelius, W. Conard, Chem. Rev., 12, 43-179; 1933

So, as wikipedia says:

The Wurtz reaction occurs through a free radical mechanism.

So starting off with a geminal dihalide as in the picture below, and then adding sodium metal you get two unpaired electrons on a single carbon.

One of these electrons binds with an already paired electron on the other carbon forming an alkene with free electrons on both sides of the double bond (this should also form a free hydrogen atom, which is not pictured below):

These electrons then pair with one another to form an alkyne (freeing another hydrogen atom, which will react with the first to form hydrogen gas):

As a disclaimer, I don't know if this is the exact mechanism this is just an educated guess on my part, but because the Wurtz reaction has a free radical mechanism, this seems the most obvious way of this occurring.

I believe after the double bond is made, the molecule remains a radical, and doesn't become a diradical

That is what I am saying. You might want to familarize yourself with the concept of oxidation levels. http://ursula.chem.yale.edu/~chem220/STUDYAIDS/oxidationlevel/Oxidationlevel.html


text:
How does bismuth's lone pair affect its properties as a metal and alloy?
Bismuth is really the only metal to exhibit a lone pair in its outermost shell, this means that it and its salts should be a Lewis base. As bismuth can act as a Lewis base, how does this affect its properties as a metal specifically in regards to alloying? For instance, if I have a 1:1 mole ratio of aluminum and bismuth in an alloy, will the aluminum and bismuth form dative bonds?

Inert pair effect makes it kinda non-issue. Also most of "metallic elements" has lone pairs all the time - lots of d electrons. Even calling an element metal/gas/etc. is pretty much misnomer 'cause they can have various forms.

If you look at the chemistry of lead(II) and the work by Tony F. Hill (and others) on platinum group metals you will see that bismuth is clearly not the only metal to have lone pairs.
What Tony Hill did was rather clever, normally metal-ligand coordination is all about the donation of lone pairs to the metal and sometimes the retrodonation of electron density onto pi acidic ligands such as carbon monoxide and cyanide.
What Tony did was to get a borane to bind to a lone pair on a low valent metal such as ruthenium, osmium or rhodium. Here is a diagram of an example of one of his compounds.

The synthesis of this rhodium compound can occur from $\ce{[RhCl2(Ph)(PPh3)2]}$, this rhodium complex can react with $\ce{Na[HB(mt)3]}$ (mt = methimazolyl) to form an intermediate $\ce{(mt)B(H)(mt)2Rh(Ph)Cl(PPh3)2}$ which then decomposes to lose a molecule of benzene to form $\ce{B(mt)3RhCl(PPh3)}$. (Chem. Comm., 2005, page 221).
I am sure that tin has a stereochemically active lone pair, if we consider the structure of $\ce{Cp2Sn}$, this is not quite like ferrocene. Instead of the two Cp rings being parallel. They are not. So I think that bismuth is not the only metal to have lone pairs on it. Even the bis Cp* complex of lead(II) (bis(h5-pentamethyl-cyclopentadienyl)-lead) is bent away from the ferrocene type structure suggesting that the lead has a stereochemically active lone pair.


text:
What is the empirical formula for kerosene?
Kerosene is a blend of hydrocarbons, so it doesn't have one chemical formula. But I'd like to find an (approximate) empirical formula giving the ratios carbon and hydrogen.
I haven't found anything online, and beggars can't be choosers. My actual interest is in RP-1. There is a NIST document (archive) on its thermophysical properties which gives a spectroscopic breakdown on pages 29–32 showing that even this highly refined kerosene is mind-bogglingly complex chemically.

On p. 28 of the NIST document linked in the question, there is a surrogate mixture formulation of 14 hydrocarbons which closely approximates RP-1. Its empirical formula is $\mathrm {C_{11.8}H_{23.0}}$, and this should be close enough for any sort of kerosene. The properties of this surrogate are given on p. 7.
This means that burning $\pu {1 kg}$ of kerosene in air produces about $\pu {3.15 kg}$ of $\ce {CO2}$ and $\pu {1.26 kg}$ of $\ce {H2O}$ with the extra mass, $\pu {3.41 kg}$, coming from oxygen. Alternately, if you have $\pu {1 kg}$ of a perfect stoichiometric mix of RP-1 and liquid oxygen, burning it would produce $\pu {715 g}$ of $\ce {CO2}$ and $\pu {285 g}$ of $\ce {H2O}$.


text:
IUPAC name of a molecule with a cyanide group and a double bond
This is the molecule:

After identifying the functional group, what should we prefer? A longer carbon chain or the one which is short but has a double bond in it.
I personally think that we should go for longest carbon chain. So the name should be 2-ethenylpentanenitrile.

Related: How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

The relevant simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds

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

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)

You have correctly identified the suffix (“nitrile”) for the principal characteristic group. Thus, according to the first criterion, the parent structure could be pentanenitrile or but-3-enenitrile since both include the nitrile group.
The next criterion for the principal chain is the greater number of skeletal atoms (i.e. the longest chain). Thus, the principal chain corresponds to the pentanenitrile part and not to the but-3-enenitrile part since pentanenitrile has a longer chain than but-3-enenitrile. Therefore, the correct name is 2-ethenylpentanenitrile.
Many people would probably still call it 2-vinylpentanenitrile; according to current IUPAC recommendations, however, the prefix “vinyl” is retained only for general nomenclature, “ethenyl” is recommended as the preferred prefix.


text:
Hexagonal water, alkaline ionized water
I am chemical engineering and i care my nutrition and health. in my opinion, it is so important that what we drink or eat daily.
 Respectfully, I have a question about alkaline ionized water bottle, Which is sold commercially. I have some information about production of electrolyzed water technology. as i study in some articles, hard water (water with minerals such as calcium bicarbonate) electrolyzed in a electrolysis cell and Hydroxide is produced. As i knew, calcium and magnesium hydroxide are insoluble in water. But we don't see any white residue in commercial alkaline ionized water bottle. These particles (as an alkaline agents) are probably filtered, my question is " if calcium and magnesium hydroxide are filtered, why this water have pH 9 or high??"
thank you
I'd be happy and it is my pleasure if you tell me this.

Both Calcium and Magnesium hydroxide are sparingly soluble in water, not completely insoluble.

As Waylander states, calcium and magnesium hydroxides do have some solubility in water.  In the case of magnesium hydroxide you can demonstrate this by simply buying some milk of magnesia and applying red litmus paper to it; the paper turns blue like there were an alkali.
Simply dissolving either calcium or magnesium hydroxide into water would raise the pH beyond the target (you get 10 for magnesium, 12 for calcium).  Additional calcium or magnesium salts may be introduced to shift the dissolution equilibrium for the sparingly soluble hydroxide, and thus control the pH.
Here and here the use of lime and magnesia for water treatment is mentioned.  Both work by making the water slightly alkaline so as to displace heavy metals as less soluble hydroxides.


text:
Is a person warmed by fire an example of endothermic reaction?
My chemistry textbook gives the example of a man being warmed by a fire as an endothermic reaction. If the man's body is the system and it is increasing in temperature, how can this be endothermic?

It's endothermic because the man is taking in heat, hence why he's increasing in temperature. Exothermic systems release energy (the fire), endothermic systems absorb energy (the man).

I wouldn't call it an endothermic chemical reaction. The fire is an example of an exothermic reaction. The man constitutes part of the surroundings for the fire (the system), and heat is being transferred between the system and its surroundings. Looking at it the opposite way, the fire could be considered the surroundings (a heat reservoir) for the man (the system), and there is heat being transferred from the surroundings to the system. But the man is not undergoing a chemical reaction.


text:
How to make snow
I want to make a DIY snowgun.
I wonder how is snow chemically made? What is the temperature needed? What is the pressure needed? What is the density of water to throw in the air? What is the humidity of the air needed? Is it easier to make with a chemical reagent?
Is there a velocity in which the water would react more efficiently? Do I need to throw the water of a certain temperature and a certain pressure directly to an atmosphere of a different temperature and different pressure?
I'm not a chemist (I'm in 12th grade) but I'm not afraid to get a complex answer that might be hard to understand.

Welcome to Chemistry! Take the tour to get familiar with this site. You should probably read about snowflakes first, which would answer most of your questions.

Google 'snow cannon'. Here's a start for you  https://en.wikipedia.org/wiki/Snowmaking


text:
Composition of Portland Cement before Hydration
I recently came across a question-

Which of following chemicals is NOT found in Portland Cement?
A. Di-calcium silicate
B. Tri-calcium silicate
C. $\ce{CaSiO3}$
D. $\ce{Ca3(PO4)2}$

I answered D. $\ce{Ca3(PO4)2}$. But according to the examiner the answer is C. $\ce{CaSiO3}$
I strongly doubt that. Any pointers to credible sources/books settling this please!!

Welcome to Chemistry.SE! Take the [tour] to familiarize yourself with the site! Why do you think that the answer is not $\ce{CaSiO3}$? It will help the answerers to answer your question!

Thanks for bringing that up...as per my understanding Tri calcium phosphate is bioactive part of Cement used to reconstruction bones, like tooth filling may be...That's completely different application from Portland Cement

Calcium phosphate is an undesireable component in portland cement. One of its origins can be the usage of meat and bone meal in cement kilns.

According to Ullmann’s Encyclopedia of Industrial Chemistry (sixth edition), Portland cement consists mainly of tricalcium silicate, dicalcium silicate, tricalcium aluminate, and calcium aluminoferrite.
$$
\textbf{Phase composition of Portland cement clinker} \\
\begin{array}{llll}
\hline
\text{Constituent} & \text{Formula} & \text{Content in wt %} \\
& & \text{Range} & \text{Average}\\
\hline
\text{Tricalcium silicate} & \ce{3CaO.SiO2} & 46{-}79 & 61 \\
\text{Dicalcium silicate} & \ce{2CaO.SiO2} & 5{-}30 & 15 \\
\text{Calcium aluminoferrite} & \ce{2CaO.(Al2O3,Fe2O3)} & 4{-}16 & \phantom{0}8 \\
\text{Tricalcium aluminate} & \ce{3CaO.Al2O3} & 6{-}18 & 12 \\
\text{Free calcium oxide} & \ce{CaO} & 0.1{-}4 \\
\text{Free magnesium oxide} & \ce{MgO} & 0.7{-}1.5 & \phantom{0}1.5 \\
\hline
\end{array}$$
Büchel, Moretto, Woditsch Industrial Inorganic Chemistry (second edition) lists the same six constituents as important clinker phases. This reference also includes a diagram for the three component system $\ce{CaO}$–$\ce{SiO2}$–$\ce{Al2O3}$ with composition ranges for various cement types, which shows that calcium monosilicate (CS) is not necessarily within the typical range for Portland cement.

Furthermore, this reference includes a table with a typical chemical analysis of a Portland cement, which shows that the $\ce{CaO}$–$\ce{SiO2}$ ratio is too high for calcium monosilicate and also indicates that small amounts of phosphate could be present:
$$
\textbf{Typical chemical composition of a Portland cement clinker} \\
\begin{array}{ll}
\hline
\ce{CaO} & 58{-}67\ \% \\
\ce{SiO2} & 16{-}26\ \% \\
\ce{Al2O3} & \phantom{0}4{-}8\ \% \\
\ce{Fe2O3} & \phantom{0}2{-}5\ \% \\
\ce{MgO} & \phantom{0}1{-}5\ \% \\
\ce{SO3} & \phantom{0}0.1{-}2.5\ \% \\
\ce{Mn2O3} & \phantom{0}0{-}3\ \% \\
\ce{P2O5} & \phantom{0}0{-}1.5\ \% \\
\ce{TiO2} & \phantom{0}0{-}0.5\ \% \\
\hline
\end{array}$$

Yes, most articles talk about the main components and active compounds. What if one of the chemicals is a non-active component in trace amount. I hope some research article talks about that...Google didn't help

Thanks for the answer. I get it now. I found some supporting content. According to the European Standard EN 197-1 , The ratio of CaO to SiO2 shall not be less than 2.0. Also there is excess free CaO found in Cement. That must be the reason why mono-calcium silicate doesnt exist in final product.


text:
How to calculate mass of water of crystallisation?
How do you work out the mass of water of crystallisation for 13.5 g of
$$\ce{Al2(SO4)3*6H2O}$$
The correct answer is 3.24 g of water but I'm unable to derive the answer.
What I have tried:
I worked out the mols of aluminium sulfate which is approx 0.038 mol by dividing 13.5 g/353.5.
Since the ratio of aluminium sulfate: water of crystallisation is 1:6, I multiplied the number of mols by 6 to result in 0.228 mol of water.
To work out the mass, I multiplied mols of water by molar mass to result in 4.101 g which is NOT the answer of 3.24 g .
What am I doing wrong and what is the correct method?

The mass of aluminium sulfate has to be taken along with the water of crystallization.
Now molar masses
$$
\begin{array}{c| c}
\text{compound} & \text{ mass (g)} \\
\hline
\ce{Al2(SO4)} & 342 \\
\ce{6H2O} & 108 \\
\ce{Al2(SO4)3\cdot 6H2O} & 450\\
\end{array}
$$
After that it's simply ratio and proportion find out the moles of hydrated salt then find out the moles of water finally get the mass of water.
$$\frac{13.5}{450} \times 6 \times 18 = 3.42$$


text:
What transition metal ions form without oxidant in water pH 7?
I am not sure how many transition metals will ionize without oxidants, where could I learn this?

They don't react in such situation at all.

@Mithoron ...unless water is intended to be the oxidant as with alkali and some alkaline earth metals.

@OscarLanzi We both know what OP meant, but on the other hand I doubt if he asked about lanthanides.

If you are counting lanthanides as "transition metals", then look at europium whose reactivity with water is similar to calcium.  Although the product listed in the reference is the trivalent hydroxide $\ce{Eu(OH)3}$, europium has a relativy stable +2 oxidation state and so the reaction msy proceed via a relatively soluble divalent hydroxide or cation ($\ce{Eu(OH)2, Eu^{2+}}$).  The combination of a highly electropositive element, possibly with formation of a relatively soluble, low oxidation state hydroxide/cation makes europium more reactive to water than most transition metals.

Very nice info for europium. I am hoping for a reference where i might learn this sort of info for any transition metal.

If only this were for pH 7: https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements


text:
Why do electrons not flow through the electrolyte?
Why do the electrons not flow through the electrolyte instead of the external circuit in a electrolytic cell connected to an external circuit? I mean, if an electrolyte is a conductor why would the electrons take the way through the external circuit rather than through the electrolyte to the other electrode of an opposite charge?
Like:

Ah, thanks @Mithoron. Thought for sure I'd seen a dupe of this before.

Oh I accepted the duplicate vote by mistake haha... Anyway, I do not see how it answers it. So think I need a little further help :)

asd, the short answer is that electrons can't carry current within the electrolyte, because they are too reactive: they would immediately react to form ions. The electrolyte is a conductor because it's an ionic conductor.


text:
Why are there two contradicting formulas for Enthalpy (Delta H):
1) Delta H = Bonds broken - Bonds Formed
2) Delta H = $H_{products} - H_{reactants}$
In the first formula, bonds break in the reactants and form in the products, so its basically saying "bonds energy of reactants" minus the"bond energy of products", which is the complete opposite of what the second formula gives. How so? And when you know to use each formula?

This is my take on this

1) Delta H = Bonds broken - Bonds Formed
2) Delta H = $H_{products}−H_{reactants}$

Equation 2 is pretty straightforward, (change in enthalpy is the enthalpy of the products minus the enthalphy of the reactants).
Equation 1 is a bit less straightforward. Let's define the system as where you are having the reaction, and the surroundings as everything else. When you break bonds, you take energy from the surroundings into the system. Then, when bonds are formed, energy is released from the system into the surroundings. Therefore, if there is more energy released in the bonds formed than there is energy gained in the bonds broken, the system loses energy. How I see it is, you are looking from the surrounding's perspective at equation 1 and from the system's perspective from equation 2, that's why you got two opposite statements (from your POV).


text:
Calculating cell-potentials
I have measured the cell potential between a number of metals e.g. Ni, Pb, Al, Cu and Zn by constructing a galvanic cell and then used a voltmeter to measure the potential.
Now, if I want to calculate the theoretical potential (if I have understood this correctly) I need to look at a standard reduction potential table and compare the two elements in question. Take Al and Cu for example. The element that will act as the anode where oxidation will occur will be the element with the more negative cell potential. In this case Al because it has a value of -1,66 V compared to +0,34 V for Cu. Then since it is a reduction table I need to flip the reaction of Al to show an oxidation with a value of +1,66 V.
The full cell potential will be 1,66 V + 0,34 V = 2,0 V
I have used this same logic and gotten results for Ni and Pb (0,13 V), Al and Ni (1,4 V) etc.
Now my teacher asks me how come you haven't gotten any negative theoretical values and to check my work again? But surely I can't get any negative cell potential values since this is a galvanic cell or am I mistaken?
She asked the same thing with the real measured potentials with the voltmeter and of course here I got a few negative readings, but isn’t that simply because I placed the positive and negative ends of the voltmeter on the opposite electrodes? So I made all negative measurements positive as if I HAD placed them correctly…
Am I wrong here, should there be any negative values theoretically?

You're right in saying you flip the reaction, but you don't flip the sign; it will still be a negative potential. Because of that, 0.34 - (-1.66) = 2V, since potential difference is finding the difference (by subtraction). It's the same value but you will have a negative in there. Think of it this way: oxidation is loss of electrons, so electrons are being released into the cell, meaning that cell will become negative.

Oh I see! So just to make sure I got it, in the case of Zn with Cu, Zn (-0,76V) will be the anode and Cu (+0,34V) the cathode. However I don't switch the sign, simply calculate E-cell as E-cathode - E-anode?

This still gives me the same potentials, but do You have a guess on what my teacher could  mean with there being a problem that I haven't gotten any negative E-cell values?  I shouldn't right?

I believe it could be that I wrote as I told you: 1,66 + 0,34 = 2,0 V and that she is hinting on why I havent used any negative values as you explained I should.. Could you help me with one last thing? If I created a table with all combinations of electrodes and the experimental values that they gave, how would I go about in ranking their reduction potentials from best to worst?

Your teacher probably does mean that. In a galvanic cell, you'll have positive and negative cells due to oxidation and reduction happening at each cell. Think of it like a standard battery, you have a positive and negative end due to the same reason. To calculate the cell potential, it's always the positive voltage - the negative voltage; subtracting a negative number just adds it instead, hence why you reached the same value either way. By the way, what do you mean by 'best to worst'?

I have constructed cells from all combinations of a few metals and gotten their potentials, e.g. Zn-Al cell (0,521 V) and Zn-Pb cell (0,321V) etc. Now I am asked to rank the metals using this information according to their reducing strengths

Rank them from the highest value to the lowest value; the highest value will have the largest reduction strength. It's called 'reduction potential' for a reason - a larger potential means a larger strength. The largest value will be the 'best' reduction cell.

So simply, if I take the values I have measured experimentally and rank them in decreasing order.. 0,896 > 0,645 > 0,525 etc. the metals reducing strength decreases from right to left?

The reducing strength will decrease from the most positive value to the least positive one.

As I understood your problem with signs of the potentials, I think it's better if you write your redox equations, starting with half-reactions. Like @Kian Stevens said, consider your Galvanic cell  is as standard battery (albeit a battery is a set of galvanic cells). So, to have a electric current, the redox reaction should be spontaneous $\Delta G^o \le 0$). That means $ E_{cell}^0 \gt 0$. Let's consider, your galvanic cell with $\ce {Al}$ and $\ce {Cu}$:
Write both reduction half-reactions:

For $\ce{Cu}$: $\ce {Cu^{2+} + 2e- <=> Cu} \pu \space\space {E_{\ce{Cu^{2+}/Cu}}^0 = \pu{+0.34 V}}\tag{1}$
For Al: $\ce {Al^{3+} + 3e- <=> Al} \pu \space\space {E_{\ce{Al^{3+}/Al}}^0 = \pu{-1.66 V}}\tag{2}$

Now you may see why you needed to switch the equation (2). If you switched the half-reaction of $\ce {Al}$, the full redox equation would have a positive $E_{\text{cell}}^0$ value.

When you reversed the equation (2): $\ce {Al <=> Al^{3+} + 3e-} {E_{\ce{Al^{3+}/Al}}^0 = \pu{+1.66V}}$ $\tag{3}$
Thus, total redox equation would be:
$\ce {3Cu^{2+} + 2Al -> 3Cu + 2Al^{3+}} {E_{\text{cell}}^0 = \pu{+2.00 V}}$

Note that followings:
(A) this would have been your voltmeter reading if you have used the standard conditions.  Actual cell potentials can be calculated by using the Nernst equation to compare your findings, if you know the solution concentrations of half-cells ($\ce {[Cu^{2+}]}$ & $\ce {[Al^{3+}]}$), and the temperature of which you performed the experiment.
(B) In equation (2), $E^0$ of $\ce {Al}$ have more negative value compared to that of $\ce {Cu}$. That means, $\ce {Cu}$ has a greater electronegativity (liking to keep electrons or pulling electron, comparatively speaking) than $\ce {Al}$, thus $\ce {Cu}$ tends to steal electrons from $\ce {Al}$ or in other words, $\ce {Cu}$ tends to oxidize $\ce {Al}$. Hence, this is an one way reaction (favoring one direction of the reaction; see total redox equation).
For more info, read the Wikipedia article on Galvanic cell.


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Why does Hydrogen Peroxide decompose?
I understand that Hydrogen Peroxide decomposes into oxygen and water over time. But why exactly does it do this? Could someone give me an explanation down to the molecular level as to why and how this happens. And also, why does MnO2 catalyse this.

https://en.wikipedia.org/wiki/Hydrogen_peroxide#Decomposition

For the your second question see https://chemistry.stackexchange.com/questions/7759/reaction-intermediates-of-mno2-catalyzed-h2o2-decomposition-reaction

One could say that hydrogen peroxide decomposes because it is unstable. But that would be cheating because this is simply restating the fact that we observe it to decompose in fancier language.
We could use some statistics about the reaction, like stating that the decomposition is thermodynamically favourable (this doesn't explain why a catalyst works but it helps to explain why the reaction happens). But, again, this is just a sophisticated way to repackage the observed fact that the reaction is observed to happen.
However, if we think what is happening at the molecular level, we can get some extra insight.
All molecules above absolute zero have some kinetic energy (that is they are bouncing around and banging into each other with a range of speeds, some slow, some fast). Temperature summarises the average amount of energy present in the compound (but the individual molecules have a range of energies, some higher, some slower than the average).
In a solution of hydrogen peroxide (or in the pure liquid, though I don't recommend ever handling the pure liquid) the individual molecules will be bouncing off each other a lot. Each collision will exchange some energy between the colliding molecules. Sometimes a Collision will involve enough energy that one or more of the O-O or OH bonds will rupture. This will generate some reactive species. Then, sometimes, the reactive species will reform in a way that causes water and oxygen to be formed (the details of the reaction are probably fairly complicated). The products oxygen and water have much stronger bonds than the hydrogen peroxide (or at least eh O-O bond in it) so the probability that they will react to reform hydrogen peroxide is very low.
The probability that peroxide will decompose this way is still pretty low, though, at least in dilute and very pure water solutions. But many substances (including some present in slightly impure water) will catalyse the reaction. The presence of manganese dioxide, for example, provides a different route to the same products (that is pretty much the definition of a catalyst). If a peroxide molecule bumps into a manganese dioxide surface, there is a much higher probability that a reaction will happen than if it hits another peroxide molecule (it probably transfers an electron from the manganese oxidising it and this can be very fast and also can precipitate more reactions that break up the peroxide and, ultimately, yields water, oxygen and leaves manganese dioxide). Human blood contains an enzyme that is designed to do the same job as peroxides and related active species are generated by other biological reactions and you really, really don't want them to build up in the blood. Catalase uses oxidation reactions involving iron atoms to do an even better job of facilitating the reaction than manganese dioxide.
So the detail of the reaction is all about molecules bumping into each other and causing bonds to break and reform in a different way. The overall reaction releases energy (which is to say that the new bonds formed are stronger then the ones that get broken and the reverse reactions are therefore a lot less likely to happen). Catalysts proved an easier route to the products than the reaction where peroxide molecules just bump into each other.

Transfering an electron to manganese is reducing it.

Clearly I can't tell the difference between to and from. Fixed. OTOH I have no idea of the detail of the actual reaction so this step might be wrong anyway. It has been studied, though, so somebody with access to JACS in 1947 can probably tell us more.


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Why does boiling oatmeal "grow"?
When cooking oatmeal (rolled oats, either long-cooking or instant) on the stove or in the microwave, when it reaches the boiling point, if left unchecked it will quickly "grow" and overflow the container in which it is cooking, even if it is a very large container.
A comparable amount of water boiling in the same pot or container would not do this.  It would boil, yes, but it would not "grow" and overflow the container.  I have not seen this happen with other grains.
Why does oatmeal do this?

Welcome to Chemistry.SE! Take the [tour] to familiarize yourself with the site!

Oatmeal swells when it absorbs water and is sticky enough to bind into a near-continuous mass. Water that is not actually absorbed into a grain is trapped by swollen soft, sticky goo. But it is still water, not chemically combined, and if you heat it to 100 C, it will volatilize with great expansion, stretching the goo as it does so, but the bubbles can't make it all the way out.
In a way, it's like a slow motion popcorn, or a speeded-up rising of yeast bread. You can do something similar with a marshmallow.
I found that if you exactly repeat a recipe, the microwave timer could be precisely set to swell the oatmeal to fill the bowl, and not overflow. Then you let it cool down and do it again.

That makes perfect sense.  Thank you.  Yes, I have discovered from doing it for many years how to cook it without boiling it over.  But always wondered why it behaved that way when other things didn't.  Awesome explanation!


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Why is ethanoic acid more acidic than phthalimide?
Why is ethanoic acid more acidic than phthalimide? $\mathrm{p}K_\mathrm{a}$ of phthalimide is 8.3 (source), while that of ethanoic acid is 4.76 (source).
Applying the logic found here would suggest the opposite. This is because both conjugate bases delocalize their negative charge, but while ethanoic acid has two oxygens to delocalize its charge among, phthalimide has two oxygens and a nitrogen (and a benzene ring); it seems that the quality of the delocalization is the same, but the quantity favors phthalimide (as the answers to the question put it). As for the other factors that affect acidity (inductive effects, hybridization, atom size/electronegativity, formal charge, etc.) none seem to come into play.

related https://chemistry.stackexchange.com/questions/10616/rationalizing-acidities-of-ammonium-ion-and-hydronium-ion https://chemistry.stackexchange.com/questions/19692/why-is-ammonia-so-much-more-basic-than-water

You seem to completely ignore fact that it's N-H bond which is breaking.

The deprotonated phthalimide anion is indeed highly delocalized, but the neutral molecule with the N-H proton reattached retains most of that delocalization.  Protonating acetate ion back to acetic acid causes a much greater loss of delocalization.

I don't get this. Even the neutral acetic acid molecule shows delocalisation, doesn't it?

Yes, that's another point I was thinking about, nitrogen donates it's electron density much more eagerly then oxygen, so energy gain between delocalisation of ftalimide and it's anion is lower.

I asked a TA about this, and they explained it the following way: in both cases, when the hydrogen is taken away, the "new" electrons go into an sp2 orbital. This is because the protonated forms of both molecules already have resonance, and therefore both already have filled p orbitals. The "new" sp2 electrons don't participate in resonance, so the electronegativity difference between nitrogen and oxygen makes the difference. Since oxygen is more electronegative, the deprotonated form of ethanoic acid is actually more stable.

Ok, can whoever downvoted please provide an alternative, better answer? Thank you.


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Tape versus 1922 silver coin
I have a 1922 silver half dollar with scotch tape on it.  The tape has been on it 50 years plus.  It’s dried out and totally adhered to the coin.  I’ve soaked it in lemon essential oil and googone but still a lot of tape left, adhered to coin.  Impossible to get off. What will melt tape but not damage silver coin?

If the coin has any numismatic value, cleaning it yourself will reduce it. If you just wish to use it as an ornament, let it soak a week in organic solvents such as methyl ethyl ketone (MEK) or xylene (a paint thinner) and then gently toothbrush it.

Odd. Is this a 1922 US half dollar? The walking liberty series was before and after 1922. But so far as I can tell no 1922 coins are known.

Out of curiosity could you provide a picture of the coin?

" soaked it in ...  googone" & "What will ... not damage silver coin?" --> too late.  Coin is damaged - from a collecting point-of view.

Circulated Coin
I'll assume that the coin is circulated so a very gentle cleaning won't be a problem. You want to use some sort of organic solvent to loosen the glue, then very gently rub the residue off. You don't want to use any sort of silverware polish or anything abrasive when rubbing the coin.
Good solvents might be olive oil or nail polish remover. After the organic solvent wash the coin with dish washing liquid, rinse with plain water,  then dry by flipping coin over on a dry cloth or paper towel.
Uncirculated Coin
If the coin is uncirculated then you would want to use an ultrasonic bath with an organic solvent to prevent scratches on the coin.
Two tone coin after cleaning
Because the surface of the coin would have been sealed  where the tape was, it is possible that the coin will have a different color where the tape was. If that is the case then you might need to use a coin dip to get the surface looking even again. Unfortunately this will leave the surface shiny which screams that the coin has been cleaned. The coin will dull over time but never really develop a natural patina again since a truly natural patina also depends on wear.

Have you tried olive oil? Smooth over surface of coin generously, leave overnight on a rag inside of plate to prevent seeping to anything you don't want oil on. The following day, use a microfiber rag to buff off the tape.
Add a little more oil sparingly if needed while buffing.
I had tape and adhesives from my old military bags get stuck to everything inside over a course of a decade; metals, plastics, everything. Olive oil did the trick soaking overnight and buffing off in the morning.

olive oil goes rancid from exposure to oxygen -- turning it into a (mild) acid. While it will help clean some oxide off your coin, one might not want to do that, since it might etch the coin and leave marks.

As MaxW noted, and as confirmed by any 'Red Book', there are no 1922 Walking Liberty half dollar coins. So you either have 1) a date typo or 2) the coin is worn enough to make the date hard to read. In the latter case, the coin may be below G-4 (Good) condition. If it is a 1921 (P, D or S mint) Walker, it is (or was: see below) worth well over spot silver value, e.g., the 1921 from the Philadelphia mint is worth 180 USD in G-4 condition (as per my 2010 Red Book). The 1921-S is the least valuable: 50 USD for a specimen in G-4 condition.  This is more or less: estimates change and a coin is really only worth what you can actually sell it for, which is typically below the Red Book listing.
As for cleaning, this is an absolute no-no unless 1) the coin is of sufficiently high numismatic rarity/value/interest and 2) the cleaning would have to be done by numismatic professionals, e.g., PCGS or NGC. Amateur cleaning causes coin values to drop like lead balloons full of radon. Could you post a photo of the coin, please?
If you ultimately go with buffing, a soft cotton cloth (or Dremel buffing bit!) plus Mothers Mag & Aluminum Polish will shine it up to ludicrous brightness.


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The difference between elementary and concerted reactions
In the IUPAC Gold Book:

elementary reactionA  reaction  for  which  no  reaction  intermediates  have  been  detected  or  need  to  be  postulated  in  order to  describe  the  chemical  reaction  on  a  molecular  scale.  An  elementary  reaction  is  assumed  to  occur in a single step and to pass through a single  transition state.
concerted reaction A  single-step  reaction  through  which  reactants  are  directly  transformed  into  products,  i.e.,  without involvement of any intermediates.

What is the difference between them?

A concerted reaction is elementary. The reverse is not necessarily true.

The difference between SN1 and SN2 reaction mechanisms illustrates this quite well.
An SN1 type of reaction can be decomposed into two elementary steps:

formation of the carbocation
substitution of the nucleophile

The rate determining step is the formation of the carbocation, but we do have two distinguishable steps and a reaction intermediate - this is where the other mechanism comes in.
Now, when SN1 is not possible due to - for example - the carbon atom in question being primary, we can have a concerted mechanism - both of the steps mentioned before occur simultaneously, thus avoiding the energetically unstable primary carbocation.
This process can not be broken down into single steps, as it is necessary that the formation and the substitution occur simultaneously. Also, there is no reaction intermediate - hence it is a elementary reaction, where two things happen in a 'concerted' way in a single step.
Also, the experimental rate law will show that the rds is different and the rate depends on both reagents.

I might be missing something, but I don't understand how this answers the question.

The language here is weird. We generally refer to the first concept as an "elementary step." By definition, this is "concerted," in the sense that all of the pieces required for this step take place at the same time.
An elementary step (except for weird barrierless cases) should correspond to a single transition state in a reaction coordinate, and therefore, it links two adjacent intermediates (or reactant/product) in the reaction.
A reaction is a process composed of one or more elementary steps. Generally if two things need to happen in a reaction (e.g., bond-forming or bond-breaking), we can ask if these happen via a single elementary step or via multiple steps. If a single step, then we call the corresponding mechanism "concerted." Mechanistic studies help to distinguish and rule out different possibilities.
For example, we think that a simple Diels-Alder is concerted since a multiple-step reaction, where the two bonds are formed in different steps, would be expected to give some mixture of stereochemistry, but we find that Diels-Alder stereochemical outcome is strictly determined by the stereochemistry of the reactants. Here, the assumption is that bond rotation is generally fast relative to elementary steps.
On the other hand, if you study some edge cases of Diels-Alder and find products that are inconsistent with a single step where all bond formation and bond breaking take place at the same time (link), you start to think about how the reaction might take place over multiple steps, i.e., a reaction via a non-concerted mechanism. These reactions would necessarily go through an intermediate where not all of the desired bond-breaking and bond-forming has taken place. If you're clever, you might be able to observe / trap this intermediate to support your hypothesis.


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Picric Acid from Salicylic Acid
When Salicylic Acid is nitrated, the end $\ce{COOH}$ group is getting knocked out. How should I reason the knocking. Is it right to say, since $\ce{HCOOH}$ is more acidic than $\ce{HOH}$, hence, $\ce{COO-}$ will be a weaker base than $\ce{OH-}$. And so, a weaker Base being a good leaving group, the $\ce{COOH}$ gets knocked out than $\ce{OH}$ group on the ring?
Or there is other reason for the knocking off of $\ce{COO-}$ group?

Substitutions in benzene rings occur via electrophilic aromatic substitution reactions. Basically, one electrophile substitutes another electrophile. The feasibility of the reaction largely depends on the electrophilicity of the incoming electrophile, but also on the leaving ability of the departing electrophile.
In ordinary electrophilic substitution of unsubstituted benzene, the leaving group is $\ce{H+}$. This is basically the substituent (in this case $\ce{-H}$) having lost an electron. If suppose, the electrophile were to attack the carbon substituted with a carboxylate functional group ($\ce{-COO-}$), the leaving group would be $\ce{CO2}$, which is nothing but carbon dioxide.
You must be aware of how stable carbon dioxide is. It has two strong $\ce{C=O}$ bonds and is the product formed from various combustion and decompress reactions as well as a respiratory by-product. This means that carbon dioxide is an excellent leaving group.
In addition, the carboxylate group is located at ortho position to the hydroxy group. Considering the fact that hydroxy group is an ortho/para directing group, this further supports substitution at $\ce{COO-}$ substituted carbon.


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Voltage applied when charging a galvanic cell

A working galvanic cell is constructed using an iron electrode in an iron(II) nitrate solution and a
  silver electrode in a silver(I) nitrate solution. When the cell runs down, it can be charged up again.
  Which of the following scenarios will occur during the charging process?
$\ce{[Eº(Ag+(aq)/Ag(s)) = +0.80 V and Eº(Fe^2+(aq)/Fe(s) = –0.41 V]}$
A. A minimum of 1.21 volts is applied to the cell and electrons move from the silver electrode to the iron electrode.
B. A minimum of 1.21 volts is applied to the cell and electrons move from the iron electrode to the
  silver electrode.
C. A minimum of 0.39 volts is applied to the cell and electrons move from the silver electrode to
  the iron electrode.
D. A minimum of 0.39 volts is applied to the cell and electrons move from the iron electrode to the
  silver electrode.
E. A minimum of 2.01 volts is applied to the cell and electrons move from the silver electrode to
  the iron electrode.

I know reduction of silver is spontaneous and oxidation of iron is spontaneous from the values given, so reduction occurs at the silver electrode and oxidation occurs at the iron electrode, so electrons flow from iron to silver.
In the reverse reaction, electrons will now flow from silver to iron. However, I'm unsure about the stoichiometry of the electron transfer occuring during charging:
$\ce{2Ag -> 2Ag+ + 2e-       -1.60V}$
$\ce{Fe^2+ + 2e- -> Fe    -0.41V}$
I multiplied the oxidation of silver by 2, as two electrons must be supplied for iron to reduce, hence getting -2.01V, and answer E. However, the solutions state A, which suggests the oxidation of silver was no multiplied by 2. Why is this?

Voltage is how much push is needed to move the electron, and is the same whether for one or one trillion electrons. Electric charge, measured in Coulombs, is what changes when more electrons are moved.
Your reasoning that moving twice the charge doubles something is correct: instead of ~1.6 × 10−19 Coulomb for one charge, it's twice that, ~3.2 × 10−19 Coulomb for two of them (no discount for large charges, but I hope this kept your interest).


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Meaning of Molecular Oxygen
Does molecular oxygen just mean the oxygen molecule? Or could it refer to a certain state of the molecule (eg. gas or aqueous)?

Molecular Oxygen means that it's bonded to another Oxygen (O2). Atomic oxygen would just give you O.


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Amonium perchlorate
I want to make amonium perchlorate, having amonium chlorate. If I have a solution of sodium chlorate and I add amonium chlorate, will precipitate into amonium perchlorate? Or it is with a solution of sodium perchlorate?

Chlorate won't magically transform to perchlorate.

having amonium chlorate | And you are still in one piece? How that?

Ammonium chlorate becomes nonexplosive when "ammonium" is spelled with only two rather than three "m"s.  -1.

Why would you want the sodium perchlorate?

Note: Ammonium chlorate is a unstable oxidizers and decomposes violently. Handle with care. Ammonium perchlorate is also known for its toxicity. You should have enough laboaratory experience for handling such chemicals.

If you think that
$$\ce{chlorate + chlorate -> perchlorate}$$
then you are wrong. There is no common ion effect occuring. You will need a perchlorate salt and an ammonium salt that have a  displacement reaction and precipitate ammonium perchlorate. This reaction will suffice
$$\ce{MClO4 + NH4Cl -> NH4ClO4 + MCl ~~~~~~~~~~~ M = Na,K}$$
Or even electrolysis of a perchlorate salt. According to this patent:

Ammonium perchlorate is generally produced by a series of steps which include the electrolysis of an aqueous sodium chloride solution
  to sodium chlorate, which is in turn electrolytically converted to
  sodium perchlorate. The sodium perchlorate is then reacted with
  ammonium and chloride ions to produce ammonium perchlorate and the
  by-product, sodium chloride.
reacting the sodium perchlorate with ammonium chloride
sodium perchlorate might be reacted with ammonia gas in the presence of carbon dioxide, to produce ammonium perchlorate. This
  alternative procedure has been generally unsuccessful because of the
  difiiculty of separating the resulting sodium bicarbonate from the
  product, ammonium perchlorate.

Process 2 is described in another patent. Read more to know the purification and the precautions needed.

... the reaction is carried out by mixing aqueous solutions of the
  reactants. The concentrations of the solutions are such that the
  resultant ammonium perchlorate precipitates out of the aqueous medium.
  The ammonium perchlorate crystals recovered therefrom contain 5% to by
  weight of the mother liquor. The product is usually contaminated with
  the Water soluble alkali salt, the byproduct of the reaction, in the
  mother liquor.

Note: direct electrolysis of ammonium chlorate would not yield ammonium perchlorate as there would be chlorate contamination. In fact, there should be no chlorate as reactants. (Here)


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Reactions between aqueous or solid LiOH
I've been working on a project to absorb CO$_2$ from fossil fuel power plants.  I found these reactions:
$$\ce{2Li(s) + 2H2O(l) → 2 LiOH(aq) + H2(g)} \tag{1} $$
$$\ce{2 LiOH(s) + CO_2(g) → Li2CO_3(s) + H2O(l)} \tag{2}$$
The first one returns an aqueous solution of $\ce{LiOH}$, and the second one requires solid $\ce{LiOH}$. I was wondering if it's possible to use the aqueous solution in the second equation, or does it have to be solid?  Why?

it would actually produce a faster reaction to have the $\ce{LiOH}$ in a aqueous solution rather than having it as a solid. all you would do is bubble the $\ce{CO2}$ through the solution and it will go cloudy. you've probably done a similar thing but with $\ce{Ca(OH)2}$ to form calcium carbonate as a test for $\ce{ CO2}$.

The problem, of course, is that the aqueous solution requires carrying a lot of water weight around to get the needed absorption capacity.  Using solid lithium hydroxide is more an engineering decision than a scientific one.

If the aqueous solution is used, does LiOH have to be the monohydrate?  Or is it possible to have aqueous, anhydrous LiOH?

well anhydrous and hydrated crystals are referring to water trapped within the crystal structure when the material is a solid. so when you dissolve it into water the struture of the lattice is broken down so it doesn't matter if it hydrated or not as any water trapped in the crystal structure will just become part of the solution.


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Reaction between aluminum oxide and SO3 in anhydrous environment
I am aware of amphoteric properties of aluminum oxide and that it can react with both strong acids and bases. I was recently asked a question whether amphoteric properties allow for reaction with oxides, which form strong acids when reacted with water, namely sulphur (VI) oxide. This question of course can be extended to oxides such as sodium oxide, which would form a strong base in water.
While I know that appropriate equations for such reactions can be easily written and balanced, I'd love to know whether such reactions can actually happen without intermediate reaction of oxides with water to form a respective strong acid/base.

If you consider the Lewis theory of acids and bases, oxides can display acidic or basic properties and react on their own, not necessarily in the presence of a solvent.

Theory or no theory, they will react.

The oxides which can dissolve in water to give a base, can react with the oxides which can dissolve in water to give acids, to form a salt, similar to an acid-base reaction. They can be thought as bases and acids respectively according to Lux-Flood concept of acid-base.
Here in this case, $\ce{Al_2O_3}$ is the anhydride of $\ce{Al(OH)_3}$, and $\ce{SO_3}$ is the anhydride of $\ce{H_2SO_4}$. So, they react to form the salt aluminium sulfate $(\ce{Al_2(SO_4)_3})$, as:
$$\ce{Al_2O_3 (base) + 3SO_3 (acid) -> Al_2(SO_4)_3 (salt)}$$
However, according to Lux-Flood definition of acid and base, acids are oxide acceptors, whereas, bases are oxide donors. Here, in this example, you can clearly see that $\ce{SO_3}$ accepts oxide $(\ce{O^2-})$ to form $\ce{SO_4^2-}$, and $\ce{Al_2O_3}$ donates oxide and forms $\ce{Al^3+}$, thus it is a base.  As, it is simply an acid base reaction, this can easily happen.
Some other examples of Lux-Flood acid-base reaction (spontaneous) are,
$$\ce{CaO (base) + SiO_2 (acid) -> CaSiO_3 (salt)}$$
$$\ce{MgO (base) + CO_2 (acid) -> MgCO_3 (salt)}$$
Thus, it is seen that, those oxides which can dissolve in water to form acids and bases, can actually react spontaneously, without water to form salts.


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Is the formula λ = h/mc wrong?
When we derive De Broglie's equation, we can do that from two ways. That's,
1st step:-
$E^2=p^2c^2+m_0^2c^4$
For a Photon $m_0=0$, So, we have ,
$E^2= p^2c^2$,i.e.
$E=pc$ -----(1)
$E=hc/λ$----(2) equating 1) with 2),
$pc=hc/λ$or,
$p=h/λ$or,
$λ=h/p \Rightarrow  \textbf{λ=h/mc}$
2nd step:-
$E=mc^2$----(1)
$E=hc/λ$----(2)
equating 1 with 2,
$mc^2 = hc/λ$
or,$λ=hc/mc^2 \Rightarrow \textbf{λ=h/mc}$
According to those ways, what's the really correct way to derive that? Is 2nd step wrong?

Could you please reformat your question so that the equations are more clear. If you need help with the syntax, check: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference?lq=1

For a photon, v=c, which makes both equations identical. Besides, for a photon, m=0, which also makes them both meaningless.

@Ivan Neretin, **for a photon , $m$(moving mass) is not equal to $0$. It is the rest mass, $m_0$ =$0$. So, we have $m= \frac{m_0}{\sqrt{1- \frac{v^2}{c^2}}} $. So, as $v=c$, both the numerator and denominator tends to $o$. You can't just tell that this ratio also tends to $0$, because you have to take limits to tell that.

What makes you do the step E = m c^2 ? You open your question with the right formula. ...

If you define mass as the mass associated with the total energy of the photon, then Step 2 is correct.  Still (imho), Step 1 is superior because it is not encumbered by having to distinguish between "mass" of the object in motion and "rest mass".

Sorry, the answer of step 1 should be λ=h/mv. According to that answer what's really correct answer?


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What is an alloy called when it melts at the same temperature as something else?
What is an alloy called when it melts at the same temperature as something else? For example, if there is an alloy that melts at the same temperature as glass then it will be susceptible to enameling and it is called an XXX-alloy.
As I recall, the term begins with the letters "eu-".

I don't think there is a term for that. Besides, glass does not really have a well-defined melting temperature. Then again, there is a word "eutectic", but its meaning is quite different.

Do you mean melting at the same point as something else on the same (binary) phase diagram? Otherwise, many alloys can be made to melt at a wide range of temperatures (simple example: lead-tin solders come (came) in many different compositions with different melting points).

Isotectic substances are those which melt at the same temperature.

Ref.: Heycock, C. T.; Neville, F. H. "On the freezing points of triple alloys of gold, cadmium, and tin". J. Chem. Soc., Trans., 1891, 59, 936-966
But be aware that this term is very rarely used.


text:
What makes a diatomic bond strong?
What are the factors affecting the strength of bonds?
Ex:- why, we need less amount of energy to break Br-Br bond than Cl-Cl bond?

In general, halogens naturally exist as diatomic molecules, bonded to each other with a covalent bond (a chemical bond that involves the sharing of electron pairs between atoms). Three of main factors affecting the strength of covalent bond are the bond length, bond order, and the electronegativity difference between the two bonded atoms if they are different (e.g., $\ce {H-Cl}$, $\ce {H-Br}$, etc.).  Two atoms with equal electronegativity would make nonpolar covalent bond such as $\ce {Cl-Cl}$ and $\ce {Br-Br}$. Since two bonds in question are single bonds, the strength of covalent bond affected by bond order will not be discussed here. In general, though, keep in mind that triple bonds are stronger than double bonds and both of them are stronger than single bonds.
The higher bond strength in $\ce {Cl-Cl}$ than that of $\ce {Br-Br}$ can be explained by simply considering the difference in their bond lengths. You can easily find the gas phase bond lengths of $\ce {Cl-Cl}$ and $\ce {Br-Br}$ are equal to $\pu {199 pm}$  and $\pu {228 pm}$, respectively by a quick search on bond lengths online [Halogens]. The respective bond energies for $\ce {Cl-Cl}$ and $\ce {Br-Br}$ bonds are recorded as $\pu {243 kJ/mol}$ and $\pu {193 kJ/mol}$, respectively.
Halogens

the strength of bond has a close link to atomic radii and the shielding affect.
in the example of halogens, a Br-Br bond is weaker than a Cl-Cl bond as  each bromine atom has an extra shell of electrons meaning that there is less of an electrostatic attraction between the pair of bonding electrons and the nuclei of the molecules. therefore, the chlorine is stronger as there's a stronger attraction between the bond pair of electrons and the two nuclei.