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Why do so many biochemical reactions require enzymes?
Studying biochemistry as part of the Great Courses, I am struck that all 10 steps in glycolysis require an enzyme.
I’d have thought that evolution would have selected for a “simpler” pathway with fewer enzyme-catalyzed steps. If a reaction A + B requires an enzyme, three molecules have to be in the same compartment and collide instead of two.
My hunch is that having enzyme-catalyzed reactions instead of more reactive molecules reacting without catalyst gives cells the ability to control their metabolism, making homeostasis and regulation possible.

Although it is more likely that 2 species will collide together as opposed to 3, the enzyme still helps. Enzymes are, if you go back to their definition, biological catalysts. They provide an alternate reaction route for the substrates, which has a lower activation energy. The increase in probability of reaction upon formation of an enzyme-substrate complex substantially outweighs the fact that 2 species are more likely to collide than 3. Without enzymes, digesting a meal would take weeks. Thankfully we can do it in hours.

Control of the reactions may matter a lot. Many biological pathways are not just doing uncontrolled reactions but are regulating their rate and enzymes are complex machines that allow that, not just catalysts.

Arevmelikyan - well I know what enzymes do, it’s just seems more than a little coincidental that 10/10 steps in glycolysis require one.

Martin - Random collisions and the evolution of enzymes for a specific job aren’t mutually exclusive.  I don’t even really see how they are related.  The molecules in the cell are still all moving about randomly, or at least chaotically.

@Mike Don´t take it badly. I find the community here refeshingly honest and scientific. Most of the times. Sometimes you land on the bad side, and it´s sometimes not your fault, but the others have got up on the wrong foot in the morning. ;)

I voted to reopen. I feel it is more focused than e.g. https://chemistry.stackexchange.com/questions/72712/if-aliens-lived-in-a-hydrogen-or-any-flammable-gas-based-atmosphere-would-the

Another, maybe more relevant question to compare with (also asking a why question about glycolysis): https://chemistry.stackexchange.com/a/98573

@Jan - I would love to hear your full-length answer take on this, and I hope there will be more votes to reopen. Meanwhile, I have added to my answer even though the question is closed (which seems weird but is allowed).

@KarstenTheis Thank you for pinging me. My answer may not be my best but there was a lot of stuff that I wanted to include.

all 10 steps in glycolysis require an enzyme

We count it as 10 steps because there are 10 enzymes. There might be other steps that are fast (like a product acting as an acid or base) that we don't count because they don't need an enzyme.

I’d have thought that evolution would have selected for “simpler” pathways.

What is simple is that everything is happening in the same reaction vessel, the cytosol (well, there are more compartments, but just a handful). This is different from the organic chemist, who works with very reactive substances in pure form, carefully mixed in separate reaction vessels with distinct solvents refluxed at distinct temperatures for hours.

Is the very fact that enzymes are required make life more probable since the cells ability to control their quantities make homeostasis and regulation possible?

Yes, that is true. It also allows for a web of reactions rather than a single pathway (you can funnel fructose or galactose into glycolysis, and you can turn pyruvate into lactate instead of decarboxylating it to carbon dioxide and acetic acid).
Stable molecules require catalysts
One requirement for life as we understand it is a way to store, replicate and access genetic information. In our living world, that requirement is met by having nucleic acids. Under physiological conditions, RNA is fairly stable (good enough for a virus) and DNA is very stable (to the degree that sometimes DNA sample from extinct organisms preserve genetic information to this day). If these molecules do not form or decay on their own, there must be enzymes to act on them.
Granted, there could be other pockets of metabolism that work without catalysis. The other requirement, though, is to capture free energy to drive reactions away from equilibrium (e.g. to make the carriers of genetic information mentioned above). This is done through coupled reactions, where e.g. the free energy from food being oxidized (with oxygen as terminal electron acceptor) is captured in proton gradients across membranes and in the phosphorylation of ADP to obtain ATP. If any of the reactions going toward equilibrium would occur without catalysis, the free energy would be lost instead of captured by a coupled enzyme reaction.
General thoughts on complexity
In any complex system (government, computer chip, industrial process etc.), as the complexity grows, the number of steps that seem inefficient or over-complicated will also grow. Many of these steps might not make sense from the present view, but were put in place (or evolved) earlier to meet a historic need. Once in place in a system that has to run continuously, there is often no path to swap it out for a more efficient way of doing things.
One example from biology is the human eye, where the nerve cells run through the optical path, giving rise to the blind spot. The octopus eye "lucked out" and evolved in a way that makes more sense, but human eye sight is just fine (there is a "software patch" in the brain to fill in the blind spot):

Source: https://external-preview.redd.it/xWn0rxCSyu9TrnXozPrsF3aq2_uILbUEJwzBv87TZo4.jpg?auto=webp&s=7a1bccf554ec19a813f9da4466c114ac84c426b7

Enzymes act as catalysts to increase the rate of reactions. In the absence of enzymes, most biological reactions would be incredibly slow, in some cases many many orders of magnitude slower than when catalyzed.
Given the large number of biological reactions that require enzymes, it seems that there might be an advantage to reactions that are slow in the absence of a catalyst, and that indeed seems to be the case. If the uncatalyzed reaction is very slow, an organism can control when the reaction happens by providing an enzyme at the desired time. When it is better for that reaction not to happen, the removal of the enzyme prevents the reaction from happening except at its slow uncatalyzed rate.
Since enzymes can be synthesized and destroyed in most organisms with a high degree of temporal and spatial control, the use of enzyme-catalyzed reactions provides cells with fairly good spatial and temporal control over metabolic reactions, thus optimizing the overall metabolism.

We speak of reactions that

build complex molecules
or break complex molecules at a specific position
they must happen at constant 37°C
all of which must be rather low energy
or be very well controlled, like photosynthesis
no fancy other reagents are available
many would never happen at all without stringent control
and/or the result could only be achieved in numerous reaction steps
you don´t want side reactions
cells are full of different compounds, all of which must be rather unreactive, or you´d get complete chaos
you want the reaction to happen where you want it, and start and stop it
you cannot control location of the educts, they diffuse in (and out of) your cell, because you have very little in the way of piping and valves

There is only one one way to get that: the self regulating production of self regulating catalysts. Enzymes. And a versatile energy source: ATP.

Interesting point about the requirement that the compounds be relatively non-reactive lest all hell breaks loose.  Thanks.

Life is all about a little compartment controlling the (bio-) chemistry inside it. From the moment the first cellular predecessors started forming in whatever the medium they had was, they had an ‘inside’ and an ‘outside’ and life evolved by taking control of the ‘inside’, protecting it against the ‘outside’ and then by some means, propagating the ‘inside’ to generate a ‘daughter inside’. They key concepts are self-assembly, autocatalysis and self-replication. To briefly introduce them:

Self-assembly describes the spontaneous assembly of a more complex structure from simple starting materials by diffusion. For example, merely adding a zinc salt of a weak acid to a solution of protonated porphyrins will produce the corresponding zincporphyrin. If the porphyrin has an amino function at an appropriate position, these porphyrins will go a step further and dimerise spontaneously in solution. A more simple example that is frequently introduced in high school or undergrad courses are the hydrogen-bonded carboxylic acid dimers that form in aprotic solvents.
Autocatalysis refers to a reaction where the reaction product is able to catalyse its own formation unspecifically. For example, consider the decomposition of cellulose acetate into cellulose and acetic acid. This reaction is catalysed by a proton source (acids), meaning that the more cellulose acetate degrades the more protons in form of acetic acid will be available and the reaction will progress faster. It is important to note that an autocatalytic reaction is typically unspecific: by adding sufficient amounts of a different acid (e.g. HCl), the reaction is also sped up and no rate increase is observable.
Self-replication is probably the most important of the three: it describes the creation of an exact copy of a template from available starting materials. Usually, a self-replicating reaction will also be autocatalytic because with each synthesis a new template is generated allowing for the synthesis of additional individual units. Probably the most well-known example is the replication of DNA where a parent strand serves as the template for an exact copy which becomes the daughter strand. Note that enzymes are perfectly able to synthesise DNA without template—however, unspecifically.

While all three concepts are essential to developing life, the third one is the one that defines life and the possibility of evolution as we know it, because if you cannot generate an exact copy of yourself if you are not passing on yourself to a daughter generation. Furthermore, it becomes essential to restrict any possible replication to the desired self-replication. This is the beginning of the complex relationship between DNA, RNA and proteins/enzymes.
Probably the initial proto-cells depended heavily on the uptake of reactive predecessors to support self-replication from the outside broth but that isn’t a sustainable long-term solution. Thus, as time passed more and more reactions were taken under control to support the proto-cell in what it was doing, more and more simple enzymes developed until finally something resulted that could live on an energy supply and a carbon/nitrogen source alone—the first cells. These have a striking advantage over proto-cells that require the uptake supplies from outside so they probably rapidly prevailed and evolved further.
With every further step in evolution, the machinery became more sophisticated and enzymes became better and more specialised at their job. But more specialised also means more complex and more complex means easier to destroy. Thus, with the intracellular machinery becoming more and more complex it became more and more necessary to regulate every single little detail of what happens within a cell lest there be some way this could be harmful to the cells survival.
For example, a lot of reactions require acidic protons to be supplied to a certain functional group of the substrate. But for a complex cell it would be devastating to generate excess acidic protons and hope they find their targets because in another corner of the same cell an acid-sensitive substrate could be in the making or a basic environment is required to abstract a different proton. In addition, even if there were only one substrate that could react with this acidic proton, that substrate can still potentially react in more than one way with an undesirable pathway being more likely. All this is inhibited if the acidic proton is kept tightly in the centre of a catalytic pocket surrounded by the enzyme’s protein backbone, out of harm’s way.
All of this has lead to quite the opposite scenario from the one you paint: typically, a cell will have an enzyme available for every single reaction it wants to perform and leave nothing to chance or random collisions. Every single reaction that was left outside of an enzyme’s control has the potential to disastrous consequences and the cell is really better off limiting these as much as it can.
To exemplify this, I like to point to one of my favourite examples: Superoxide dismutase. This is an enzyme that catalyses the disproportionation of superoxide $\ce{O2^.-}$ generating diatomic oxygen and hydrogen peroxide. Superoxide is a highly reactive oxygen species that is generated in one-electron reduction of diatomic oxygen. This decomposition reaction is extremely rapid even in the absence of an enzyme: superoxide will disproportionate to the same products so rapidly that it is used as a chemical oxygen generator. However, even that rapid reaction is not fast enough for cells as even the short half-life of superoxide in water might cause irrecoverable damage. Thus, highly complex enzymes evolved and most species have a handful of them to combat superoxides. Furthermore, superoxide dismutases are remarkable in that they have one of the largest catalytic effeciencies (measured as the catalysed reaction’s rate constant divided by the uncatalysed reaction’s rate constant) $k_\text{cat}/k_\text{background} \approx \pu{7e9 M^-1 s^-1}$ – for a reaction whose background rate is already estimated at $k_\text{background} \approx \pu{10^5 M^-1 s^-1}$!
The product of superoxide dismutase, peroxide, is another highly harmful compound that has a high tendency to oxidise anything in proximity so it is also immediately broken down by peroxidases—but at least that doesn’t come as an equally big surprise since peroxide solutions are actually shelf-stable so peroxide has the far more obvious potential of harming half the cell.
Having considered all this dire need for controlling what actually happens in the cell’s inside, it should be less of a surprise that enzymes perform one step only and that each individual step of a biochemical pathway will have an associated enzyme—although it should be mentioned that if a reaction is close to equilibrium then enzymes will often indiscriminately catalyse both the forward and the backward reaction.
In the previous paragraphs I have not at all spoken of the idea of regulation: that a cell can turn on or turn off biochemical pathways as it sees fit. Bluntly, you don’t want to be breaking down your precious storage of glucose-6-phosphate (not actually stored as glucose-6-phosphate) into carbon dioxide and ATP if your cell is already oversaturated with ATP. The multitude of biochemical reactions catalysed by one specific enzyme each allows simple up and down regulation of individual pathways by means of inhibition or enhancing individual enzymes—another immediate advantage for the cell.
Now you might still want to argue why enzymes aren’t collected together to perform multiple steps at once or one after the other. Well—they are. Some of the most interesting molecular assembly lines are polyketide synthases. These multi-enzyme complexes typically consist of individual subunit domains which perform any of these steps: $\ce{C-C}$ bond formation in an aldol reaction typically using malonyl-CoA or methylmalonyl-CoA as a chain extender (ketosynthase), reduction of the resulting ketone (ketoreductase), elimination of water to create a double bond (dehydratase) and reduction of the double bond to give a saturated product (enoylreductase). Among these domains, there is an acyl-carrier protein (ACP) which serves as a robot arm transferring the growing chain from one domain sub-enzyme to the next. After having passed a domain, an acyl transferase will move the growing molecule on to the next one.
Multi-enzyme complexes are of course more difficult to synthesise and maintain as they are agglomerates of multiple individual enzymes. Thus, these complexes will typically only be assemblied where it is required, where an intermediate cannot be liberated or where a liberated intermediate cannot be sufficiently re-recognised.

I'm glad the question was reopened to allow you to post this thoughtful and thorough answer!


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How ro interpret the given cell representation?
I have encountered cell emf having a reactant and product in the subscript. But this is the first timeI have witnessed 3 elements/compounds/ ions. How do i interpret this and what would be the solution? If possible do give the reaction equation.
EDIT:- This is what I tried:-

I made out the first part of question but stuck at comparing whose EMF will be higher.

Please read https://en.wikipedia.org/wiki/Silver_chloride_electrode and if there are any questions remaining, please [edit] you question to be more specific.

Half reactions
The species in the subscript describe the half reaction going on at the electrode. It contains all the species except for the electrons. You can figure out from the oxidation states whether species are on the reactant or product side. For example,
$$E_\ce{Ag/Ag+}:\ \  \ce{Ag+(aq) + e- -> Ag(s)}$$
Your examples are a bit more complicated. The first has species $\ce{Ag, AgCl, Cl-}$. The half reaction for that is:
$$\ce{AgCl(s) + e- -> Ag(s) + Cl-(aq)}$$
You should be able to figure out the one for bromide. The one with iodide is strange (not balanced). Iodide would react with AgCl:
$$\ce{AgCl + I-(aq) -> AgI + Cl-(aq)}$$
So then we could just write:
$$\ce{AgI(s) + e- -> Ag(s) + I-(aq)}$$
But in that case, it is not clear what the standard set of concentrations would be. Maybe a typo?
Cell potentials
The (silver/silver cation) standard potential is for silver cations at 1 M concentration (you would have to take a soluble silver salt like silver nitrate). The (silver/silver halide) standard potential is for the halide ion at 1 M concentration, resulting in a much lower concentration of silver cation. All you have to do is calculate that concentration, and get the (silver/silver cation) potential at that concentration from the standard potential.
For example for chloride, the silver cation activity of a saturated solution (with chloride activity of 1 in solution - I am using activities so the units work out) is:
$$a_\ce{Ag+} = \frac{K_\ce{AgCl}}{a_\ce{Cl-}} = K_\ce{AgCl} = \pu{1e-10}$$
For the half reaction $\ce{Ag+(aq) + e- -> Ag(s)}$, the potential is:
$$E^\circ_\ce{Ag/AgCl/Cl-} = E^\circ_\ce{Ag/Ag+} - \frac{RT}{zF} \ln{Q}$$
With Q = 1/$a_\ce{Ag+} = 1/K_\ce{AgCl}$, the potential comes out as:
$$ E^\circ_\ce{Ag/AgCl/Cl-} = E^\circ_\ce{Ag/Ag+} + \frac{RT}{zF} \ln{K_\ce{AgCl}} = E^\circ_\ce{Ag/Ag+} - \pu{0.59 V}$$
So the reduction potential is lower (by 0.59 V) because the concentration of silver cation in the presence of 1 M chloride is lower (by a factor of 10 billion). It is even lower for 1 M bromide, and the lowest for 1 M iodide because of the respective solubility products.

[OP] ... $E_3$ ...

In the OP's answer, the dissociation reaction is denoted by $E_3$, but it is not a redox reaction, so it should not have a reduction potential. You could solve this problem by calculating the free energy for each step (with a common second half reaction as a reference), and then going back to cell potentials. However, I think the path to a solution given here is easier to follow.
It is odd that under standard conditions, not all concentration are 1 M. Because the aqueous silver ions are not part of the half reaction for the saturated silver halides, however, this is just fine.

Thanks i now got the equation. But still I am stuck at comparing emfs as asked in question. I have added the try.

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Trouble understanding equation from density functional theory
In the review by Oliviera et. al., I do not understand how the authors get equation 14:
\begin{equation}
E[\rho]=\sum_i^Mn_i\langle\Psi_i|-\frac{1}{2}\nabla^2+v_{ext}(\vec{r})+\frac{1}{2}\int\frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|}|\Psi_i\rangle+E_{xc}+\frac{1}{2}\sum_\beta^N\sum_{\beta\neq\alpha}^N\frac{Z_\alpha Z_\beta}{|\vec{R_\alpha}-\vec{R_\beta}|}\tag{14.}
\end{equation}
 To me the energy should be in this form:
$$E = \langle\psi|T + V_\mathrm{ext} + V_\mathrm{ee}|\psi\rangle$$
How did they manage to put some of term out of the bracket? I also have trouble retrieving (16) as they cite, are they using a Taylor expansion?
References:

Oliveira, A. F.; Seifert, G.; Heine, T.; Duarte, H. A. Density-functional based tight-binding: an approximate DFT method. J. Braz. Chem. Soc. 2009, 20 (7), 1193–1205. DOI: 10.1590/S0103-50532009000700002.

Could you please duplicate equations (14) and (16) in your question, as well as add some context and define all the variables in case the full-text article becomes unavailable?

I think they pull E_xc[\rho] out of bracket because it is a functional. V_ee is composed of coulombic and exchange interaction right, the coloumbic term can be calculated represented easily with the 1-RDM (?) operator, whereas the exchange is more complicated so they pulled it out for simplicity.

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What type of plastic (PET, HDPE etc.) is appropriate to store alcohol based hand sanitizer?
I’ve been tasked with distributing hand sanitizer to public safety staff from a bulk supply in gallon jugs. I need a portable squeeze bottle, but I’m having trouble determining what type of plastic (PET, HDPE etc.) is best to store this alcohol based product. I want to make sure to buy a squeeze bottle that will hold up. I read a product review that said silicone leak-proof travel bottles will break down with alcohol. Am I safe with anything but silicone? Researching with Google has not provided much in the way of answers for this scientific lay person.

Welcome to chemistry.se! I'm afraid these types of questions are not on topic for our site, please consult your (local) expert(s). Also please do not use slang and/or abbreviations in your posts.

There are quite a few chemical compatibility lists available from reputable sources, e.g. https://www.hmcpolymers.com/uploads/files/resources/hmc-pp-chemical-resistance.PDF (for PP), https://www.calpaclab.com/chemical-compatibility-charts (for HDPE/LDPE and a summary for others) and https://www.plasticsintl.com/chemical-resistance-chart (large summary table including PET). I know links aren't sufficient, but there's far too much data to reproduce here.
In short, PET, HDPE, LDPE, PP, PPS and PTFE are all suitable for storage of ethanol or isopropanol, the two most common hand sanitizing agents. Most travel cosmetics bottles or reused cosmetic/household bottles will be one of these. You should be able to check at least what the main active component of your hand sanitiser is and what the material of your bottles is, even if you're reusing bottles, and look them up in one of these lists. Check for synonyms and categories (ethanol/ethyl alcohol/aliphatic alcohols) if you can't see what you're looking for. If there are more ingredients listed on the label or SDS, you could check them, too, but I don't think it's likely that they would present other compatibility problems.
You may need to give slight consideration to factors besides chemical compatibility, such as temperature (esp. for LDPE) and possibly UV (for PET). However, for indoor room-temperature storage, again, almost anything will do.


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Finding the volume needed to make a new pH

In the titration of 20.00 mL of 0.500 M HCl by 0.500 M NaOH, calculate the volume of 0.500 M NaOH required to reach a pH of 2.0.

I have tried to change the 20.00 mL into the amount of substance (0.01 moles), then I turned the 2.0 pH into concentration (0.01 M). I then tried to find the total volume, which came out to be 1 L, which creates 980 mL of NaOH. Plugging it back in though doesn't give 2.0 pH. So, I was wondering if anyone knew how to find the volume, or if there was a formula that I could use to solve it.

The hydrogen ion concentration of 0.01 M is a great first step. Adding the NaOH does two things: 1) It adds hydroxide, which lowers the amount of hydrogen ion by reacting with it. 2) It adds water, which lowers the concentration of all solutes.

$$\ce{NaOH + HCl -> NaCl + H2O}$$
Let $x$ Litres be the amount of $\ce{NaOH}$ needed.
$$-\log[\ce{H+}] = 2$$
$$-\log\left(\frac{(0.5 \times 0.02 - x \times 0.5)}{(0.02 + x)}\right) = 2$$
$$x = \frac{49}{2550}$$
$$x \approx 0.0192 \; \text{(3 sig. fig.)}$$
Thus, $\pu{0.0192 L}$ of $\ce{NaOH}$ is needed.


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How much energy do you need to heat and melt a sample of ice from -5°C to 50°C?
A 67.8 g sample of ice initially at -5.00 oC is heated until the final temperature is 50.0 oC.  Calculate the amount of energy gained by the sample. ANSWER IN kJ for sig figs.
The specific heat capacity of ice is 2.09 J/g oC, the specific heat capacity of water is 4.184 J/g oC, the specific heat capacity of steam is 2.03 J/g oC, the ΔHfus = 3.35 x 10^5 J/kg, and the ΔHvap = 2.26 x 10^6 J/kg.

I need help with finding where you are stuck please.  Can you show what you've tried?

Knowing that

The mass is $m=67.8$ g;
The specific heat capacity of ice is $c_{\text{ice}}=2.09$ J/(g°C);
The specific heat capacity of water is $c_{\text{water}}=4.184$ J/(g°C);
The latent heat of fusion is $\Delta H_{\text{fusion}} = 3.35\cdot10^5$ J/kg$=335$ J/g.

To go from $T_1=-5$ °C to $T_2=0$ °C you need $c_{\text{ice}}\cdot m\cdot(T_2-T_1)=708.51$ J.
To transform from ice to water you need $\Delta H_{\text{fusion}}\cdot m=22713$ J.
To go from $T_2=0$ °C to $T_3=50$ °C you need $c_{\text{water}}\cdot m\cdot(T_3-T_2)=14183.76$ J.

So, in total you need $708.51$ J+$22713$ J+$7085.1$ J = $37605.27$ J $\approx 37.6$ kJ.

Hi, welcome to chem.SE. We do not as a community answer homework questions such as the one you have answered here.

Sorry, I just tried to help. Should I delete my answer ?

I would say, leave it for future students as a solved example. You can ask them next time if they tried to solve the problem or not. Thanks


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Total Dissolved Solids and Alkalinity
I need to calculate total dissolved solids (TDS) of various groundwater samples from chemical analyzes. I have concentrations of ions/molecules in mg/L. I’ve found various formulas to calculate TDS but the ions/molecules I have never match.  I don’t have a lot knowledge in chemistry.
Here are the three equations I have found. (As i said, I found many different equations and I don't know which one I should use):
Eq. 1 : TDS=Na(+) +K(+) +Ca(2+) +Mg(2+) + Fe^(2+) +HCO3(-) +Cl(-) +SO4(2-) +NO3(-)
Eq.2 : TDS=0.6*Alkalinity (as CaCO3) +Na(+) +K(+) +Ca(2+) +Mg(2+) +Fe(2+) +Cl(-) +SO4(2-) +NO3(-) [1]
Eq.3 : TDS=cations +anions +SiO2
Problems/questions I have:
•I don’t have the concentration of HCO3(-).
•I have the concentration of CaCO3. In natural waters, alkalinity (mg/L CaCO3) is HCO3(-)+CO3(2-) +OH(-) +H(+). At pH under 8.3, could I simply convert the concentration of CaCO3 in HCO3(-)? If yes and if the pH is over 8.3, is there a way to calculate the concentration of both HCO3- and CO3(2-)?
•I don’t have the concentration for NO3-. I only have the concentration for NO3-+NO2. Is there a way to isolate the concentration of NO3(-)?
•I don’t have the concentration for SiO2. I only have the concentration for Si. Could I simply use the concentration of SiO2?
•The pH of my water samples varies from 6 to 9.
•Also, I don't have the samples so I cannot titrate the water or measure electrical conductivity.
Thank you!
References:
[1] Federation, W. E., & American Public Health Association. (2005). Standard methods for the examination of water and wastewater. American Public Health Association (APHA): Washington, DC, USA.

I am sorry this is not the way it works. Do you remember the mathematical requirement that if we have 9 unknowns we need 9 independent linear equations. Nobody solves such equations for TDS. Here is the step-by-step process https://www.wikihow.com/Calculate-Total-Dissolved-Solids

Sorry, english is not my first language. I meant that I don't know which equation is better. Also, I don't have the water samples, so I cannot use electrical conductivity or titrate the water samples.

No problem with English and no need to say sorry for that. Without conductivity you $cannot$ measure TDS easily. None of the equations are useful or officially accepted ones. Could you clarify what is the context of your problem? Also clarify what information do you have.

Can you get hold of this book "Environmental Laboratory Exercises for Instrumental Analysis Frank M. Dunnivant". It will explain everything to you.

I would like to calculate TDS of groundwater samples to compare it to values of TDS measured in wetlands. Then, I would like to establish if there is a connection between groundwater and wetlands.
I foud the second equation in the Standard methods for the examination of water and wastewater so I think it is possible to calculate TDS from cations and anions. And thanks, I'll try to get the book!

In order to implement equation 2, you need titration data. I haven't seen this equation but it must have a charge balance.  By charge balance I mean the positive charges must equal the negative charges in number. Your reference is highly reliable. Please get hold of the book I suggested as well. It talks about mass and charge balance. Whoever gave you the data ask them if CaCO3 came from titration or not?

yes CaCO3 comes from titration! So I can use it to calculate the HCO3(-) concentration and CO3(2-) concentration. And yes, all the samples I have are balanced.

I think you can try the second equation and make sure you double check with experts before publishing it.


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How do I decide the electromeric effect on pent-2-ene?
There are two possible case
π electrons shifting to C2
In this case, nucleophile attacks C3
It is stabilized by +I effect of two ethyl groups
π electrons shifting to C3
In this case, nucleophile attacks C2
It is stabilized by +I effect of a methyl group and a propyl group.
How do I decide which one is more stable, and ultimately, which one is favoured?

Here what you should consider first is hyperconjugation, rather than inductive effect (a brief explanation by me).
You're right that $\ce{Nu-}$ will either attack on $\ce{C-3}$ or $\ce{C-2}$, and hence two carbocations would be formed. Now, we'll analyse the stability of these carbocations on the basis of hyperconjugation which is as follows,

carbocation at $\ce{C-2}$ will have a total of 5 HC structures
carbocation at $\ce{C-3}$ will have a total of 4 HC structures

Based on the above difference, it is easy to conclude that $\ce{Nu-}$ will attack on $\ce{C-2}$ carbon. Whereas based on inductive effect, those carbocations are nearly equally stable.

Note: HC is short for "hyperconjugative structures".

Oh thanks, but in my book it says that +I effect of c2h5>ch3 ,therefore electrons shifting to C2 is favoured more.any idea of what it's trying to say?

@DatBoi: your book had stated right order. It is bcoz of the fact that, ethyl group has two carbons, whereas methyl has only one. However, as you know that inductive effect isn't effective till next to neighbour C, hence the above diff. is near negligible. Also, in my this answer, I gave a brief explanation about the extent of stabilizing effect. So, any considerable diff. in hyperconjugation will overrule small diff. in inductive effect.

@YusufHasan: I think you've already noticed that I'd gave an explanation (short) on this in a previous answer. So, should I link that one in my post? it would be very effective, IMO.

Yusuf?! Who's that?

@DatBoi: he is Yusuf Hasan. Actually, he added some suggestion just after your comment, and then deleted it ahead of my comment.


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Is there any software that can do geometry optimization using machine learning?
DFT is a computational tool that is used in optimizing and calculating the electronic structure properties of molecules.
Are there any machine learning codes that can do something similar in a shorter time? Is there any software that does this?

I believe there was a NDDO-SEMO method that employs machine learning - you might want to try that out.

There is one Python package that I know: AMP (Atomistic Machine-learning Package). It is based on ASE (Atomic Simulation Environment), a homogeneous interface to a lot of computational chemistry packages.
Basically, after you obtain geometry trajectories through ASE (using any computational chemistry package you like), an approximation to the potential energy surface (PES) is machine-learned. It can be used as an ASE interface.
You can theoretically do anything ASE is capable of with this, from saddle point optimization to quantum dynamics.
I am not an expert in this field, but they seem to apply ideas from convolutional neural networks to molecular structures for describing the chemical environment surrounding atoms.
Their method is described in Khorshidi, A.; Peterson, A. A. Amp: A Modular Approach to Machine Learning in Atomistic Simulations. Computer Physics Communications 2016, 207, 310–324.

If you're interested in metal complexes, there's a free software, Python-based tool called molSimplify. From their site:

Geometry optimization with density functional theory (DFT), a general
  procedure to obtain the ground state structures of a complex, is
  computationally demanding in terms of time and can also easily fail.
  Two main failure modes are 1) the expected geometry cannot maintain
  stable during the DFT simulation (e.g., ligand dissociation) and 2)
  the electronic structure of the optimized geometry is bad, which
  indicates the system of study is out of the domain of applicability of
  DFT. Either case can only be identified after a simulation completes,
  leading to a massive waste of the computational resources (and your
  time!).
To address this challenge, we built machine learning models to
  classify the simulation outcomes and readily achieved a good
  performance on the out-of-sample test data.

It's really handy when you have dozens of structures to create. It gives a good initial geometry for further optimizations, avoiding boring work to assemble each one by hand and fiddling with bond angles. For more details, see their paper Duan, Chenru, et al. “Learning from Failure: Predicting Electronic Structure Calculation Outcomes with Machine Learning Models”. Journal of Chemical Theory and Computation, vol. 15, no 4, abril de 2019, p. 2331–45. ACS Publications, doi:10.1021/acs.jctc.9b00057.


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How does the cyclisation of the 3,6-dimethylocta-2,6-dien-4-yne occur?

The given molecule is treated with $\ce{H+}$ and an acid, that is all. I know that the alkene part reacts with the $\ce{H+}$ first, but have no idea on how to proceed further. I don't see how a cyclisation takes place as the final answer given is 3,4,5,6-tetramethyl-3-cyclohexen-1-one. Any help on how this happens?

I know you didn't draw this, but please don't draw your molecules like that, ever... the bond angles are all wrong!

Bent triple bond and inconsistent angles make me think you might want to find a better textbook. Also, please don't approve edits that make the quality of your question worse.

Also name of the final product should be 3,4,5,6-tetramethyl-2-cyclohexen-1-one (if that is the preferred one).

First add water to the triple bond a ketone would be formed as the molecule is symmetrical hydroxy group can be attached yo any of the two carbons after that it would undergo tautomerism to form a ketone. Now the second step cyclisation occurs as follows:

It seems that you just proposed a pathway to get to the answer provided in the question. I can see what you did with the aqueous acid and the tautomerism, but what about the reagents in the 2nd step? Why would they cause this cyclization, how do they operate here? I guess you should elaborate on that

The double bond is an internal neucleophile to attack at the  electrophilic site of the carbonyl compund  this triggers cyclisation then in the second step the addition of weak acids is to ensure that there would be no substitution in the compound (concentration of hydroxide in the water would be decreased)

That's not quite the reason, the acid catalyst is meant to activate the ketone for an intramolecular electrocyclization called the Nazarov cyclization reaction. You can check the mechanism here Double bonds are nucleophilic, but they don't just start attacking carbonyls in an alpha beta unsaturatec situation.. This is more due to the thermally allowed rotations


text:
Procedure vs. synthetic method?
I am genuinely confused on the difference between a procedure/flow chart and the synthetic method for a lab report and what exactly a synthetic method includes, and I can't find any information online. I would really appreciate the help if anyone has knowledge about it.
This is what the directions say, but it almost sounds the same as what a procedure would cover.

Thank you very much!

With respect to version 1 of your post: Please type out the text in the image so that it's accessible by search engines. Thanks.

Procedure/flow chart is a short, often diagrammatic, summary of a reaction sequence with reagents, solvents, temperature & yields. It is expected that a competent synthetic chemist would be familiar with the reactions shown and would be able to understand how the product was made.
Synthetic method contains all the details of the reaction. Sources and quantities of reagents and solvents, times and temperatures of reaction, workup and purification procedures, characterisation of product.


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Is there anything we lose by using Google Colab for DeepChem?
DeepChem does not run on windows, so I am considering to try Google Colab. Is there any key feature I would be missing by using Google Colab remotely from Windows rather than the original DeepChem on Linux?

Guys, instead of giving a brand new user 3 downvotes without writing a single comment, why don't you try editing the question to make it better? In essence it was an appropriate question for ChemSE, just could have benefited from a bit of improving touches.

When you are using Google Colab, it runs on Google's servers, so there is no such thing as Colab on Windows or Colab on Linux. Besides that, the most obvious problem is that the calculations using Colab have limited resources, CPU time, etc if you use a free account.

@Buttonwood I frankly do not understand why you answering this to me. One can try to run it on Windows, indeed.

@Greg So it was an misunderstanding by mine « it runs on Google's servers», ok.

@Greg since you say that "there is no such thing as Colab on Windows" I edited the question to say "remotely FROM Windows" ... without the caps lock on "from". This indicates that the user is starting from Windows, but remotely connecting to the Google servers. Is it better? Feel free to edit it further if you can!

For those who closed this, why don't you tell the brand new user, what details need to be added for clarity? It seems like quite a perfectly good question to me.

DeepChem's project web page indicates further documentation and access to the source files used.  Just click the «fork me on github» sign in the top right hand corner leads you to this page.  From the notes there, it becomes obvious that many tasks the program suite handles are managed with Python which you should install in first place (e.g., Python's Python, or Anaconda's Python) then offering a platform independent use of deepchem in Linux, Mac, or Windows (see this report).
Once you have this installed, you just have to solve the «must have» dependencies», in Python's Python with python package installer pip3 on the CLI
pip3 install joblib pandas sklearn tensorflow pillow deepchem
and maybe the ones listed as «soft requirements» such as RDKit.
A nice thing is that a package manager like pip will check if additional requirements have to be adjusted, and if so, will offer to adjust your system accordingly «on the fly».  Just be aware that you might need a considerable chunk free memory on permanent memory for the modules deepchem relies.  Don't forget the inofficial package (wheel) index for Python modules in Windows by Christopf Gohlke which may assist you in solving the dependencies, too.

I am not sure it answers the original question, but definitely a useful attempt to do so.


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What will be the major product of the reaction between (2R,3R)-2,3-dimethyloxirane and triphenylphosphine?
Problem

My Attempt

But answer is given B. Where did I go wrong?

First, note that with the mechanism you have drawn you should expect to get a mixture of cis and trans isomers. In the intermediate you drew, rotation about the C–C bond will be fast and will lead to loss of any stereochemical information that was previously there.
The only way to get exclusively trans-2-butene is for the reaction to occur in a single step without a discrete intermediate:

However, that's not the case. The mechanism of epoxide deoxygenation by triphenylphosphine isn't a concerted pathway, but is stepwise.
Triphenylphosphine opens the epoxide ring first;  because this epoxide is symmetrical, it doesn't matter which carbon you attack. This gets you to a charge-separated intermediate 1a (a "betaine"). These species tend to combine their positive and negative charges if they can, and here they certainly can, but first the C–C bond in the middle has to rotate to get the negatively charged oxygen and the positively charged phosphorus on the same side (1b). Note that this bond rotation brings the right-hand side methyl group to the front of the molecule.
Cyclisation to the oxaphosphetane 2 then occurs, and collapse of this to give triphenylphosphine oxide and cis-2-butene is concerted and stereospecific, exactly like the final step of a Wittig reaction.

See also the Barton–Kellogg reaction for a similar mechanism, just with sulfur instead of oxygen.

First thing come to my mind was Wittig reaction, and then show you have beaten me! :-) +1 for you.

Think about how the PPh3 first reacts with the epoxide - it does a nucleophilic attack on one of the carbons from below the epoxide ring to give the intermediate shown below left (I have drawn the P substituents as Me for clarity). This then rotates to form the oxaphosphetane (4-membered ring with P-O bond shown right) putting the Me groups cis to each other. The oxaphosphetane then eliminates triphenylphosphine oxide in a [2+2] process with the Me groups locked in the cis configuration hence the product alkene.

Better diagrams are here if you are still confused.


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How do you work out whether reduction would occur based on reduction potentials
I'm having a bit of trouble with standard reduction potentials. Specifically, when you're given a table of half reactions and reduction potentials...is the potential of the reverse reaction (the oxidation potential) just the negative of the reduction potential? So, say you were given the two reactions :
$\ce{Cl2 + 2e- -> 2Cl- }$ $     E = 1.36$V
$\ce{F2 + 2e- -> 2F-  } $ $  E = 2.87$V
and the question asked whether F- would reduce Cl2 to Cl-, so
$\ce{2 F- +  Cl2 -> 2Cl- + F2 }$
...how would you work this out?

You should be able to make a reasonable guess at the outcome without looking at the numbers. I´ve added the formula for the reaction in question.

I have the answers...it says that no, F- will not reduce Cl2 to Cl-....but I was more asking for an explanation about the reversing the given half reactions.

First, using observations, since the E value for fluorine's reduction is higher, fluorine more preferentially undergoes reduction compared to chlorine. Furthermore, if you know your periodic table, fluorine being higher in the group more readily gains a valence electron (with fewer electron shells and hence a stronger force of attraction) and would be reduced preferentially and oxidise chloride ions.
Now using the numbers
We know the following formula: $$E^o_{cell}=E^o_{cathode}-E^o_{anode}$$
And also that oxidation is at the anode and reduction at the cathode.
Now, if and only if $E^o_{cell}$ is positive would we have a spontaneous reaction.
All that is left is to consider the two cases and see which would work.
Case 1: $\ce{2 F- +  Cl2 -> 2Cl- + F2 }$
Reduction Equation (Chlorine): $\ce{Cl2 + 2e- -> 2Cl-  }$,  $E^o_{cathode} = 1.36\,V$
Oxidation Equation (Fluorine): $\ce{2F- -> F2 +2e-  }$,  $E^o_{anode} = 2.87\,V$
Then $$E^o_{cell}=1.36\,V-2.87\,V=-1.51\,V\lt0\,V$$
Hence this reaction would  not be spontaneous and not take place in this direction.
Case 2: $\ce{2 Cl- +  F2 -> 2F- + Cl2 }$
Reduction Equation (Fluorine): $\ce{F2 + 2e- -> 2Cl-  }$,  $E^o_{cathode} = 2.87\,V$
Oxidation Equation (Chlorine): $\ce{2Cl- -> Cl2 +2e-  }$,  $E^o_{anode} = 1.36\,V$
Then $$E^o_{cell}=2.87\,V-1.36\,V=1.51\,V\gt0\,V$$
Hence this reaction would be spontaneous and take place in this direction.


text:
Computing energy of Guassian basis set
Actually, I'd like to compute decay rate using :
$$ \Gamma = 2\pi \ | \langle ab|V|cd \rangle|^2 \delta\ (\epsilon_f-\epsilon_i) $$
where a, b, c, d are the MOs and V is coulomb interaction, and $\epsilon_f$ and $\epsilon_i$ are final and initial energy of the system.
I am interested in computing this value regarding to a special atom (for example atom A) which means MOs should be expandes in terms of basis set:
$$ \Gamma = 2\pi \ \sum_{\mu, \nu,\lambda, \sigma \ \in A  }C_{\mu,a}\ C_{\nu,b}C_{\lambda,c}C_{\sigma,d}|\langle \mu\nu|V|\lambda \sigma \rangle|^2 \delta \ (\epsilon_{\mu}-\epsilon_{\nu}-\epsilon_{\lambda}+\epsilon_{\sigma}) $$
where $\mu, \nu, \lambda, \sigma$ are states in Gaussian basis sets, and $\epsilon$ are corresponding energies.
It seems a straightforward calculation, but I am completely confused because I don't know how I can calculate the energy of basis sets! I think the energy of the basis set is not meaningful.
Somebody, suggests using a numerical atomic orbital basis set to compute this kind of decay rate. (for numerical atomic orbital, the energy of states is meaningful) but I also don't know how to compute this kind of basis set!
Is there any suggestion to resolve this problem?

I'm not sure how the partitioning to atom $A$ will work. Would all AO basis functions $\mu, \nu, \lambda, \sigma$ have to be centered on $A$? Just the ones on one side of the integral? (The same is true for the answer by @jezzo)

In the summation, one just considers the terms which localized on atom A.

First, these Coulombic interaction integrals can be simplified to $ \langle \mu\nu|\frac{1}{r_{12}}|\lambda \sigma \rangle $, which is the common two electron integral in the AO basis (your atom-centered gaussian basis functions).
You are correct that energies are only well defined for eigenfunctions of the Hamiltonian, and thus basis set functions (the choice of which is arbitrary) do not have a well-defined energy. Therefore, the most straightforward workaround for this is to calculate the two electron integrals in the AO basis, forming the rank 4 tensor of two electron integrals, $g(\mu,\nu,\rho,\sigma) = \langle \mu\nu|\frac{1}{r_{12}}|\lambda \sigma \rangle $. Then, you must (approximately) solve the Schrödinger equation with some level of electronic structure theory (e.g. CCSD, MP2, etc.) to solve for the molecular orbital coefficient matrix, C. The a$^{th}$ column of the matrix C is the weightings of the AO basis (atom-centered gaussian basis functions) to give the a$^{th}$ molecular orbital with well-defined energy $\epsilon_a$.
Finally, you should rotate $g(\mu,\nu,\rho,\sigma)$ with the coefficient matrix C to yield $g(a,b,c,d)$:
$$g(a,b,c,d)=C_{\mu a}C_{\nu b}g(\mu,\nu,\rho,\sigma)C_{\rho c}C_{\sigma d}$$
Note that, in this form, this change of basis scales as $N^8$, where N is the number of basis functions used, but can be implemented efficiently (by doing one rotation at a time) to scale as $N^5$. See the Crawdad programming tutorial on MP2 (site seems to be down right now) for more discussion on tensor rotations. Although it was done for MP2, it works identically for you.

I am not wondering about rotating rank 4 tensor of two-electron integrals. The main problem is the delta function which makes sure energy conservation in the system.  To implement energy conservation in the system, the energy of the basis set is required (because by using the molecular orbital coefficient matrix, the two-electron integral is transferred to Gaussian basis space) but as I mentioned it is not a meaningful concept.


text:
How does the derivative of concentration make sense?
First of all, I would like to make clear that I don't know very much about chemistry, and I'm not very sure this is the right place on Stack Exchange for this question, so please let me know if it needs migration.
I was reading a PDF about differential equations that mentioned equilibrium reactions and where the law of mass action was used. I looked it up on Wikipedia then looked up reaction rate because I was unfamiliar with the concept, and went to its formal definition.
How does the derivative of a concentration with respect to time even make sense? I seems to me that the concentration of a substance is  discontinuous in time and increases or decreases in steps.
Imagine a reaction
$$\ce{aA + bB <=> cC}$$
where $A,B,C$ are reactants and products and $a,b,c$ are stoichiometric coefficients. Now, we see that:
$$\ce{\frac{a}{N_A}A + \frac{b}{N_A}B <=> \frac{c}{N_A}C}$$
where $N_A$ is Avogadro's Number. These are the smallest stoichiometric coefficients for this equation. This reduced equation happens $N_A$ times in the above equation, considering it has an efficiency ($\eta$) of $100\%$. And this reduced equation takes some short time $\Delta t$. Before this reaction, the concentration of $A$ is $[A]_0 = \frac{\nu_A}{V}$ and after it, $[A]_{\Delta t} = \frac{\nu_A'}{V} = \frac{\nu_A - a/N_A}{V}$. I see no in-between that could link these two steps. How could we even define the concentration during the reaction? The only way I see it makes sense it to define the concentration on the interval $(0, \Delta t)$ as $[A]_t = [A]_0$. There is clearly a jump at $\Delta t$. And this doesn't happen only once. The graph of concentration vs time looks like a very compressed and scaled down floor function.
What am I missing?

You start with molecular dynamics, then approximate with stochastic PDEs, then average and approximate with PDEs. Eventually, you assume spatial homogeneity and approximate using ODEs. Which is why you may have read that the reaction is taking place while under magnetic stirring. If the number of molecules involved is small, then you use stochastic models, not ODEs.

You seem to be assuming that, since chemistry happens one molecular interaction at a time, there cannot be a continuous equation describing the rate. But chemists work on large quantities of molecules at a time (1 mole is ~10^26 molecules) this means that most reactions are very well approximated by a continuous mathematical function (unless you can measure rates to 26 decimal places).

So, everything is just approximated? Do chemists consider the graph of concentration vs time as the best linear fit?

Regarding concentrations as continous is an extremely good approximation, unless perhaps if you try to model biochemical processes on a single cell level. (or are interested in the molecular/atomistic mechanism of the reaction obviously)

Additional Information There is another aspect to consider. For an elementary reaction the forward rate, $r_f$, would be given by the equation: $$r_f = k\ce{[A]^a[B]^b}$$ where $k$ is a constant. However in reality most reactions are not elementary and the coefficients $a$ and $b$ must be determined experimentally as well as the rate constant $k$.

There are two ways to look at this. In the first, we note that all thermodynamic and kinetic analyses rely on the assumption of very large populations of molecules. The curve of concentration vs time is made of discrete points, but the large population ensures that the change in concentration resulting from addition or subtraction of one molecule is so small that it can be treated as infinitesimal. Thus, we can create a smooth curve by interpolating between points, and the result will be continuous and differentiable. From a practical standpoint, any experiment to determine the concentration will result in discrete points separated by much much more than the change of one molecule, so we are implicitly assuming that the behavior between points can be treated as a well-behaved continuous curve.
The second, and perhaps more mathematically rigorous, view is that we never differentiate the actual concentration v time "curve". Instead, we are claiming that the change in concentration over time can be represented by a function which is continuous and differentiable. The only difference between this and the first view is semantic. Does $[A](t)$ refer to the actual concentration of A or to a function that is a proxy for the concentration? It doesn't matter.
Either way, the key point is that traditional kinetic and thermodynamic analyses assume a population large enough that the discrete behavior can reasonably be treated as equivalent to a continuous function.
[On a side note, there are those who would extend your argument to the $dt$ part of the derivative, as there is no proof that time is infinitely divisible, but that's a discussion for another time and place]

I will try to present an answer, to the question, 'How does the derivative of concentration make sense?' simply from the perspective of advanced mathematics, as the question directly relates to the concept of 'derivative'.
Now, yes, delta, a measure of a finite change in concentration with time does NOT exactly equal the first derivative of concentration (relating to instantaneous change). Clearly, an approximation, but here, courtesy of the Calculus of Finite Differences, is the mathematical link between the first difference (delta) in an interval of change 'h' (not necessarily approaching zero) which can be derived, by the way, from the so-called Taylor series:
$\ce{f(x + h) - f(x) = hf'(x) + {h^2}/2!f''(x) + {h^3}/3!f'''(x) + ...}$
Source: An online available ebook, Page 13 at Calculus of Finite Differences - Charles Jordon.
Note, if one divides the equation through by 'h' and takes the limit as h approaches zero, we have the calculus definition of the first derivative as other terms involving h disappear.
So, the incremental change in the concentration can be approximated by a series expansion involving terms of higher derivatives, and as a very rough approximation, one could just use the first term only.
The appropriateness of this approximation is arguable depending on reaction particulars.

Let us assume that we can view molecules as point particles. We uniformly discretize space into "small" boxes of volume $\Delta x \Delta y \Delta z$ and assume that this volume is "sufficiently small", though not "too small" — otherwise we can no longer view molecules as point particles.
Take a snapshot. We have many adjacent boxes and many (though fewer) particles. Most boxes will be empty. Some will have $1$ particle. Fewer will have $2$ particles. Even fewer will have $3$ particles. Take the total mass of the particles in each box, divide by $\Delta x \Delta y \Delta z$ and call it concentration. Note that the concentration at most boxes will be zero.
Unfreeze time. Let a "tiny" amount of time elapse and take another snapshot. The concentration at most boxes will still be zero, but since the particles are moving, different boxes will be empty. Hence, the concentration at a box does indeed undergo discrete changes over time. It jumps up and down. The set of admissible concentrations should be discrete, as each particle has an integer number of protons, neutrons and electrons.
However, over a "small" — yet "sufficiently long" — period of time, each box will be "visited" by "many" particles. Hence, though the concentration at each box undergoes discrete changes, averaging over a period of time, the concentration will appear to be continuous, somewhat smooth and slowly-varying. This concentration can be differentiated with respect to time. Differentiating before averaging is problematic, however.


text:
How to calculate the emission wavelength of a substance?

A light photon (700nm) gets absorbed by a substance*, this substance than emits the light photo.
How to find out the wavelength emitted by this light photon?
*substance can be anything, I've just used the beryl mineral as an example

@Poutnik What if the substance is very simple, like 1 hydrogen atom? Will it still be hard to get a general mathematical formula to calculate it? or maybe a software?

@Poutnik, so if I understand you correctly, are you saying that the only way to get accurate emission wavelengths for a complex substance like minerals, is to do a real world experiment and record the results?
(i have 0.01% chemistry knowledge, so i hope you don't mind my silly questions)

I have moved comments into an answer.

There is no general way.
For a particular case, there is need a good quantum chemistry model of energy levels, and a knowledge of character of the emission. It could be an elastic excitation, releasing the same wavelength. It can be fluorescence or phosphorescence with longer wavelength.
For very simple cases, like a hydrogen atom, it is easy and googleable. See the Rydberg constant :

$$\frac{1}{\lambda} = Ry \cdot {1\over h} \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{m_\text{e} e^4}{8 \varepsilon_0^2 h^3} \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
  where $n_1$  and $n_2$ are any two different positive integers (1, 2, 3, ...), and $\lambda$ is the wavelength (in vacuum) of the emitted or absorbed light.
$$\frac{1}{\lambda} =  R_M\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
where $R_M = R_\infty/(1+m_{\text{e}}/M)$ and $M$ is the total mass of the nucleus. This formula comes from substituting the  of the electron.

The more complex the real scenario is, the less are usable theoretical models and chemists must rely either on experimental data, either on empirical models not supported by underlying theoretical principle.
There is an example of mercury melting point. Experimentally, it is a piece of cake, some $\pu{-38.8 ^{\circ}C}$. Theoretically, physical chemists waited more then 50 years until there were recently powerful enough computers available. It took combination of extensive relativistic quantum chemistry computations and random calculations to reach result within $\pu{\pm 1 ^{\circ}C}$ error. The prior non-relativistic calculations predicted the melting point with the error more than $\pu{100 ^{\circ}C}$ !!


text:
Addition to equally substituted carbons
I was wondering whether there is a trick in the following problem:

Since both carbons are equally substituted, shouldn't the products be the same with or without the use
of peroxides?

Equally substituted but not equal enviroments. The Ph group makes a significant difference to the stability of both radicals and cations. This will result in regioselectivity for both reactions.

The use of HBr with peroxide leads to an anti-Markonikov's product(known as the peroxide effect).The carbons are both substituted once, but not equally! One has a methyl group while the other side has a phenyl group. The chemical basis for the Markovnikov's  rule is the formation of a a stable carbocation during the addition process.
So, with just HBr a nomal Markovnikov's addition takes place i.e the Br group gets attached adjacent to the phenyl group while the H on the next carbon(as a benzylic carbocation is more stable).Final product: Ph-CHBr-CH2-CH3.
With the peroxide however , the opposite happens leading to the formation if Ph-CH2-CHBr-CH3.


text:
Anti-Markovnikov addition of HCl
The following reaction has been troubling me for a while now.

Since peroxides won't give the anti-Markovnikov product when used with HCl, I cannot think of any reagents that would result in the product shown in the image above. There doesn't seem to be any Wagner–Meerwein rearrangement either. This left me wondering, what reagent(s) should be used to form the product shown above.

One way to do it would be do first do a hydroboration-oxidation reaction, which leads to the anti-Markovnikov addition of $\ce{H2O}$. So you will get an $\ce{-OH}$ group instead of $\ce{-Cl}$. Then, we can replace the $\ce{-OH}$ group with $\ce{-Cl}$ by treating it with $\ce{PCl3}$ or $\ce{PCl5}$.

MsCl/Et3N will also accomplish this and is easier to handle than PCl3

1.Using 1%alkaline KMnO4 (Bayers reagent)
2.By pinnacole pincolone rearrangement we get aldehyde at terminal position
3.using suitable reducing agent reduce( Aldehyde)it to OH
4.Use PCl3 or PCl5  to replace OH by Cl


text:
When is a compound sterically hindered?

In a nucleophilic substitution reaction , which one of the following undergoes complete inversion of configuration ?
(A) $\ce{C_6H_5CCH_3C_6H_5Br}$
(B) $\ce{C_6H_5CHCH_3Br}$
(C) $\ce{C_6H_5CHC_6H_5Br}$
(D) $\ce{C_6H_5CH_2Br}$

Clearly, the question was asking which of the following compounds underwent $\mathrm{S_N2}$ most easily, i.e the case in which the nucleophilic approach was least sterically hindered.
I eliminated (A) and (C) immediately. I then went with option (B). We had pushed the phenyl group far away from the carbon to which the bromine was attached. It would thus not sterically hinder the nucleophilic.
However, the correct answer was (D) . The explanation: The PhEt group is bulkier than the PhMe group, and hence causes more steric interference.
So , what is the fallacy in my argument? Does pushing bulky groups away  have any effect on steric interference? If so, then how far away should we push them? If not, then why not? I guess my key doubt is, under what circumstances would approach of the nucleophile be sterically compromised?

I think it has to do with the fact that benzylic halogen undergo a better SN2 reaction due to the stabilisation of the SN2 transition state. Hence is more reactive than normal primary halides.Try referring here http://www.mendelset.com/category/keywords/sn2-mechanism..

Please don't add spaces before punctuation marks... also, do A and C have two phenyl groups on the central carbon? More common to write that as $\ce{(C6H5)2C(CH3)Br}$...

@orthocresol Correct, they do. I just copied the question, as it was, from my book.

@boomavijay Would it not prefer $S_N1$ ?

Yes it would prefer an SN1 over an SN2. What i meant to say was that it can also undergo an SN2 and is better than a normal primary halide. See the first question here. http://www.mendelset.com/category/keywords/sn2-mechanism, @Aspirant

@Zhe option B...

@Zhe A Walden inversion , not necessarily a change from $R$ to $S$ , or vice versa.

@Zenix Ah, good. Thanks. I misread the methyl as another phenyl...

@Aspirant Walden inversion is defined for optically active compounds only.


text:
Can pheophytin be converted back to chlorophyll?
Some fossil leaves are "greenish" when first exposed but rapidly darken to an "olive brown" color when they come into contact with air. If the fossil leaf contains pheophytin, can the addition of magnesium ions to an extract of the fossil change the pheophytin back to chlorophyll?

What is a fossil leaf, exactly?

Well, according to the Wikipedia pheophytin can be produced by treating chlorophyll with a weak acid. The question is then whether that reaction is reversible.


text:
Is my understanding of cyclic voltammetry correct?
So recently I've been roped into doing some CV and I'm trying to wrap my head around the fundamentals of the science behind it (and electrochemistry more generally! Bare in mind I'm very much not an electrochemist...). This is where I'm at:

We have our three electrode setup, two of which are submerged in the same electrolyte solution: working electrode (WE), reference electrode (RE - Ag/AgCl in this example - contained in its own little unit with its own electrolyte solution; the contained RE unit is however submerged in the bulk electrolyte solution along with the WE and CE) and counter electrode (CE - Pt wire in this example).
When we cycle through the voltages, we're measuring the electric potential of the working electrode relative to the reference electrode (which, handily, will always remain at 0V and thus the difference in volts between the WE and RE - the voltage - will always equal whatever the electric potential of the WE is).
In order to polarise the WE (ie, either pump electrons into it or remove electrons from it) we use an external power source. We attempt to do the same with the RE but pumping electrons into it simply results in Ag being reduced, and removing electrons from it results in Ag being oxidised, and thus no net polarisation of the RE is possible (hence the 0V).
Polarising the WE negatively (eg, -0.4V, -0.6V, etc) = pumping it full of electrons, which means those electrons are relatively high in energy and readily available to reduce anything at the WE surface.
Polarising the WE positively (eg, +0.5V, +0.8V, etc) = removing electrons from it, which means the electrons that remain are relatively low in energy and thus electrons in things at the surface of the WE are likely to 'jump' into the WE, in the process oxidising whatever it is they're jumping from.

With all that in mind, here are my questions:

When we're pumping electrons into/out from the WE, do these electrons come from/go to the Pt CE? Given Pt is very inert, this means something in the electrolyte solution must be being oxidised/reduced?
What is the electric potential of the CE throughout the process? Say we've polarised the WE to +0.6V (ie, forcibly removed electrons from it), the RE remains at 0V (as it always does) - does the CE become -0.6V? After all, the electrons being removed from the WE have to go somewhere - presumably some would go to the RE as well as the CE, so the polarisation of the latter won't be the exact inverse of the WE?
The RE is contained in its little unit with its own electrolyte solution. I've seen in diagrams that a porous frit separates the RE's electroyle solution from the bulk electrolyte solution that the WE and CE are submerged in. What can traverse this barrier? I assumed it would only be certain ions (so it would effectively act as a salt bridge as seen in normal galvanic/electrolytic cells), but the word porous implies that water as well as certain ions can pass through it?

Any and all help is appreciated! If you're not too sure on the questions I'd be very happy if you just weighed in on the CV understanding part, and if you only know one/some of the questions then that'd also be great :)

Besides your specific questions you may find this introduction useful;  Faulkner, L. R., "Understanding electrochemistry: Some distinctive concepts", Journal of Chemical Education 1983, 60, 262.

In addition to the comment by @porphyrin, have a look at «A Practical Beginner’s Guide to Cyclic Voltammetry» in J. Chem. Educ. 2018, 95, 2, 197-206 (https://pubs.acs.org/doi/10.1021/acs.jchemed.7b00361) which is an open access publication.

You have very good questions but most of your points are discussed on here: https://www.researchgate.net/post/What_is_happening_on_the_counter_electrode_of_an_electrochemical_cell

@porphyrin Thank you! I'll have a read through.

@Buttonwood Ah that looks very handy - thank you.

@M.Farooq Thank you for the link. I'm interested in your statement 'the rate of sweep felt by the working electrode should be the same as the counter electrode'. My understanding is that during a CV, an external power source (ie, potentiostat) forcibly polarises the working electrode, and it does this by forcing electrons into it - electrons which 'come' from the redox reactions at the counter electrode. Does your statement that the same sweep must be 'felt' by both electrodes just mean that for whatever happens at the WE, there must be an opposite but equal redox activity at the CE?

@Technocratic, Yes, exactly this is my current understanding. The number of electrons entering the WE, must equal the number of electron leaving the counter electrode. This is the beauty of electrochemistry. It is exact and quantitative.

The suggested J.Chem. Ed articles are must reads. You can also get hold of Laboratory Techniques in Electroanalytical Chemistry, Revised and Expanded
By Peter Kissinger, William R. Heineman. Very famous and way more friendlier than AJ Bard's book.


text:
Reduction of only double bonds
In the conversion shown below I need to find B . Initially I thought of protecting the carbonyl group using ethylene glycol and then carry out reduction of double bonds using hydrogenation but it is required to be a single reagent and cannot be multi step. What are the other ways in which it can be done?

You're overthinking. Usual reagents are selective enough here, so no protecting is needed.

Hydrogenation under relatively low pressure will do this without touching the ketone

You can do a normal hydrogenation reaction(H2,Pd) under low pressures as suggested by @Waylander. Alternatively you can use the Wilkinson's catalyst which also is a way to do catalytic hydrogenation but is capable of only reducing the double bond.


text:
Can Na2CO3 remove temporary hardness of water?
This was asked in my test and I remembered that all compounds that can remove permanent hardness of water can remove temporary hardness of water also. But my teacher marked it wrong and I cannot find the answer anywhere.

I guess you are right..

You teacher apparently has no clue of basic chemistry. You are right. Sodium carbonate is added in most laundry detergents to soften it.

$$\ce{Ca(HCO3)2_{(aq)} + Na2CO3_{(aq)}   -> CaCO3_{(S)} +2NaHCO3}_{(aq)} $$
$$\ce{Ca(Cl)2_{(aq)} + Na2CO3_{(aq)}   -> CaCO3_{(S)} +2NaCl}_{(aq)} $$

Definition of hardness per a source:

The simple definition of water hardness is the amount of dissolved calcium and magnesium in the water. Hard water is high in dissolved minerals, largely calcium and magnesium. You may have felt the effects of hard water, literally, the last time you washed your hands.

Adding $\ce{Na2CO3}$ introduces the carbonate ions:
$$\ce{Ca++/Mg++ + CO3(2-) -> CaCO3/MgCO3 (s) }$$
which results in a precipitate of insoluble carbonates.
So, yes it is a temporary solution to address water hardness.
[EDIT] I recommend relatedly to hard water treatment with Washing Soda, a very interesting experiment of adding Na2CO3 (aq) to MgSO4(aq) as the entire solution, at times (it depends on respective concentrations),
can convert to a wet hydrate creating so-called Magnesium alba. Cited reaction per Wikipedia:

If magnesium chloride (or sulfate) is treated with aqueous sodium carbonate, a precipitate of basic magnesium carbonate—a hydrated complex of magnesium carbonate and magnesium hydroxide—rather than magnesium carbonate itself is formed:
$\ce{5 MgCl2(aq) + 5 Na2CO3(aq) + 5 H2O(l) → Mg(OH)2·3MgCO3·3H2O(s) + Mg(HCO3)2(aq) + 10 NaCl(aq) }$

where the magnesium mix oxide/carbonate is known, but not uniquely in name or precisely in chemical composition, as Magnesium alba. Now, MgSO4 aqueous per a source:

Magnesium sulfate salts can form hydrated compounds with up to seven ... Magnesium sulfate hexahydrate is formed by coordinating 6 water molecules to Mg2+ ...

So, Magnesium sulfate forms $\ce{MgSO4(H2O)6.H2O}$ with six coordinating water molecules (which could be created upon concentration of the aqueous MgSO4 solution, further extracting water from the mix). Further, if Magnesium alba similarly also extracts water, with excess MgSO4, it may be understandable as to why the entire solution can apparently turn into a voluminous wet precipitate, somewhat surprisingly.

As my answer was, yet again, baselessly down-voted (Maurice?), I might as well add some interesting chemistry.

according to this equation : $$\ce{5 MgCl2(aq) + 5 Na2CO3(aq) + 5 H2O(l) → Mg(OH)2·3MgCO3·3H2O(s) + Mg(HCO3)2(aq) + 10 NaCl(aq) }$$,the water still hard

Having prepared Mg(HCO3)2, I can share that over a course of days in a closed container, a precipitate of MgCO3 develops, so the cited reaction may be technically correct, but in time, it should not be a substantial source of ones hard water (more like problematic deposits).

Yes, you can add sodium carbonate to remove temporary hardness of water. Temporary hardness is caused due to presence of calcium/magnesium bicarbonate. Treating them with sodium carbonate also known as washing soda helps precipitation of insoluble calcium/magnesium carbonate ultimately softening of water. Overall reaction is:
$$\ce{M(HCO3)2 + Na2CO3 -> MCO3  + 2NaHCO3 ~~~~~ M = Ca,Mg}$$

The above reaction is not accurate with all magnesium ions in the presence of Na2CO3. See my expanded answer with references. I also recommend a very interesting chemistry experiment.


text:
Does lower activation energy definitely mean higher rate?
Does lower activation energy definitely mean higher rate? For example the decomposition of $\ce{H2O2}$. Can we say it has the highest rate with the catalyst making the reaction's activation energy the lowest?

The lower the energy of the intermediate, the faster the transition from the reactants to the products, at a given temperature.

So this means the statement is true? And could you please give me a source/citation for this?

The rate constant k = A*e^(-Ea/RT) this is the Arrhenius equation, where Ea is the activation energy. Consider that the temperature is kept constant. This means that the only variable is the activation energy Ea, and as we decrease the activation Ea, the rate constant k becomes higher.

It is not really related to this question but does the amount of catalyst used affect activation energy? From its relationsheep with rate, I found this on web: Increasing the amount of catalyst used will not increase the rates of reaction beyond a certain point.

The catalyst provides an "alternative route" for the reaction, this route has a lower activation energy, so the amount of catalyst doesn't affect activation energy of that "alternative route". I would expect there to be some asymptotic relationship where when you add more catalyst, the rate of the reaction keep increasing, but more slowly, since each molecule of the reactant would have a molecule of catalyst nearby, so yes I'd expect there to be a limit.

The rate constant would be rather $k=A1.\exp{-E1/kT} + A2.\exp{-E2/kT}$. A1 is frequency of collisions leading to possible reactions without catalyst, A2 with catalyst. E1,E2 are then the respective activation energies. With more if catalyst, A1 gets lower, A2 gets bigger.

@Mathrix - RE: "Increasing the amount of catalyst used will not increase the rates of reaction beyond a certain point." --  Be very careful when making (or interpreting) such far reaching statements. You could have 1000 kg of catalyst and 1 gram of reactant, or 1 gram of catalyst and 1000 kg of reactant. In the first case the reaction will happen more of less instantaneously.  In the second case there is a huge excess of reactant so the reaction seems to proceed at a constant rate. The reaction would have to run for a long time to see a decrease in the concentration of the reactant.


text:
What's the definition for mole before Avogadro's constant?
I have been searching for about 3 days how Avogadro's constant is found, and as all of us know the value is determined by division of charge of one mole of electrons (Faraday's constant) by the charge of one electron.
What confuses me is that, Faraday's constant is 96500 (it's not accurate value)
and its unit is C/mole. If the mole is defined by Avogadro's Constant
and Avogadro used Faraday's constant to determine his constant,
what was the definition of mole for Faraday?

did you look here? https://en.wikipedia.org/wiki/Avogadro_constant#History Not sure what you're really trying to figure out. It would take a book to discuss how weight/mass and the understanding of atoms developed scientifically. It wasn't a direct line of thought.

A gram equivalent times the stoichiometric coefficient contains the same amount of substance as a mole. To define the mole, you don’t need Avogadro’s constant, just atomic weights and a balance.

These constants have been named in honour of these people. They did not even know the existence of mole concept.
In order to search how early precursors to Avogadro's number's were found, read about Loschmidt constant and then move to Avogadro's story. There were many $independent$ methods to determine Avogadro's constant. Loschmidt was trying to estimate size of the "air molecule" in 1865.  Number of particles in air in a given volume.
The determination of Faraday's constant does not require the knowledge of Avogadro' ideas or his laws. This is a modern misconception. However, it does require the concept of gram equivalents (precursors to atomic mass or formula weights). In Faraday's time, measuring current/voltage and masses of substances was not a big deal. By electrolyzing a solution of silver salt, you can deposit a known mass of silver on the electrode. I am showing a page how Faraday's constant was found. As you can see we never needed to invoke Avogadro's law or numbers and constant. Faraday did such neat experiments! Clever and simple.
Electro-chemistry: General theory, Volume 1, pg 4, By Robert Alfred Lehfeldt, T. S. Moore. Longmans, Green and Company, 1904.


text:
What is the rate expression for decomposition of $\ce{H2O2}$ in the presence of a catalyst?
What is the rate expression for decomposition of $\ce{H2O2}$ in the presence of a catalyst?
As far as I know for catalase, it is like this: $$\ce{Rate = k[H2O2][Catalase]}$$
I don't know what it is for the catalysts $\ce{MnO2}$ and $\ce{Fe2O3}$. Another question in my mind is that even if it's similar to that of catalase, since these are powdered and so do not have a concentration, how can I express it?
It is kind of emergency so any help would be greatly appreciated.
$$\ce{2H2O2 + Catalyst ->2H2O + O2}$$
I found on web that both reactions are probably (still not sure even though I saw it) first order reactions.
Edit: Is it possible that there shouldn't be anything to be added and it will only affect the $\mathrm{k}$?

Don't use MathJax or LaTeX in tiles. Don't use \ce{} for rate constants, use \mathrm{} instead.

For catalase, it is most likely Michaelis-Menten kinetics.
At typically low enzyme concentration, this would be first order in catalase. For hydrogen peroxide, it is first order at low concentrations (<< Km) and approaches zero order at high concentration (>> Km). This behavior is also called saturation kinetics.

I don't know what it is for the catalysts $\ce{MnO2}$ and $\ce{Fe2O3}$. Another question in my mind is that even if it's similar to that of catalase, since these are powdered and so do not have a concentration, how can I express it?

For catalysts in the solid phase, it is the surface area (compared to the volume of the solution) that matters. You might see similar saturation kinetics, indicating that at some concentration of reactant, all the surface area is used so that an increase in reactant concentration no longer translates into an increase in rate.

So can we say that its effect can be seen as a change in rate constant?

If the amount of catalyst is always the same, you could define an effective rate constant that incorporates information about the catalyst (like giving distinct rate constants for different solvents or pH values instead of including the hydrogen ion concentration explicitly in the rate equation). However, often we use the rate expression to learn about the mechanism, and then it would be counterproductive.


text:
Is this bond cleavage correct?
Is this bond cleavage of cyclopentadienone correct?

The ring becomes aromatic and stability "should" increase. But my book and data from experiments suggest that the molecule isn't that polar as it would have been from this cleavage. Why isn't this cleavage favorable?
Doesn't aromaticity counter the appearance of a positive charge on oxygen?

The cleavage that you have shown is NOT favourable. The dipole of the carbon double bond O points towards the oxygen , not the carbon. The bond tends to shift onto the oxygen giving  it a negative charge , while the carbon a positive charge. This, however is anti-aromatic so the molecule tends to do a Diels-Alder reaction with itself.
The bond will not cleave the way as you have mentioned . Just for the sake of aromaticity, a carbon double bond oxygen will not cleave in such a manner where the carbon contains the negative charge and the oxygen a positive charge. An oxygen with positive charge is extremely unstable(as it has a high electronegativity) and will show a strong inductive effect which will pull the electrons from the carbon(which is way less electronegative than oxygen) thus reverting to carbon double bond oxygen structure. Aromaticity can't be obtained as oxygen would just pull those elctrons from the carbon atom.
However let us consider the case of cyclopropenone. Here, the molecule becomes aromatic due to the shifting of the double bond onto the oxygen (oxygen gains a negative charge while carbon , positive).


text:
What do speciation curves really represent?
So I'm just reading about speciation diagrams in my textbook...and I'm very confused.
I get how you work out that the crossover point in when [A-] = [HA] mathematically...but I don't understand it in relation to what an equilibrium constant represents. My understanding was that equilibrium constants represent a ratio that is ALWAYS true for that reaction. So, for instance, at the beginning of this diagram where the pH is low (lots of protons)...I get that this drives the equilibrium to the left (hence the high mole ratio of the protonated form)...however, shouldn't the ratio of acid/base form still equal the acid dissociation constant (Ka)? Sorry if this makes little sense but I'm really struggling to reconcile this with what I've learnt about equilibrium constants in general.

shouldn't the ratio of acid/base form still equal the acid dissociation constant (Ka)?"

No,
$$K_\text{a} = \frac{\ce{[H+]}\ce{[CH3COO]}}{\ce{[CH3COOH]}}$$
You are forgetting that the $\ce{H+}$  concentration ($\ce{[H+]}$) is also present in the equilibrium expression.

Yes, sorry, I do understand that. I meant to ask, is that ratio still satisfied at every point in that graph? Because I thought when you 'disturbed' an equilibrium, the reaction shifts - according to Le Chatelier - to maintain this ratio (the Ka value).

Yes that ratio is satisfied (assuming activities can be approximated by concentrations, but never mind that for now). Le Chateliers principle still holds.

ok...i think I'm starting to grasp it thanks :)

Forget about equilibrium constants when you try to read speciation curves. They are derived from equilibrium calculations though. If you are interest you should derive the mathematical relations yourself by defining what is the concentration ratio (Search speciation diagram acetic acid). It will satisfy you if you can draw then in Excel yourself.
Reading it is very simple. The y-axis represents the fraction of two molecular species obtained from acetic acid and the x- axis is the pH.
Say at pH 1, almost 100% of the solute exists as CH$_3$COOH, and almost 0% is CH$_3$COO$^-$
Say at pH 6, more than 95% of solute exists as  CH$_3$COO$^-$ and less than 5% exists as CH$_3$COOH.
Similarly try to read the what is happening at the pKa of acetic acid.

I'll add that the pH where the two curves intersect is the pKa of acetic acid.

Yes I understand how to read the graph...but I don't want to ignore the equilibrium constant because that's what's causing my confusion. Is it not correct to say that the equilibrium constant ratio should always be satisfied...and is it satisfied here?

You really have to look at the derivation. Believe it or not it not take more 3-4 steps. One should not talk about curves qualitatively. Lets us start from defining what your y-axis is from acetic acid and for acetate ion. Write an algebraic expression for both.

ok. It's the mole ratio...so for acetic acid its n(acetic acid)/n(total) where n(total) = n(acetic acid)+n(acetate ion)?

@MatH - Yes that's it. I just noticed that the Y-axis is labeled "molar ratio" which is really incorrect. What is graphed is rather the "mole fraction."  So the blue curve isn't $\frac{\ce{[HA]}}{\ce{[A-]}}$ but rather $\frac{\ce{[HA]}}{\ce{[HA] + [A-]}}$

Right, so you can write that in terms in concentrations  as molar conc (acetic acid)/ molar  conc. (total). Also write the equilibrium expression Ka for acetic acid.

Ka = [CH3Coo-][H3O+]/{CH3COOH]

Correct, now you have two equations, now we want to eliminate [HA] and [A-] from [HA]/([HA]+[A−] ) for acetic acid so that we can find an expression which contains [H+] and Ka only in this ratio. What if we rearrange the Ka expression and write [HA]=[A-][H+]/Ka. Substitute this value and see what you get.

If you rewrite the equilibrium constant expression, you get:
$$\frac{K}{\ce{[H+]}} = \frac{\ce{[CH3COO-]}}{\ce{[CH3COOH]}}$$
So the ratio of the two species is equal to the ratio of equilibrium constant to proton concentration. Accordingly, the ratio changes with pH, and is equal to one when the pH is equal to the pKa.
At each pH value in your graph, the acid/base reaction is at equilibrium (so the the equilibrium constant expression with the concentrations plugged into it will have the same value (= K).
You might think the graph should be a straight line. It is a curve because the x-axis is showing the pH and not the proton concentration, and because the y-axis is showing the fraction of the species and not the ratio. If you plot log(ratio) against pH, you get a straight line.

Dear Karsten, I am afraid this explanation is partially incorrect. The y-axis is wrong in student's textbook.The graph of the so called alpha functions of acetic acid is not a plot of ratio of conjugate base over acid concentration.

As $$\alpha=\frac{1}{1+\frac{\ce{[A-]}}{\ce[HA]}}$$

@M.Farooq "...because the y-axis is showing the fraction of the species and not the ratio", i.e. Poutnik's alpha.


text:
Is the specific heat of salt the same as Na + CL when dissolved?
I had this question when my friends were discussing how salt influences boiling time.  When crystal salt is dissolved into water, it turns into a single Na and a single Cl atom.  If knew the total heat capacity of the salt I want to add to my water, could I be confident, that when dissolved, total heat capacity I added to the water is equal to the undissolved salt's heat capacity?
On one hand my intuition tells me that heat capacity should remain the same.  Salt is salt.
On the other, doesn't the salt lose a rotational vibration mode when is dissolves into its two monatomic elements?
I am not trained in chemistry, so sorry if this is a naive question.
Thanks!

Salt does not dissolve to to $\ce{Na}$ and $\ce{Cl}$ atoms, but to hydrated $\ce{Na+}$ and $\ce{Cl-}$ ions, so it's heat capacity is very different to the solid $\ce{NaCl}$.
And yes it looses vibration modes of the crystal lattice, less the rotation, as there is nothing to rotate in the lattice.
There are also vibrations of coordination bonds to water. Dissolved ions also have translational degrees of freedom, what is missing in solids.
Even for the identical substance, its heat capacity in solid and liquid state differ. E.g. liquid water has about 2x bigger heat capacity than ice ( 4.2 versus 2.1 J/g/K ).


text:
What element did Rutherford use to obtain alpha particles in his experiment of gold foil?
What element did Rutherford use to obtain alpha particles in his experiment of gold foil?

Both are incorrect from the web. Always take the info from public websites with a grain of salt. It was radium emanation, which is nothing but radon gas. A good example which proves how much junk is out there on the web. Am-241 is completely off the hook.

Initially, I stated in the comments(now deleted) that the source for alpha-particles was either Am-241 or radium metal. Both are incorrect as @M. Farooq said. The actual source was called radium-emanation which was basically radon gas or Rn-222 which was several times more radioactive than uranium. Researching a little bit more on the history of Rutherford's gold-foil experiment gave me following references. Have a look if you want to know more about the "inspiration/motivation" of this experiment:

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4123660/
Boltwood, Bertram B. “Radioactivity.” Proceedings of the American Philosophical Society, vol. 50, no. 200, 1911, pp. 333–346. JSTOR, www.jstor.org/stable/983927.
https://history.aip.org/exhibits/rutherford/sections/exploring-radioactivity.html
https://arxiv.org/ftp/arxiv/papers/1202/1202.0954.pdf
http://publications.iupac.org/publications/ci/2011/3301/8_radvanyi.html

It is an understatement, that 222Rn with halftime 3.x days is several times more radioactive than 238U with billions of years, 235U with hundreds of millions, or even 226Ra with about 1600 years.... :-)


text:
Is it correct to say that the pKa is the pH at equilibrium?
I've heard this said when looking at the charge of certain groups on a molecule (with different $\mathrm{p}K_\mathrm{a}$ values) in a solution of certain $\mathrm{pH}.$ But is that really correct to say?
Based on speciation diagrams, I would have thought that the $\mathrm{p}K_\mathrm{a}$ is always constant and that $\mathrm{p}K_\mathrm{a}$ is simply the $\mathrm{pH}$ at the point where the concentrations of the acid and its conjugate based are equal.

To say that "The pKa is the pH at equilibrium" isn't really true. Acid/Base reactions in aqueous solutions happen very rapidly and thus the solution reaches some equilibrium quickly. The pKa is a particular equilibrium where the protonated form of the acid and it unprotonated form are present in equal concentrations in the solution.

$\ce{}$No$\ce{}$

$\mathrm{p}K_\mathrm{a}$ and $\mathrm{pH}$ are apples and oranges.* We can't equate them on a graph. The correct wording should rather be that the $\mathrm{pH}$ at which the concentration ratio of an acid and its conjugate base is equal to $1,$ then the $\mathrm{pH}$ is numerically equal to $\mathrm{p}K_\mathrm{a}.$
* It will be another painful story if one begins to argue on the units of $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}.$ They are dimensionless. Just a subtle point $\mathrm{p}K_\mathrm{a}$ is constant as long as the temperature/pressure is constant. It is not always a constant.


text:
it is possible for the hydrogen bond to form between the hydrogen chloride molecule, HCl?
Recently I'm doing my work, and I can't find what it is the answer. Can you guys help me?

HINT What are the three most common states of matter?

Hydrogen bond is formed by three most electronegative elements: fluorine, oxygen, nitrogen.

@MaxW I believe he tries to find if the hydrogen bonds significantly affect the properties of the $\ce{HCl}$.

See  Chloronium, H2Cl+, in Extragalactic Sources - cdms.astro.uni-koeln.de.  But I suppose in water environment, it would be very unstable, with proton having  preference for oxygen.


text:
Is the number of tetrahedral voids always equal to double the number of octahedral voids in any crystal?
I came accross the question whether the number of tetrahedal voids is always equal to twice the number of octahedral voids (in the case where both the voids are present). The key was given as the statement was true. Could anyone justify this?
I know that it is true for ccp and hcp but can this be generalized for all types of crystals?

Well alpha polonium has a primative cubic structure. What types of voids would that crystal have?

@MaxW I asked that in the case when the crystal had both the voids would the condition be valid (I would edit the question)

Probably true for all closed packing lattices, not sure if can be generalized

@Zenix Then in the spinel structure tetrahedral voids and octahedral voids are occupied then in that case  would there be any complexity in deciding the validity of this statement or is it's answer given is wrong

The title reads "Is the number of tetrahedral voids always equal to number of octahedral voids in any crystal?"; Didn't you mean double the number of octahedral voids?

Just what are the structures you want to generalize to?

@GuruVishnu yeah that was a typo I edited it

@IvanNeretin I just wanted to gather information about generalizing the statement  I  had the doubt for it's validity in ionic compunds such as spinel, hematite where the cations occupy the voids. After filling the voids then how can we decide the no.of tetrahedral and octahedral voids

After filling the voids, it is no longer useful to think of the structure in terms of voids. Sure, there are smaller voids remaining, but they no longer have distinct shapes like tetrahedral or octahedral.

https://chemistry.stackexchange.com/questions/131930/is-the-number-of-tetrahedral-voids-always-equal-to-double-the-number-of-octahedral-voids-in-any-crystal

Note that, in case of tetrahedral voids, each sphere is in contact with three tetrahedral sites above, and three tetrahedral sites below, hence, there are two tetrahedral sites associated with each sphere. Thus,
Number of tetrahedral voids $(N_{th}) = 2 \times$ (Rank of unit cell)

Or alternatively, if you divide a FCC lattice into 8 octants, each octant will have a tetrahedral site at it's body centre.

For octahedral sites, each sphere is associated with one octahedral sites, thus,
Number of octahedral voids $(N_o) =$ Rank of unit cell
For FCC, besides the body center, there is one site at the center of each of the 12 edges. It is surrounded by 6 atoms, three belonging to the same cell unit cell, and the three belonging to two adjacent unit cells.
For clarity see:

For HCP,

thanks for your answer but my question was was about generalizing the statement I know that it holds for CCP and HCP I think all these are associated with CCP and HCP

@Adithya I haven't personally encountered any source discussing octahedral and tetrahedral voids for end centred cubics or body centred cubics...

Generally tetrahedral and octahedral voids is formed in HCP and CCP structure, and don't form in other crystal structure like primitive cubic and BCC.