text:
What is the rate equation of decomposition of H2O2 with catalysts?
What is the rate equation of this reaction with the presence of catalysts $\ce{MnO2}$, $\ce{PbO2}$ and $\ce{Fe2O3}$? (asking for each of them, separately, not altogether)
$$\ce{2H2O2 -> 2H2O + O2}$$

TL, DR: There's no equation.

The rate of the reaction between a solute and a solid catalyst cannot be defined in the usual way, namely with an expression which is proportional to the concentration of the catalyst. Such a rate is proportional to the surface of the catalyst. And the surface of a powder is a parameter difficult to know with precision

How can I find it? I am working on experiments that show the difference at $\ce{E_a}$ between two different catalysts. In order to do this, do I not need the rate equation to find the value of $\ce{k}$, so that I can use Arrhenius equation? What can/should I do?

For particular dosing of the catalyst, measure the rate for various temperatures and H2O2 concentration. You can then get both factors for Arrhenius equation and reaction order for H2O2.

@Poutnik. Even if you choose the same amount of catalyst and H2O2, the rate of reaction depends on the dimension of the grains. The reaction goes quicker with fine grains than with coarse grains.

@Maurice sure, but I have meant the same catalyst lot. By other words, keeping constant as many degrees of freedom as possible, varying just [H2O2] ( the reaction order ) or temperature ( Arrhenius eq. parameters ).

The decomposition of H2O2 is complex and has been the source of several studies.
On iron oxide surface, here is a cited work `Catalytic Decomposition of Hydrogen Peroxide on Iron Oxide: Kinetics, Mechanism, and Implications’. Some selected quotes:
"As depicted in Figure 2, the decomposition rate of H2O2 appears to be independent of the goethite particle size. On the other hand, the decomposition rate is directly proportional to the iron oxide concentration (Figure 3)."
Also:
"The large surface area measured for the granular particles is equivalent to the external surface area of the colloidal particles, as reported in the literature. This might explain the observed lack of size effect in the present study....
However, the experimental results showed that the observed kinetics and rate constants were not affected by various concentrations of H2O2, and the dissolved iron concentration was always low (e0.2 mg/L)."
Table 2 and Table 3 outlined the reaction mechanism.
I would expect similar results with H2O2 for other transition metal oxides, but MnO2 is claimed to avoid a radical based path (per Wikipedia), and the decomposition reaction can be energetic (and possibly explosive with 30% H2O2 in a closed vessel).
Here is another example at "Active sites and mechanisms for H2O2 decomposition over Pd catalysts". Some extracts:
"We conclude that both Pd(111) and OH-partially covered Pd(100) surfaces represent the nature of the active site for H2O2 decomposition on the supported Pd catalyst reasonably well. Furthermore, all reaction flux in the closed catalytic cycle is predicted to flow through an O–O bond scission step in either H2O2 or OOH, followed by rapid H-transfer steps to produce the H2O and O2 products. The barrier for O–O bond scission is sensitive to Pd surface structure and is concluded to be the central parameter governing H2O2 decomposition activity."
Also:
"A mean-field microkinetic model was developed to describe the experimentally measured reaction rates, reaction orders, and apparent activation barrier."
And:
"The decomposition of H2O2 has been studied both in the vapor phase (47) and aqueous phase (48) (thermal, noncatalyzed), and over a variety of materials including metal oxides (49⇓–51) and metal ions in solution (52, 53). Based on these studies, we have compiled an encompassing network of 17 elementary reactions involving four closed-shell species (H2O2, H2O, O2, H2) and four surface intermediates (O, H, OH, OOH), which are shown in Table 1. Elementary reactions are classified as follows: adsorption/desorption, O–O bond scission, dehydrogenation, and hydrogen transfer."
Also:
"The microkinetic modeling results suggest that both the close-packed Pd(111) and more open Pd(100) facets can contribute to the total H2O2 decomposition activity. Furthermore, on both Pd facets, all reaction flux is predicted to go through an O–O bond breaking step (in either H2O2* or OOH*), followed by successive H-transfer steps to O*/OH* adsorbates. The relevant surface coverage of O*/OH* is then a function of the ability of the Pd surfaces to generate the O*/OH* fragments through O–O bond breaking [which can vary strongly with surface coverage, as seen from calculations on the OH*-modified Pd(100)], and the availability of H-donating species (H2O2* and OOH*) to reduce O*/OH* to H2O* through the rapid H-transfer reactions."


text:
Determining a molecular formula of the catalyst for polymerisation of butadiene
I've got the following question in a workbook which I need to solve:

A $\pu{50 g}$ sample of an experimental catalyst used in the polymerisation of butadiene is made up of $\pu{11.65 g}$ of $\ce{Co}$ and $\pu{25.7 g}$ of $\ce{Cl}$. If the molar mass of the compound is $\pu{759 g/mol},$ find the molecular formula of the catalyst.

I attempted to first find the percentage by mass of each given element by dividing their masses ($\pu{11.65g}$ and $\pu{25.7 g},$ respectively) by the mass of the entire catalyst $(\pu{50 g}).$ From there, I divided both percentages ($23.3$ and $51.4)$ by the smallest one, and multiplied the results $(1$ and $2.2)$ by $5$ to get two whole numbers. The formula I ended up with was $\ce{Co5Cl11}.$
Now that made sense to me, but somehow the correct answer is $\ce{Co3Mo2Cl11}$. I don't understand how I'm supposed to figure out that $\ce{Mo}$ is required? Could somebody please shed some light on this for me?

At least you should know there is a third element in the catalyst other than $\ce{Co}$ and $\ce{Cl}$. You should include that in your calculations. Nevertheless, this question is very poorly written. It is missing one important information.

I agree that the question is poorly written. Does this mean there is no way to determine that the third element is Mo? Or are you inferring that this question impossible because there is too little data?

It is still possible to solve, but cannot pinpoint $\ce{Mo}$ is the third element. If you show your work, we may able to help you. Otherwise, this may be closed as a homework question.

You say "$\pu{11.65g}$ of $\ce{Co}$ and $\pu{25.7g}$ of $\ce{Co}$." -- It seems that one of the masses should be for $\ce{Co}$ and the other for $\ce{Mo}$.

@MaxW This was a typo introduced by copy-pasting elements during third-party editing. OP included 11.65 g  of Co and 25.7 g of Cl (fixed now).

Despite the absurd lack of data, this problem attracted my attention, probably because it closer resembles a real-life challenge rather than a textbook problem.
It looks like it's an adaptation of the problem from Schaum’s Outline of Theory and Problems of College Chemistry [1, p. 39]:

3.37 What is the empirical formula of a catalyst that can be used in the polymerization of butadiene if its composition is $23.3\%$ $\ce{Co},$ $25.3\%$ $\ce{Mo},$ and $51.4\%$ $\ce{Cl}.$
Ans. $\ce{Co3Mo2Cl11}$

Here we are dealing with zero knowledge as to what the remaining element(s) are, but with a couple of assumptions we actually can come up with an answer from the textbook, and an answer on top of that with a justification.
Let's dive in and assume we are dealing with an unknown compound with the following formula:
$$\ce{Co_xCl_y}\sum_{i=1}^N \ce{El}_{i,z_i}$$
where we account for $i$ other elements $\ce{El_$i$}$ with the respective coefficients $z_i.$
The first assumption required to advance with the solution is to treat the compound as stoichimetric. e.g. $x, y, z_i \in\mathbb{N}.$
This would immediately allow us to find the exact values for both $x$ and $y$:
$$y : x = \frac{m(\ce{Cl})}{M(\ce{Cl})} : \frac{m(\ce{Co})}{M(\ce{Co})} = \frac{\pu{25.7 g}}{\pu{35.45 g mol-1}} : \frac{\pu{11.65 g}}{\pu{58.93 g mol-1}} = 3.67\tag{1}$$
To satisfy $x, y \in\mathbb{N},$ and taking into account the "hint" decimal part $.67$ (which implies a triple factor to an integer):
$$x = 3, 6, 9, \ldots, 3n~(n\in\mathbb{N})\tag{2}$$
To pinpoint the exact allowed value of $x$ (and $y,$ as $y = 3.67x),$ we can use the total molar mass:
$$M = x × M(\ce{Co}) + 3.67x  × M(\ce{Co}) + \sum_{i=1}^N z_iM_i\tag{3}$$
$$\pu{759 g mol-1} = x × \pu{58.93 g mol-1} + 3.67x × \pu{35.45 g mol-1} + \sum_{i=1}^N z_iM_i\tag{3a}$$
$$\sum_{i=1}^N z_iM_i = (759 - 189x)~\pu{g mol-1}\tag{3b}$$
Since molar mass is a positive number, $759 - 189x > 0$ and the only $x$ to satisfy this criteria would be $x = 3.$
Accordingly, $y = 3.67 × 3 = 11,$ and at this point we are left with the following formula:
$$\ce{Co3Cl11}\sum_{i=1}^N \ce{El}_{i,z_i}$$
and the remaining sum
$$\sum_{i=1}^N z_iM_i = (759 - 189 × 3)~\pu{g mol-1} = \pu{192 g mol-1} \tag{3c}$$
The second assumption is that we are dealing with the common oxidation numbers of the elements, no exotic stuff.
With this in mind, and knowing that cobalt catalyst used in polymerization are vastly only $\ce{Co^{II}}$ and $\ce{Co^{III}}$ species, we can deduce the possible charge of the remaining sum:
$$(\ce{Co3Cl11})^{q-}\left(\sum_{i=1}^N \ce{El}_{i,z_i}\right)^{q+}$$
Since it's likely only $\ce{Cl-},$ the $q$ can adopt only the values defined by $\ce{Co^{II}}:\ce{Co^{III}}$ ratio.
For the total of three cobalt(II,III) atoms $q = 2,3,4,5.$
This will assist us in proposing missing element(s), and one can start to iterate over numbers.
The third assumption is that $i = 1$ and there is only one extra element with single coefficient $z_1$.
The table below summarizes possible outcomes:
$$
\begin{array}{ccccc}
\hline
z_1 & M_1/\pu{g mol-1} & \ce{El_1} & M(\ce{El_1})/\pu{g mol-1} & \text{Formula Example} \\
\hline
1 & 192 & \ce{Ir} & 192.22 & \ce{Co^{II}2Co^{III}Ir^{IV}Cl11} \\
  & & & & \ce{Co^{II}Co^{III}2Ir^{III}Cl11} \\
2 & 96 & \ce{Mo} & 95.95 & \ce{Co^{II}3Mo^{II}Mo^{III}Cl11} \\
  & & & & \ce{Co^{II}2Co^{III}Mo^{II}2Cl11} \\
3 & 64 & \ce{Cu} & 63.55 & \ce{Co^{II}Co^{III}2Cu^I3Cl11} \\
  & & & & \ce{Co^{II}3Cu^{I}Cu^{II}2Cl11} \\
4 & 48 & \ce{Ti} & 47.87 & \ce{Co^{?}3Ti^{?}4Cl11} \\
\hline
\end{array}
$$
Starting with $z = 4,$ there doesn't seem to be an appropriate set of oxidation numbers for the element to comply with the charge balance, so I'd say the candidates for the third elements are only iridium, molybdenum and copper.
Judging from the application (catalysis) as well as from the typical oxidation states, I'd actually propose iridium cobalt chloride $\ce{Co3IrCl11}$ as the answer, yet $\ce{Co3Mo2Cl11}$ would also be possible (AFAIK, molybdenum(II,III) isn't a common composition for halide salts), and $\ce{Co3Cu3Cl11}$ would be a stretch both due to higher deviation from the declared molar mass and presence of copper(I).
References

Rosenberg, J. L.; Epstein, L. M. Schaum’s Outline of Theory and Problems of College Chemistry, 8th ed.; Schaum’s outline series; McGraw-Hill: New York, 1997. ISBN 978-0-07-053709-5.

Thanks andselisk for the detailed answer. Even though this is a little above my level, I can make sense of most of it, so I'll mark it as correct. I think that perhaps this question was just worded incorrectly and simply doesn't contain enough information, however.

Fun answer, but what about $\ce{Co^{II}2Co^{III}Mo^{II}2Cl11}$?

@user5713492 Thank you, good point, I appended this one too.

It is no use to calculate the mass ratio of the atoms in the catalyst, as Doofilator did. What is important is the number of moles. The trouble is that the data are wrong. As MaxW said it previously, the following data is wrong : "11.65 g of Co and 25.7 g of Co". Cobalt cannot be mixed with Cobalt. And worse ! There is nowhere any information about a third element.
I have tried to rewrite the whole problem thinking that the 25.7 g are Chlorine atoms. With this assumption, the given masses correspond to 0.197 mol Co and 0.725 mole Cl. The ratio Cl/Co would be 3.68, which could be explained with 0.197 mol of CoCl2, but it remains 0.725 - 2·0.197 = 0.331 mol of Chlorine. And this Chlorine must be combined to some other element.
Of course the third element weighs 50 g - 25.7 g - 11.65 g = 12.65 g.
If the third element is Molybdenum, the sample contains 12.65/95.94 = 0.132 mol Mo. The ratio Cl/Mo would be 0.331/0.132 = 2.5. This may correspond to Mo2Cl5.
So the sample may be made of CoCl2 and Mo2Cl5, both in a ratio that must be calculated. I let it to Doofilator. This ratio can be obtained from the molar mass of the compound.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Is there any aqueous solution (or simple method) to remove trace level of sulfur dioxide/nitrogen monoxide and nitrogen dioxide in a gas stream?
I need to generate pure gas from a gas stream which is free of $\ce{SO2}$/$\ce{NO}$/$\ce{NO2}.$ The air is already pretreated by activated carbon and purafil. However, I can't find any solution for removing trace level of this pollutants. I find that many solutions of capturing for large scale like flue gas. However, it is not suitable for trace level (1–10 ppb).
I know $\ce{O3}$ can react with $\ce{NO}$ to $\ce{NO2}$. However, as I mentioned, I need the gas stream free of $\ce{NO}$ and $\ce{NO2}$ simultaneously. So, converting $\ce{NO}$ to $\ce{NO2}$ is not the answer.

Can you describe what are you doing? Which pure gas and how it is being generated? Is this a science project?

You may have the gas mixture bubble in a NaOH solution, or wash it through a column with flowing NaOH solution at countercurrent. SO2 and NO2 will be dissolved. And if some O2 is also present, NO will soon be oxidized into NO2, and then dissolved.

If you cool it what will liquify first and last?

As I mentioned, I need the gas stream free of NO and NO2 simultaneously. So, converting NO to NO2 is not the answer. Thanks. @Maurice

I need to purify the gas stream to zero ppb of SO2/NO/NOX. The gas is pre-treated by purafil and activated carbon. The flow rate is around 5 lpm. @DrMoishe Pippik

You can convert NO in NO2 with O2, and then wash the gas with NaOH solution : 2 NaOH + 2NO2 -->  NaNO2 + NaNO3 + H2O

You could also use a gas mask containing some active charcoal, which has pores big enough to adsorb impurities like NO2.

Removal of $\ce{SO2, NO2}$ from solution
As mentioned by Maurice, you can remove $\ce{SO2, NO2}$ by bubbling the mixture bubble through NaOH or KOH solution, or washing it through a column with flowing NaOH or KOH solution at countercurrent.
Removal of $\ce{NO}$ from solution
You can bubble the mixture through $\ce{aq.FeSO4}$ solution.
I remember doing this in eudiometry experiments.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How to simplify these system of equilibrium equations

What is the pH when $0.025$ mol $\ce{NiCl2}$ is added to $\pu{1.00 L}$ of a $\pu{0.100 M}$ solution of $\ce{HCN}$, if the formation constant for $\ce{Ni(CN)4^2-}$ is $K_f=\pu{1.0e22}$, and the $K_a$ for $\ce{HCN}$ is $\pu{4.9e-10}$?

My approach is as follows:
1) $\ce{\frac{[H+][CN-]}{[HCN]}=4.9*10^{-10}}$
2) $\ce{\frac{[Ni(CN)4^{2-}]}{[Ni^2+][CN-]^4}}=1.0*10^{22}$
3) $\ce{[H+][OH-]}=10^{-14}$
4) $\ce{0.025=[Ni^{2+}] + [Ni(CN)4^{2-}]}$       .............  (mass balance for $\ce{Ni^{2+}}$)
5) $\ce{[CN-] + 0.05 + [OH-] + 2*[Ni(CN)4^{2-}] = 2*[Ni^{2+}] + [H+]}$ ........... (charge balance)
6) $\ce{0.1 = [CN-] + [HCN] + 4*[Ni(CN)4^{2-}]}$ ............. (mass balance for $\ce{CN^{-}}$)
Since the equilibrium constant for the formation of the complex ion is very large, I assume that $\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$, so $\ce{[Ni(CN)4^{2-}] = 0.025 M}$, from the mass balance for $\ce{Ni^{2+}}$.
However, I am not sure what is the next step and what other assumption to make.

If $\ce{[Ni(CN)4^{2-}] = 0.025 M}$, then what does that imply for mass balance of $\ce{CN-}$ species?

One information is missing : What is the volume of the HCN solution ? If the volume is greater that 1 cubic meter, the effect of NiCl2 is negligible.

@Maurice the volume of the system is 1.00 L

@MaxW it seems to be that [CN-] + [HCN] = 0?

"I assume that $\ce{[Ni(CN)4^2−]≫[Ni^2+]}$"
$$.$$
That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex with nickel, and there is very little free cyanide.

@KarstenTheis thank you. How may I make a better assumption to solve this problem?

Since the equilibrium constant for the formation of the complex ion is very large, I assume that $\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$

From the comments:

[comments:] That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex with nickel, and there is very little free cyanide.

OP replied:

How may I make a better assumption to solve this problem?

In the absence of nickel, the pH is around 5 and HCN is the major species, with the concentration of cyanide in the micromolar range (standard weak acid problem). When you add nickel, the concentration of cyanide gets lowered further. Because the free cyanide concentration contributes little to the total, we can combine the two reactions to see whether there is more complex or more HCN and to determine the pH:
$$\ce{4HCN + Ni^2+ <=> Ni(CN)4^2- + 4H+}$$
The equilibrium constant is
$$K_\mathrm{combined} = \pu{5.8e−16}$$
So for this reaction, reactants are favored. HCN and nickel concentrations hardly decrease compared to the initial concentrations. If we call the concentration of the complex $x$, we get:
$$K_\mathrm{combined} = \frac{x (4x)^4}{[\ce{HCN}]^4 \cdot [\ce{Ni^2+}]}$$
Solving for $x$, we get $\pu{2.24e−5}$, confirming that reactants are favored. From this we can calculate the pH as
$$\mathrm{pH} = -\log 4x = 4.05 $$
Checking the solution
Upon adding nickel, the pH turned a bit more acidic. That makes sense as a micromolar concentration of nickel complex formed, releasing some more protons from HCN. We can calculate the cyanide concentration from either one of the given equilibrium constants, and verify that it is miniscule. Then, we can verify that all reactions are at equilibrium (at pH 4, hydroxide is a minor species as well, and has little impact on the exact solution).

Very nice answer!

My approach is the following. There is exactly enough $\ce{Ni^{2+}}$ ion and CN- ions for making up the ion $\ce{Ni(CN)4^2-}$. As the stability constant of this complex is huge, and HCN is an extremely weak acid. I would admit that the rare HCN molecules remaining in the solution have no effect on the pH of water, which must be 7.

If most of the original $\ce{Ni^{2+}}$ is converted to $\ce{Ni(CN)4^2-}$ then the solution must be very acidic.

I had a different approach to the problem which I have highlighted below
Notes :

Initially, I have considered the entire reagents to have reacted and
then considered the equilibrium to set in
I have considered two equilibria simultaneously, one involving the dissociation of the Nickel complex and also the hydrolysis of the Cyanide ion because it is a strong conjugate base of HCN which is a weak acid

Hope this answers your question.

Why the downvote? Is there something wrong with my approach?

The reactants are $\ce{NiCl2}$ and $\ce{HCN}$. Assuming complete reaction, the equation would be $$\ce{4HCN + NiCl2 <=> Ni(CN)4^{2-} + 2Cl- + 4H+}$$

I downvoted because you calculated a pH that is basic. Something must be wrong with your approach because, as you said, HCN is a weak acid and there is no base otherwise. I suspect that if you calculate all of the concentrations, your charge balance will be off.

Yeah I realized that later. Really sorry

@Karsten Theis yeah the pH is coming basic because I considered only the cyanide ion contribution. If I would have considered the H+ like MaxW said, I'm pretty sure the numbers would work out. But thanks for the heads up

There is no harm in posting an incomplete answer that results in an incorrect numerical results. In fact, others can learn from it when comparing the different answers.


text:
Solvent Effects in Friedel–Crafts Reaction
While reading about the factors affecting the product of Friedel–Crafts acylation with acetyl chloride, I came across a reaction involving napthalene in two different solvents: in nitrobenzene substitution took place at the beta position; in carbon disulphide substitution took place at the alpha position.
What is the reason for this?

Alpha substitution gives the kinetic product while beta substitution gives the thermodynamic product. But exactly how the choice of solvent influences the pathway of substitution is quite interesting...

It should be related to shapes and effective volumes of the substrates and the solvents. Though I am aware that this is not a in deep explanation.

@Alchimista could you please elaborate. I'm just a beginner in organic so a thorough explanation of the concepts would be really beneficial for me

See this earlier answer and keep in mind that running the reaction in refluxing nitrobenzene (b.p. 210.9° C) will pump a lot more energy into the reaction than when the reaction is run in refluxing carbon disulphide (b.p. 46.3° C).

Now there is an answer

On Friedel-Crafts-acylation reactions, two types of Friedel-Crafts-acylation mechanisms, namely an ion pair mechanism and a dipolar ion mechanism, have been proposed (Ref.1).  Normal acylations are presumed to proceed via the ion Pair mechanism, which seems to be more important in sterically hindered reactions:

The product complexes with aluminum chloride at the end. This complex liberates the final product after addition of water in the final step as shown in following diagram:

The Friedel-Crafts acetylation of naphthalene has been extensively reported and frequently studied as of today (Ref.1). Basically, two isomeric ketones, 1-acetylnaphthalene and 2-acetylnaphthalene, can be prepared by the reaction of unsubstituted naphthalene with acetyl chloride and aluminum chloride. A variety of solvents have been used for this acylation, and the solvent effect on naphthalene acylation has been studied intensively (Ref.1). The product ratio of 1-acetylnaphthalene to 2-acetylnaphthalene is found to be solvent-dependent. For example, when the reaction was performed in non-polar solvents such as $\ce{CH2Cl2}$, $\ce{CH2ClCH2Cl}$, $\ce{CHCl3}$, or $\ce{CS2}$, the 1-acetylnaphthalene formation is preferred and when reaction is done in polar solvents such as nitrobenzene or $\ce{CH3NO2}$, the reaction has proceeded to give 2-acetylnaphthalene,  exclusively. This phenomenon has been explained considering reversibility of the reaction:

The aluminum chloride-acylaromatic complex is insoluble in non-polar solvents carbon disulfide, but very soluble in polar solvent such as nitrobenzene. The 1-acetylnaphthalene-$\ce{AlCl3}$ complex can form rapidly (kinetic product) and it would precipitate from the solution, preventing further reaction. On the other hand, the 1-acetylnaphthalene-$\ce{AlCl3}$ is soluble in the polar solvents such as nitrobenzene. Thus, the reaction proceeds further and 1-acetylnaphthalene would deacetylate, and the slower 2-acylation reaction may occur to give more stable thermodynamic product (Ref.2).

REFERENCES:

P. H. Gore, “The Friedel-Crafts Acylation Reaction and its Application to Polycyclic Aromatic Hydrocarbons,” Chem. Rev. 1955, 55(2), 229-281 (https://doi.org/10.1021/cr50002a001).
Gary Warren Keen, "The acetylation of naphthalene and the formation of (chlorovinyl)naphthalenes therein," The Ph.D. Dissertation, Oklahoma State University, Stillwater, OK, 1976.
Wenpeng Li, Haibo Jin, Suohe Yang, Xiaoyan Guo, Guangxiang He, Rongyue Zhang, “An environmentally friendly acylation reaction of 2-methylnaphthalene in solvent-free condition in a micro-channel reactor,” Green Processing and Synthesis 2019, 8, 474–479 ( https://doi.org/10.1515/gps-2019-0015).


text:
Synthesis methods of inorganic materials
On our exams we are asked to determine which synthesis method is best suited for a given compound. We have discussed direct synthesis methods such as high temp. and ampull, and indirect synthesis methods such as solvothermal, sol-gel, and ion exchange.
I thought that the direct high temperature synthesis was mostly used for compounds with three or more elements such as $\ce{BaTiO3}$. However, in our exam the compound $\ce{Fe3O4}$ is said to be synthesized by the direct high temperature method.
For example, one question states:

Propose and motivate two methods (each) of synthesis of $\ce{NaAlSiO4}$ and $\ce{Cu(CN)(pyr)2}$

I believe one method for $\ce{NaAlSiO4}$ should be direct high temp. because that method is applicable for aluminosilicates but have no idea what the second one could be. Is there a specific way to determine the synthesis methods of different compounds just by looking at them?
All help is truly appreciated!

I'd say, no, it's impossible to say which method exactly would work the best, you can only make general assumptions such as organic ligands or hydroxo-salts won't survive a high-temperature synthesis. Metal oxides, on the other hand, can be synthesized by a variety of low- and high-temp methods and their combinations (see e.g. Pechini method).

The synthesis method depends on your preferred starting material. What do you propose for both of them?

@Karl For Cu(CN)(pyr)$_2$ I don't really have any idea on what the methods could be. For NaAlSiO$_4$ the starting materials would be NaAlO$_2$ and SiO$_2$. How does that help determine the method?

Your list of methods has a few missing, e.g. precipitation from solution and transport reaction in a temperature gradient.

Yup. Both NaAlO2 and SiO2 have very high melting points. Do you have an oven and crucible that goes that high? I say no. So you need to get the necessary temperature down. Ideas? And the starting materials for the other one are rather obvious, think simple!

@Karl I only listed the methods we have discussed in class. So for NaAlSiO$_4$ I should not use high temperature methods but low? Maybe, Cu(CN) and (pyr)$_2$ then?

You are not doing this systematically. High/lo temp. is totally insufficient to categorise those methods. You cant do the alsilicate in the melt. You cant do it at STP in solution, because everything is dead insoluble. Solid state would take centuries. How can you get oxides in solution?


text:
How does ammonium chloride increase the solubility of magnesium hydroxide in water?
In my textbook it is written that $\ce{Mg(OH)2}$ is slightly soluble in water, and it forms a milky solution which is called "milk of magnesia".
Its solubility can be increased by the addition of $\ce{NH4Cl}.$ But how? There seems to be no reason.

Is it magnesium(II) hydroxide when that is the only known magnesium hydroxide?

I don't know mag( II ) hydroxide...someone has edited my question...I only typed magnesium hydroxide.

There is a strong reason. But it is the chemical reaction, not just a better dissolution.
$\ce{Mg(OH)2}$ is a base with the limited solubility, defined by $K_\mathrm{sp}=[\ce{Mg^2+}][\ce{OH-}]^2$
$\ce{NH4+}$ ion, created by $\ce{NH4Cl}$ dissolution, acts as a  weak acid:
$$\ce{NH4+ + H2O <<=> NH3 + H3O+}$$
with $\mathrm{p}K_\mathrm{a}=9.25$
$\ce{OH-}$ ions formed by dissolution of $\ce{Mg(OH)2}$ are eliminated by recombination
$$\ce{OH- + H3O+ <=>> 2 H2O}$$
what supports the dissolution by keeping the product of ion concentrations below the $K_\mathrm{sp}$.
Effectively, there is ongoing equilibrium:
$$\ce{Mg(OH)2 v + 2 NH4+ <=> Mg^2+ + 2 NH3 + 2 H2O}$$

This is a well known problem in qualitative analysis. When you add $\ce{NH4Cl}$ to a solution containing $\ce{OH-}$ ions, you produce the reaction:
$$\ce{NH4+ + OH- -> NH3 + H2O}$$
The result is that the concentration of $\ce{OH-}$ decreases. If this operation was done in a saturated solution of $\ce{Mg(OH)2}$, the solubility product is no more obtained. A greater amount of $\ce{Mg(OH)2}$ can pass into solution. As a consequence, the solubility of $\ce{Mg(OH)2}$ is increased.


text:
Halogenation of 4‐methylcyclohex‐1‐ene

My problem is with the methyl group. If it wasn't there, I had no problem.
When the halonium bridge is made does the methyl group shift to give a more stable transition state or not?

I would believe that Br is attached mostly on the Carbon 4, and Br or OH on the Carbon 3.

Why? I would expect a mixture close to 1:1

In order for a methyl shift to occur, there would need to be an empty orbital resulting from a carbocation intermediate for the methyl group to occupy.

Note that in the bromonium intermediate shown above, neither carbon in the three-membered ring has an entire empty orbital available. Hence methyl shift doesn't occur and the mechanism continues as expected, resulting in halohydrin formation.

That's not quite true; alkyl shifts can occur with simultaneous loss of a leaving group, cf. pinacol/semipinacol-type rearrangements. The C–LG σ* orbital would play the same role as the empty orbital which you are speaking of.

@orthocresol Purely out of curiosity, would a methyl-shifted product be more favorable than the above scheme? I’m not far enough in my organic chemistry curricula to know the correct answer.

The methyl would have to shift over two carbons. Or more precisely, you would need a hydride shift followed by a methyl shift. I find that rather unlikely, so I agree with Waylander (comment on another answer) that there is no selectivity in this reaction.

methyl groups will not shift in a non-classical carbocation mechanism. This reaction will first have a $TS$ where the $\ce{Br2}$ molecule makes a 3 membered ring with the $\pi%$ bond and the water molecule  will attach itself to $\ce{C3}$, as that is a more stable electron-deficient site due to the inductive effect of the methyl group.

Your answer would be fine for cyclohexene but you omit regiochemistry. An inductive effect of methyl is not sufficient. A Reaxys search gave no hits but 7 for direct epoxidation with no sign of facial selectivity. See https://chemistry.stackexchange.com/questions/86341/why-does-water-attack-the-more-substituted-carbon-in-oxymercuration for a guideline on how to think about such a problem.


text:
How do bonds form without violating the second law of thermodynamics
It might be obvious that entropy related to reactions always increases and I understand (somewhat) why this happens. My question is why if you have two hydrogen atoms in a vacuum without being bonded to a molecule they would form a bond. Does that not show a movement from chaos to order and thus a decrease in entropy?

Note that 2 isolated hydrogen atoms will not form a stable hydrogen molecule, unless they are able to dissipate the energy, released by the bond formation. It can be done e.g.  passing mechanical energy to neighbour molecules, or emitting a photon.
In a macro scale, a system can decrease its entropy, if its Gibbs energy decreases, as it means the dissipated heat causes the bigger entropy increase of the surrounding, than is the entropy decrease of the system, so the total entropy increases.
$$\Delta G = \Delta H - T \cdot \Delta S$$
If $\Delta G_\mathrm{sys} \lt 0$ then $\Delta S_\mathrm{tot}=\Delta S_\mathrm{sys} + \Delta S_\mathrm{surr} \gt 0 $
as $\Delta S_\mathrm{sys} + \Delta S_\mathrm{surr}=\Delta S_\mathrm{sys} - \frac{(\Delta G_\mathrm{sys} + T \cdot \Delta S_\mathrm{sys})}{T}=-\frac {\Delta G_\mathrm{sys}}{T}$


text:
Modifications and alterations in the state of chemical equilibrium
Let us consider that we have a mixture of gases, considered ideal, that is in balance. Like, for example,
$$3\ce{NO} (g) \rightleftharpoons \ce{N2O} (g) + \ce{NO2}(g)$$
If we consult the thermodynamic tables, we can appreciate that the reaction is exothermic and exergonic.
This thermodynamic state of chemical equilibrium can be altered depending on various parameters, such as temperature or pressure.

Modification of the temperature: If we increase the temperature, the system moves in the endothermic direction, that is, towards the formation of products. This rule is well known, but I would like to be able to justify it by mathematical reasoning. I have thought about Van't Hoff's equation, but since I do not know the value of the equilibrium constant at a certain temperature, I have not been able to use it.
Modification of the volume. This is where I have the most doubts. I need to determine which way the system is moving. I know that the system moves to where it is least bothered, in this case, to the right. But, again, I wonder if this can be justified mathematically. I seem to remember my physical chemistry teacher doing it based on the definition of K (considering partial and constant pressures of fugacity)
If a non-inert (reactive) gas is added to the equilibrium mixture Here I know that there are several cases to distinguish:
3a) If the total volume is not modified
3b) If the total pressure remains unchanged
3c) If the temperature remains unchanged
3d) If the intensive variables can be modified
But, really, this one I have no idea how to reason with.
If we add an inert gas, any noble gas, I'm not sure what to do either.

I know the question is a long one. But, I have tried to reason out everything I know about the subject (basically, Le Châtelier's principle). But, I need to justify it on the basis of mathematical reasoning, since my teacher has told me that, in some cases (which are not many), Le Châtelier fails.

Le Châtelier is an universal principle. Did your teacher give one valid counterexample?

What do you want to get ? What do you mean by mathematical reasoning ? By the way you have chosen a poor example. I am pretty sure that NO will never react to produce N2O and NO2, whatever your thermodynamics calculations.

@Maurice - at high pressures, this reaction is well known to occur. See, for example, https://www.ncbi.nlm.nih.gov/pubmed/12653205

@Karl  The addition of nitrogen gas, at constant temperature and pressure, to an equilibrium mixture containing N2, H2 and NH3

@aprendiendo-a-programar I read about that one now, too. If you add nitrogen at constant pressure, you lower the partial pressure of hydrogen so much that the reaction runs backwards. Quite trivial, and totally expected from LeChatelier.

nother example cited by Cheung (2004) is the dimerisation of NO2. Consider an equilibrium mixture of NO2 and N2O4 in a sealed container. When we raise the temperature, we may expect that the position of equilibrium to shift to the left since the backward reaction to form NO2 is endothermic. However, there is also an increase in the total pressure of the system when the mixture is heated. Again, by LCP, the pressure increase would likely shift the position of equilibrium to the right to the side with less moles of gaseous molecules. @Karl


text:
How would I synthesize ethyl 2‐oxocyclohexane‐1‐carboxylate from cyclohexanone?

I thought the 1st step would be deprotonation of a H atom to form the enolate. Then an aldol reaction, but the product I end up with is the alcohol, not the ketone.

Well, you need suitable reagent, to get such product...

Read this: https://pubs.acs.org/doi/abs/10.1021/ja01853a065.

You could use ethyl chloroformate (EtOC(O)Cl) to make the desired product, I would suggest that you react the cyclohexanone with sodium ethoxide in cold ethanol before adding the electrophile to it.

As you will see if you look at the reference mentioned by @Mathew-Mahindaratne,   the standard conditions for this transformation is NaH/KH in THF at reflux with diethyl carbonate (EtOCOOEt) with an acetic acid workup.
The Organic Syntheses procedure here uses dimethyl carbonate, but will be equally applicable for diethyl carbonate.

Diethyl carbonate can be sluggish in this reaction. The classic procedure is to use the more reactive diethyl oxalate instead and then decarbonylate carbon monoxide from the codensation product by heating over crushed soft glass. This reaction was discovered by taking the melting point of a solid condensation product in a soft glass melting point capillary.


text:
Will aluminum continue to corrode after its surface is oxidized?
After doing some reading, it is common to read that aluminum is corroded by oxygen, but that the thin layer of aluminum oxide protects the internal metal from further damage. How much does this layer of aluminum oxide protect the underlying layers? Does it completely seal off the aluminum or is there still a slow degradation compared to a metal like iron?

It passivates, so reaction stops with very thin layer.

Yes, the aluminum oxide coating is generally protective, but not always. For example, in the presence of $\ce{NaCl}$ (like in spray from ocean water), it is reported that aluminum alloys in coastal communities undergo significant corrosion.
Here is a reference to quote:

When salt air and salt water come into contact with aluminum they can cause both the chalky, white coating of aluminum oxide and unpleasant pitting. Fortunately, there’s an easy way to protect aluminum from salt water and prevent unsightly corrosion: a powder coating.

A more precise discussion of chemical intermediates is outlined in a paper by Foley and Nguyen [1].
I also recall a comment on the slow removal of the passive $\ce{Al2O3}$ layer in the presence of vinegar.
So, I do not recommend leaving a salt and vinegar dressing in contact with an aluminum foil-wrapped sandwich for days before eating.
References

Foley, R. T.; Nguyen, T. H. The Chemical Nature of Aluminum Corrosion V . Energy Transfer in Aluminum Dissolution. J. Electrochem. Soc. 1982, 129 (3), 464–467. DOI: 10.1149/1.2123881.

No. Aluminium stops corroding once a strong oxide layer is formed.
Aluminium is a fairly reactive metal. The reason why it usually survives in air and water under normal conditions is because the aluminium oxide layer on the surface is very strong and normally protects the metal from further attack. Even if the layer is disrupted, it reforms quickly preventing any further reaction. Aluminium oxide is a very strong compound and impervious to water and oxygen. In bulk form it is known as corundum and appears in nature as ruby and sapphire, just about the hardest non-diamond minerals.
But there are substances that can disrupt that layer and expose aluminium to attack from air and water. Mercury and its salts and gallium metal both do this. Once they penetrate the surface layer, aluminium will be destroyed fairly rapidly and will crumble to a weak bulky mess. See examples on youtube from NurdRage and NileRed.


text:
Reaction mechanism of lithiated benzene and oxygen in acid
According to (1) addition of O2 in acid to lithiated benzene makes phenol.
I'm trying to deduce the mechanism, and somehow it doesn't feel right but I'm not sure.

(1) Snieckus, V. Directed ortho metalation. Tertiary amide and O-carbamate directors in synthetic strategies for polysubstituted aromatics. Chem. Rev. 1990, 90 (6), 879–933. DOI: 10.1021/cr00104a001.

Addition of oxygen followed by acid work up, not oxygen in acid

Your proposed mechanism doesn't quite work; loss of H2O would lead to a six-electron oxygen species (that should already set off alarm bells), and to get back its octet, that oxygen would have to pick up the equivalent of a hydride, not a proton (from water). Based on the original paper https://pubs.acs.org/doi/abs/10.1021/jo00380a034 that Snieckus cites there isn't any attempt to generate singlet O2, so I imagine that the mechanism is (at least partly) radical, but I don't have a concrete suggestion in mind.

In addition, @Waylander's comment is a very salient point which you should pay more attention to, imo: it is oxygen followed by acid, not at the same time (that would be rather pointless, as adding any kind of acidic solution - $\ce{O2}$ or not - to an organolithium would simply protonate it). So we presumably need to react $\ce{ArLi}$ with $\ce{O2}$ (with a radical mechanism) to get to a phenoxide $\ce{ArO-}$, which upon workup forms the phenol $\ce{ArOH}$. Not trivial...

@orthocresol: There is the possibility that the aryl hydroperoxide generated during workup oxidizes TMEDA. but I am not convinced.


text:
Why are moles of compounds have only 6.022x10 23 atoms?
I am a bit stuck on a basic chemistry concept. I understand that the Avogadro number indicates the number of chemical entities it has (ion, molecule or atom) What I don't understand is that when you add the masses of all of the elements in a compound, that it will equal to 6.022x10 atoms of that substance - why wouldn't there be more? For example shouldn't H2O have 3 X avogadros number (2X Hydrogen moles of 6.022x1023 and 1X mole of O).
Thanks!

Avagadro's number signifies the number of representative particles, so in the case of water, you still have Avogadro's number of representative particles (water is treated as your representative particle in this case).

1 mol of water does indeed have $3 \times 6.022 \times 10^{23}$ atoms.

No one mole of water contains three moles of atoms, two moles of hydrogen atoms and one mole of oxygen atoms

One mole of a compound does not have $\ 6.02· 10^{23}$ atoms, as you say.
One mole of a compound has $\ 6.02· 10^{23}$ molecules
One mole of water has $\ 6.02· 10^{23}$ molecules H2O, and $\ 3 · 6.02· 10^{23}$ atoms.
One mole of H2O has $\ 6.02· 10^{23}$ atoms O, and $\ 2·6.02· 10^{23}$ atoms H.

The answer might benefit from mentioning the new definition of mole and Avogadro's constant. Also, note that $N_\mathrm{A} = \pu{6.02214076E23 mol-1}$ (exactly), so, strictly speaking, $\pu{6.02E23}$ is just some number that is not Avogadro's constant. It's better to either stick with the exact value (as currently defined), or use the symbol $N_\mathrm{A}$ for the constant, e.g. "One mole of water has $N_\mathrm{A}$ oxygen atoms and $2N_\mathrm{A}$ hydrogen atoms." Since the question is focused on this very topic, one must be precise.


text:
In UV photoelectron spectroscopy why are filled orbitals seen at a higher binding energy?

I understand that the partially filled Ni intersects the Fermi level as there is a change from filled to unfilled orbitals. But what is the explanation for the lower binding energy of the filled 3d Cu orbital. Surely there would be greater electron-electron repulsion resulting in higher kinetic energy of the electrons due to more shielding, which corresponds to a lower binding energy?


text:
Can the solubility of a compound be used to estimate the degree of separation of a solute from a solution?
I'm confused exactly on how to use the concept of solubility to get the amount of separation from a certain solute in the solution.
This arises from attempting to solve a problem regarding this matter. The problem described is as follows:

$400$ grams of anhydrous sodium sulphate ($Na_{2}SO_{4}$) is dissolved
  in a liter of hot water. The solution is then let to cool carefully
  until reaching $20^{\circ}C$ to remain supersaturated with respect to
  the formed decahydrate $Na_{2}SO_{4}\cdot 10H_{2}O$. Then a small
  crystal of the latter salt is added to the solution, separating the
  excess of $Na_{2}SO_{4}$ dissolved, remaining a saturated solution. It
  is known that the saturated solution is equivalent to $19.4$ grams of
  $Na_{2}SO_{4}$ by $100\,mL$ of water. What amount of the decahydrate
  would had been separated?

The given alternatives are:
$\begin{array}{ll}
1.&249\,g\,\textrm{to}\,259\,g\\
2.&318\,g\,\textrm{to}\,327\,g\\
3.&689\,g\,\textrm{to}\,698\,g\\
4.&721\,g\,\textrm{to}\,730\,g\\
5.&890\,g\,\textrm{to}\,899\,g\\
\end{array}$
What I assumed is that:
$400\,g\,Na_{2}SO_{4}\times\frac{142+180\,g\,Na_2SO_{4}\cdot 10 H_{2}O}{142\,g\,Na_{2}SO_{4}}-100\,mL\frac{19.4\,g}{100\,mL}\approx 807\,g$
Therefore that would be the grams of sodium sulphate decahydrate but I'm not sure if that would be the ammount. I'm confused why do the alternatives features a range?. How does it appear that?. Can somebody explain how exactly to get to that given range?.

I don't understand this part -- "Then a small sack of the latter salt is added to the solution, separating the excess of Na2SO4 by 100mL of water."

@MaxW I am also a bit confused about the intended meaning of the author. Since it is not indicated the ammount of sodium sulphate which was added, I don't know the reason for the inclusion of that information.

Then a small sack of the latter salt is added to the solution,  so some $\ce{Na2SO4\cdot 10H2O}$ is added to the solution to nucleate precipitation separating the excess of $\ce{Na2SO4}$ of course $\ce{Na2SO4}$ was originally added to the solution but  its is of course the decahydrate that is precipitating by 100mL of water. this is the part that absolutely baffles me...

@MaxW Does ignoring this portion (about adding the decahydrate) of the problem would help into solving it?. Its been days and I still can't get a clear answer. Needless to say that it seems that wikipedia data for the solubility of that decahydrate is not correct. Perhaps do you have that value or help me with this problem?.

@MaxW It turns out that you were right. This question was incorrectly typed thus sounded too strange. I went to the original source and found there was a missing piece of information and corrected the typo which caused confusion. As it stands now can this problem be solved?.

$400 g$ $\ce{Na2SO4}$ contains $400/124$ = $3.225$ mole of $\ce{Na2SO4}$. As one mole $\ce{Na2SO4}$ will produce $1$ mole decahydrate $\ce{Na2SO4}· \ce{10H_2O} $ , one deduces that $3.225$ mole $\ce{Na2SO4}· \ce{10H_2O} $ weighs $3.225 · 322 g = 1038g $. But for doing this salt, $1038 - 400 = 638$ g water is necessary. Remains in solution $1000 - 638 = 362$ g water
As $100$ mL water may dissolve $44$ g decahydrate, $362 $ g water will dissolve $160$ g decahydrate. The excess will be deposited in the container, namely $1038 g - 160 g = 878$ g

The information it is a bit condensed. Can you split the steps?.Why do you subtract the grams of the sodium sulphate decahydrate to the weight of sodium sulphate to obtain the water?.The part where I'm more stuck is is why do you subtract $1000-638$? what's the meaning of $362,g$ of water? what does it represent?. For the second step where you say $100,mL$ may dissolve $44,g$ of decahydrate and then $362,g$ water will dissolve $160,g$ decahydrate how did you obtained this number?.Is it from doing this?, $362,g\textrm{water}\times\frac{44,g}{100,g\textrm{water}}=159.28,g$

I'm not very sure about value of the solubility of sodium decahydrate in water at $20^{\circ}C$. Can you check this data please?. Why in the given alternatives does it appear two values for the amount of separation for that salt?. What was the intended meaning?. None of the answers seem to match with what you obtained but only the fifth option does seem to get very close. Can you help me with this part?.

Hi. Sorry, but can you attend my questions in the comments? I'm still doubtful about those.

The original question was incorrectly stated. I went to the original source and found that mistake and corrected it. If you look at your arithmetic the division is not right. Can you fix that according to the updated information?.

Assume:

Water has a constant density of 1.000 g/ml from $20 ^\circ\pu{C}$ to $100 ^\circ\pu{C}$.

Given:

$400\pu{g}\ \ce{Na2SO4}$ contains $400/142.04 = 2.8161$ mole of $\ce{Na2SO4}$.
$1000\ \pu{ml}\ \ce{H2O} = 1000\ \pu{g}\ \ce{H2O} = 1000/18.015 = 55.509\ \pu{moles}\ \ce{H2O}$
$19.4\ \pu{grams}$ of $\ce{Na2SO4}$ per $100\pu{mL}$ of water $= 19.4/142.04$ or $0.13658\ \pu{moles}$ of $\ce{Na2SO4}$ per $100/18.015$ or $5.5509$ moles of water. Thus in solution there are $0.13658/5.5509=0.024605\ \pu{moles}$ of $\ce{Na2SO4}$ per mole of water.

Let:

$x =$ number of moles of $\ce{Na2SO4\cdot 10H2O}$ formed
$2.8161 - x =$ number of moles of $\ce{Na2SO4}$ still in solution.
$10x = $ number of moles of water in in $\ce{Na2SO4\cdot 10H2O}$
$55.509 - 10x =$ number of moles of water still in solution

Thus:
$2.8161 - x =(55.509 - 10x)\times0.024605$
$2.8161 - x = 1.3658  - 0.24605x$
$1.4503 = 0.75395x$
$x = 1.9236$
Grams $\ce{Na2SO4\cdot 10H2O} = 1.9236\times322.20 = 620\ \pu{grams}$

The assumption about the density of water isn't valid of course, but it is within about 4%. 620*1.04 = 651 which is still not within one of the mass ranges given.

The issue which still confuses me is why on earth ranges are given as an answer? Suspiciously all the alternatives are separated by a margin of 9 grams. Does it exist a justification for this? You used the density of water to get the mass in one liter and moles, but im stuck on the justification of the steps. What exactly is happening in lets say a beaker? You dissolve the salt and some part is transformed into the decahydrate form?  But when you add a crystal what does it change there?

More decahydrate is formed? Does adding this crystal acts as a magnet to get all the possible decahydrate from the supersaturated solution? Can you illustrate this process to me because im confused on why you did those steps. The confusion i also have is the thing about supersaturated solution. In this case the precipitate is the decahydrate form or is it the anhydrous form? What's in the solution is the decahydrate form or is it the anhydrous form? Can you help me with this part please?

@ChrisSteinbeckBell - The cooled solution is super saturated with  $\ce{Na2SO4}$. The added crystal of $\ce{Na2SO4\cdot 10H2O}$ is added to nucleate precipitation of  $\ce{Na2SO4\cdot 10H2O}$ from the solution. After the precipitation is complete the solution is saturated with  $\ce{Na2SO4}$

@ChrisSteinbeckBell - I don't know why the choices are given as ranges.

@ChrisSteinbeckBell - I used the density of water but I could have worked the problem using volumes. However since no temperature was given for the hot water there would have been an implicit assumption that the water's volume didn't change. Stating that the density is constant makes clear where an poor assumption must be made.


text:
Which bond is stronger?
Are ionic bonds stronger than covalent bonds?
I googled it and found various answers, some saying it is stronger and some saying it’s not.

1 covalent bond> 1 ionic bond. But ionic comounds have higher boling points than covalent compounds as we need to break an infinite number of ionic bonds due to its lattice structure whereas a finite number of bounds need to be broken in the case of covalent compounds.

there can be no general answer here. Some covalent bonds are very strong and some ionic bonds are very weak even when you can agree a consistent definition of strength.

https://chemistry.stackexchange.com/questions/11048/are-metallic-ionic-bonds-weaker-than-covalent-bonds

Dora read all the answers except one. I just invite you to consider that often ionic bonds are considered weaker because they can be broke by high dielectric constant media, like water. Also in practice we deal with ionic solids rather than isolated C+A- so reticular energy must be taken in account. The two types of bonds can overlap. Note the final part of @matt_black answer.

Generalisations are impossible. There are many examples of strong and weak bonds in both ionic and covalent compounds
There are a number of problems with the question. One is that there bonding is not an either/or concept: there is a something of a continuum between "pure" ionic and "pure" covalent bonding. There is also a lot of confusion about what sort of bonds make a solid. Many "covalent" solids are made of molecules with covalent bonds but the solid exists because of weaker forces between the individual molecules.
Even if we ignore molecular solids the bulk properties are determined by the weak bonds between molecules, there are a wide variety of different behaviours. It is often assumed that ionic bonding is "stronger" because many familiar ionic compounds (like sodium chloride) are high melting-point solids. But many compounds with networks of covalent bonding are also refractory solids, like diamond. Some even show both weak and strong bonds in the same compound (graphite has weak bonds between the planes of carbon hexagons but the hexagonal planes are very strongly bonded). So, both ionic and covalent examples exhibit very strong bonds.
But the opposite is also true. Ionic compounds with large, floppy ions are often liquids suggesting that "weak" bonds are possible in ionic compounds. Examples like 1-butyl-3-methylimidazolium hexafluorophosphate ([BMIM]PF6) are liquids at room temperature.
Also some covalent compounds have such weak bonds that they can only be observed at very low temperatures. Some argon covalently bonded compounds have only been observed in frozen argon.
The point being that you can't generalise. Some ionic compounds have strong bonds and others have weak bonds. Some covalent bonds are very strong, some are very weak.

Part of the reason for many answers is that "stronger" can be different things. If we consider a typical two-electron bond between two atoms A and B, it can break in three ways:
1) 1 electron can stay with A and one with B: $\ce{A-B -> A. + B.}$
This is called homolytic bond cleavage.
2) Both electrons can stay with A: $\ce{A-B -> A- + B+}$
This is called heterolytic bond cleavage.
3) Both electrons can stay with B: $\ce{A-B -> A+ + B-}$
This is another example of heterolytic bond cleavage.
If you look at tables of "bond strength", they usually report the enthalpy change for the homolytic cleavage reaction indicated above.
That value is typically larger for ionic bonds than covalent bonds, because ionic bonds form when both electrons have more density on the same atom in the bond. In order to break the bond homolytically, one electron has to go up substantially in energy to stay with the more positive atom.
In contrast, in a covalent bond, the electrons are shared nearly equally between the atoms, so, while bond breaking still takes a lot of energy, it is easier to end with one electron on each atom.
If, on the other hand, we look at the heterolytic cleavage, ionic bonds are much easier to break, since they already have both electrons mostly on one atom. The major energetic cost of breaking the bond is just the energy required to separate the charges from one another. For the covalent bond, we have the same cost of charge separation, but additionally we need to force both electrons onto one atom.

Usually ionic bonds are stronger than covalent bonds. But there are exceptions. Quartz SiO2 for example is made of covalent bonds, and it melts at very high temperature (> 1400°C). I am afraid there are no general rules.

Covalent bond is defined as the sharing of electrons between two atoms (non metals). Ionic bond is defined as the transfer of electrons from the valence shell of anion(s) to the valence shell of cation(s).
On a covalent bond, there are two types of forces that held the atoms together; (1) intramolecular forces and (2) intermolecular forces. Intermolecular forces are quite weak compare to the intramolecular forces. Also, the atoms on the molecules are not properly held by the intramolecular forces. Covalent bonds can be observed usually on gases, liquids and low melting solids.
On the other hand, anions and cations are held together by electrostatic forces that are usually strong. Ionic bonds can be observed on the solids at room temperature which have high melting points.
Ionic bond is stronger than covalent bond, however, other compounds demonstrate different properties of the bond due to various factors, such as environmental conditions, electronegativity, bond energy, orbitals, etc.

So basically, it varies?

Yes. The strength of ionic bonds depends on the charge and radius of ion, while the strength of covalent bonds depends on the distance of the atoms and difference on the electronegativity.

Please wait a bit before accepting an answer as it might turn out that it is wrong or not particularly good, as in this case.

This is incredibly broad, fuzzy, oversimplified, but only where it isn't completely wrong. @dora you should look at other answers, or wait for them; this will not do you any good.


text:
Adiabatic compression of an ideal monatomic gas. What would the change in entropy be?
Two moles of an ideal monatomic gas initially at 300K and a pressure of 2 atm is compressed adiabatically using a constant external pressure $P_{ext}$ until equilibrium is reached. The temperature of the gas is then 400K
a) calculate w, q, $\Delta U$, and $\Delta H$ for this process
b) calculate the final equilibrium pressure $P_f$ of the gas (note $P_f =P_{ext}$)
c) calculate $\Delta S$for this process.
a) I calculated $\Delta U$ by $\Delta U= nC_v\Delta T$. Since it is an ideal gas, the $C_v= \frac{3 }{2} R$. From that I got $2494.2$ J for $\Delta U$. Then, since it's adiabatic, q = 0, thus w = $2494.2$ J. I calculated $\Delta H$ using $nC_p\Delta T$. As $C_p$ would be $\frac{5}{ 2} R$, $\Delta H$ is $4157$ J.
b) I used the ideal gas law $V_2= \frac{nRT_2}{P_2} $and subbed it into $w= -P_{ext}(V_2-V_1)$ and as the question says$ P_f  (P_2)= P_{ext}$ I managed to eliminate some variables to end up with a $P_{ext}$ of 3.68 atm.
c) I don't know how to do this. I'm confused as to what formula to use for this. We can't use the $\frac{q_{rev}}{T}$ one since it doesn't state in the question if this is a reversible process. There are 2 other ones I learned, but they can only be used when it's isothermal or constant volume/constant pressure. They are:

You wrote "it doesn't state in the question if this is a reversible process".  Actually, you know it can't be reversible, because reversibility requires P_ext = P_int, while in this case, P_ext is a constant, yet P_int must be increasing.

Ah I see, however we still can't use Qrev/T since that formula can only be used for reversible processes. So I'm still wondering how to calculate delta S?

Go to https://en.wikipedia.org/wiki/Entropy and, from the Contents section, go to sec. 6.2 ( "Cooling and heating").  At the end of that section you will see more general equations that allow you to calculate delta_S  for processes involving an ideal gas in which either both T and V, or T and P, change.  You should also understand that, if your process were reversible, the fact that it's also adiabatic means delta_S would always be zero, because q_rev = 0.

Entropy is a state function so as long as you end up where you need to end up, you can choose how to get there, i.e. combine two processes for which you know the entropy change. https://memegen.link/custom/state_functions.../are_path--independent.jpg?alt=https://iili.io/H0DOYb.jpg


text:
Electrolysis of aluminum in a solution of magnesium sulfate
So I recently started looking into electrolysis to produce chemicals and I have only produced a few chemicals from it like copper hydroxide and copper sulfate. However I can’t seem to figure out what this wight precipitate that is produced in a electrolysis between  a aluminum cathode and a aluminum anode in a solution of magnesium sulfate.I am relatively new at chemistry and I am only in high school so I can’t figure out this precipitate. I would also like to know if there is a new dissolved salt being formed. Thanks in advance all ideas are much appreciated.

Magnesium sulfate ionizes to Mg++ and SO4-- ions in water. When current is passed thru a solution of these ions, sulfate ion will be electrolyzed at the anode, giving O2 and SO3 which becomes H2SO4. The pH at the anode will decrease as acid is produced and attacks the aluminum. A thick layer may build up. This is the basis for anodization of aluminum https://en.wikipedia.org/wiki/Anodizing. I think any thickening of the aluminum oxide layer is quite adhesive, but some may fall off.
The cathode (negative electrode) attracts the Mg++ ions, and when they get their electrons, the magnesium atoms react with water to give H2 and Mg(OH)2. The magnesium hydroxide may stick to the cathode at first, but will eventually thicken and pieces/powder will fall off. The liquid around the cathode will become alkaline.
So the precipitate will be mostly Mg(OH)2, but there may be some aluminum oxides or hydroxides in it too.

Aluminum is not a good idea for carrying out electrolysis, because this metal is covered by a thin layer of aluminum oxide (or alumina) that is insulating. The current passing between these electrodes must be really weak, because the resistance of this layer is high. So that you need a high tension to break this layer and make an electrolysis. But, as soon as aluminium is in contact with water, a new aluminum layer is formed. The majority of the electric energy is wasted and lost for heating the solution. This is non-sense. The powder that you get is probably aluminum oxide coming from the broken alumina layer I told you about earlier. And magnesium is not involved in this reaction.

I have also seen a white precipitate in a galvanic cell scenario involving aluminum and magnesium sulfate.
My guess is $\ce{Mg[Al(OH)4]2}$ and not $\ce{Mg(OH)2}$ which can form a problematic white suspension (not observed).
An extract from Wikipedia:

There are many mixed oxides containing aluminium where there are no discrete or polymeric aluminate ions. The spinels with a generic formula $\ce{A^2+B^3+_2O^2-_4}$ that contain aluminium as $\ce{Al^3+},$ such as the mineral spinel itself, $\ce{MgAl2O4}$ are mixed oxides with cubic close packed $\ce{O}$ atoms and aluminium $\ce{Al^3+}$ in octahedral positions.


text:
Converting cyclic compounds to their corresponding aliphatic compounds
Is there any reaction available to convert cyclic compounds to their corresponding aliphatic compounds?
eg. Converting cyclohexane to n hexane

One way I can think of is converting a cyclic compound to a ketone using wacker process, then cleaving the ketone with conc. $\ce{HNO3}$ and reducing the aldehyde obtained using clemmensen or Wolff kishner reduction. Not a straightforward way but just my two cents.

Nothing straighforward, you're looking at multiple steps some with poor yields

This question have been asked and have been answered click on the link to view the answer: https://chemistry.stackexchange.com/questions/15170/converting-cyclic-compounds-to-linear-compounds-possible

The reaction you want to know about is "Cleavage reaction" in particular oxidative cleavage.
In this dihydroxylation, if the solution is warm or acidic or too concentrated, oxidative cleavage of the glycol may occur.
Then they can be reduced to aldehydes by reducing agents.Reduction of carboxylic acids to aldehydes is difficult because aldehydes are more reactive than carboxylic acids toward most reducing agents. Almost any reagent that reduces acids to aldehydes also reduces aldehydes to primary alcohols.
Lithium tri-tert-butoxyaluminum hydride, LiAlH(O-t-Bu)3 is a weaker reducing agent than lithium aluminum hydride. It reduces acid chlorides to aldehydes because acid chlorides are strongly activated toward nucleophilic addition of a hydride ion.
Under these conditions, the aldehyde reduces more slowly and can be isolated. Therefore, reduction of an acid to an aldehyde is a two-step process: Convert the acid to the acid chloride, then reduce using lithium tri-tert-butoxyaluminum hydride.

And now we can reduce it by the Wolff Kishner reduction.

Ethylene glycol, diethylene glycol, or another high-boiling solvent is used to facilitate the high temperature (140–200 °C) needed in the second step. I am assuming you can convert the cyclohexane to cyclohexene by any radical halogenation and then dehydrohalogenation.

How do I add citations or references?

For the published work such as a textbook we prefer to use ACS style. I could assist you by editing the citation in your post; which textbook you've used? Looks like Wade's Organic chemistry, however, knowing the edition would be nice. Also, I'm not an organic chemist, but this doesn't seem to answer the question IMO (or at least only the last sentence addresses it).


text:
R and T conformations in relation to Activators and Inhibitors; Km of reaction in relation to uncompetitive and non-competitive inhibitor
Do
-Activators
-Competitive Inhibitors
-Non-Competitive Inhibitors
-Uncompetitive Inhibitors
all change the conformation of the targeted enzyme from T to R (activator), and from R to T (the rest) respectively?
Also, why does the Km of a reaction decrease with an uncompetitive inhibitor, while remaining constant with a non-competitive inhibitor?
I have tried to find a coherent answer to this all across the internet, but to no avail. Many thanks!

Allosteric inhibitors changes the conformation of enzyme enzyme for preventing the substrate binding, and allosteric activators modify the enzyme conformation so that  substrate binding can take place.
Competitive inhibitors don't change the enzyme conformation, as these inhibitors compete with the substrate to bind at enzyme's active site.
Non-competitive inhibitors change the enzyme conformation, as they inhibit enzyme activity by binding to site other than active site and prevent the substrate binding by changing the active site's structure so that substrate can not bind to enzyme.
Uncompetitive inhibitors inhibit enzyme activity by binding to enzyme substrate complex and these inhibitors change the conformation of the enzyme which reduces the affinity of the substrate for the enzyme's active site.
Km decreases in uncompetitive inhibition because inhibitor binds to the enzyme-substrate complex and it becomes incredibly difficult for the substrate to leave the enzyme. This results in increased affinity and Km represents an increase in affinity.
(Via: https://courses.lumenlearning.com/boundless-microbiology/chapter/enzymes/
https://en.wikibooks.org/wiki/Structural_Biochemistry/Enzyme/Reversible_Inhibitors)


text:
Controlling rate of endothermic reaction
In an experiment, I have two endothermic reactions taking place simultaneously:
$$
\begin{align}
\ce{A + B &-> C + D}\tag{1}\\
\ce{E + G &-> Z + F}\tag{2}
\end{align}
$$
$\ce{C}$ and $\ce{Z}$ also take place in a reaction to form the final product of the process $\ce{J}$. In my experiment, I have to use a single heating source, and cannot change the concentration of the chemicals in $\ce{A + B}$ and $\ce{E + G}$ such that the production of $\ce{J}$ is as fast as possible. How can I find the optimum temperature for this reaction, and will it be useful?
It is my theory that
$$\frac{\text{rate}_1 + \text{rate}_2}{2}$$
can be found when both reactions take place in water heated by a single source, and the activation energies are known. However, there is a need for another factor other than concentration and heat which needs to ensure that the products $\ce{C}$ and $\ce{Z}$ are produced at such rates (independently and at different rates) which will ensure an efficient process.

Usually endothermic reactions don't produce one product out of two reactants. Here you seem to have two such endothermic reaction simultaneously. It is hard to believe. And why is it a zero order reaction ? Please be more explicit !

Do you know quantitatively the kinetics of the reactions involved, and the heats of the reactions?!


text:
Why does aluminum chloride have a higher lattice energy than aluminum fluoride?
From the table below (source: McMurry's Chemistry [1, p. 212]), it is evident that $\ce{AlCl_3}$ has a higher lattice energy than $\ce{AlF3},$ even though $\ce{F}$ is smaller than $\ce{Cl}$. Why is this so?

However, the melting point of $\ce{AlF3}$ $(\pu{1291 °C})$ is larger than that of $\ce{AlCl3}$ $(\pu{192.4 °C}).$ I think the huge melting point difference is because $\ce{AlCl3}$ is more covalent, while $\ce{AlF3}$ is more ionic. That being said, does lattice energy account for the covalent and ionic character of compounds?
References

McMurry, J.; Fay, R. C.; Robinson, J. K. Chemistry, 7th ed.; Pearson: Boston, 2015. ISBN 978-0-321-94317-0.

I think this solves the problem: https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.quora.com/What-is-trend-of-lattice-energy-in-periodic-table&ved=2ahUKEwjD5cifztjmAhUpDmMBHS3jDicQFjALegQIAhAB&usg=AOvVaw00vEFxX5wpb7NcPQP2l5J2

Please don't forget to properly cite the data sources (I added a reference to the textbook I happen to have, you probably have a different edition as the table style is slightly altered). Also, I corrected units typography; feel free to visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

See this site: https://www.chemguide.co.uk/physical/energetics/lattice.html

Lattice energy is a function of the radius ratio, which when tends to 1, improves the packing efficiency of the molecule. Here the radius ratio of the chloride ions and aluminium ions is much closer than the fluoride ions.
You could think of this from the born habers cycle also. Where the electron affinity of chlorine is higher, which leads to overall higher lattice energy. However, the radius ratio logic is much more convincing as the habers cycle logic would require the other parameters like the heat of formation also

"… packing efficiency of the molecule": what molecule are you talking about? Also, it's Born–Haber cycle, not "habers cycle".


text:
Is there a software that can predict the shape of a custom protein, given its amino acid sequence?
I know that the internet is full of enzyme and protein simulators which can show active site and 3D structure (like bioblender), but the problem with these programs is that they use predefined information and coordinates supplied by several databases (such as PDB ) , but actually I didn't find a software that can predict the shape of a custom protein with custom amino acid sequence. So, do you know such software?

It's not a trivial task, so the shortest answer is: there is no such software. Actually, a more complex answer would imply homology modeling and molecular dynamics, but the procedures are quite elaborate, and there are not "black box" pieces of software for that

+1 to @The_Vinz, very good answer. Would you consider committing to the Materials Modeling Stack Exchange? https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=ODMyMDc3YzliMTY4MzI1YzcyMjczOGNlYWQ3Yjk1MTkxM2Q3NGIyYzM1MzcwYzdmNmE3ZjhiYmYzNzA5Njg2N7YXbfH40JIpwIJ67zouTLfOm4fPVJM-PNoOsMCnOzzc0

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

There might one day be such software but I do not think it has been written yet, it is a very difficult thing to do. It might be better to obtain some of the protein and then use NMR or crystallography to determine the 3D shape.


text:
Competing half reactions in electrolysis of aqueous solution of sodium hydroxide
If you run a direct current using pure nickel electrodes through a solution of NaOH in distilled water, you will separate the water into hydrogen and oxygen.
Are there any other reactions that could possibly take place that would release anything other than hydrogen and oxygen?

You can certainly exploit overpotential

Unless you get rid of water, likely no. @Zenix please have a look at the Meta post regarding mobile links.

@andselisk Had I known that earlier... I will take care in future, sorry for the inconvenience.

@Zenix No prob, if this is more convenient for you, I guess it's OK to occasionally use mobile links for the comments. Just make sure you don't use them in the answers (comments are usually temporary, answers are meant to stay).

@andselisk for example in Castner Kellner, usage of Hg cathode helps in liberation of $\ce{O2}$ instead of $\ce{Cl2}$. Though there's no need to remove water (present in central compartment). But I must make it clear that I am currently not in touch of these things.

@Zenix Yeah, overpotential would have an effect if only we had, say, sodium chloride. Since it's NaOH, I don't think there is much that can deviate from traditional water splitting.


text:
Why is volume inversely proportional to pressure?
If temprature is directly proportional to volume (Charles's law) and
temperature is directly proportional to pressure (Gay-Lussac's law), then why is pressure and volume are inversely proportional?

Given you know and understand Charles' and Gay-Lussac's laws, it's not about chemistry, rather, simple ratios:
$$
\begin{cases}
T \propto V\\
T \propto p
\end{cases}
\implies
p \propto \frac 1 V
$$
which, as Zenix commented, is a math form of Boyle's law.


text:
Why is it important to remove the used up electrons in oxidative phosphorylation?
I get that oxygen is the terminal electron acceptor in oxidative phosphorylation.  But why is it essential to accept these electrons and remove them as $\ce{H2O}?$
Oxygen is essential for cellular respiration, but after learning about cellular respiration it seems oddly unimportant.  Why is having an efficient garbage collector (oxygen picking up the used electrons) so crucial to cellular respiration, and to ATP production?

Where would you put the said electrons otherwise?

I get that oxygen is the terminal electron acceptor in oxidative phosphorylation. But why is it essential to accept these electrons and remove them as H2O?

A redox reaction is made of two half reactions. Neither one of them is more or less important; the combination of both determines the energetics of the reaction.

Why is having an efficient garbage collector (oxygen picking up the used electrons) so crucial to cellular respiration, and to ATP production?

The electrons are not "used" as in an old car or worn-out clothes. There as shiny and new as before. Without oxygen, the reaction would not happen in the first place (electrons are transferred from less tightly bound to more tightly bound). The Gibbs energy of proton transport matches what is necessary for ATP synthesis, which in turn matches the Gibbs energy of the redox reaction. With a different electron acceptor (specificities of the enzymes involved aside), the proton gradient might be too weak to make ATP.

Oxidative phosphorylation consists of electron transport chain and chemiosmosis. During electron transport chain, electrons transfer causes the energy production and forms an electrochemical gradient. Then, during chemiosmosis, the energy produced during this electrochemical  gradient is used to synthesize ATP.
So, why oxygen is needed in electron transport chain. If oxygen is not present there to accept electrons, the electron transport chain will stop taking place, and ATP formation by chemiosmosis will stop. And lack of ATP will lead to cell death.

For some reason it seems less dramatic than i expected lol.  It seemed like I must have had something wrong.  But thanks for the response, I wasn't confident in that really being the right answer.  Oxygen is needed to prevent the ETC from getting clogged with used up electrons, thus continuing ATP production.  Which is definitely important.  Oxygen is one hell of a garbage truck.  Thanks again!


text:
Ionization trends in period 6
I know that there is a drop of Ionization energy in the second period with Nitrogen and Oxygen, but why does that not apply to bismiuth and polonium, which are in the same groups but different period.

Trends are, well, just trends. Basically all such ‘rules’ of things up/down/sideways/across the periodic table have problems somewhere.

Bi and Po have 6 shells however Nitrogen and Oxygen have 2 shells...
Due to bigger radius of bismuth and polonium there is no significant drop in the enthalpy.
Plus they both have almost similar mass number(difference of just 0.002)


text:
Why oxidation potential of an electrode equals negative reduction potential of the same electrode?
My textbook states that:

$E^\circ_\mathrm{ox} = -E^\circ_\mathrm{red}$ (of same electrode)

I know that electrons flow from anode half cell (oxidation) to cathode half cell (reduction).
But if $E^\circ_\mathrm{ox} = -E^\circ_\mathrm{red}$ in each electrode, then potential in each electrode of the galvanic cell should be 0 volts (oxidation cancels reduction).
Hence potential difference in the whole circuit is 0 volts (no electricity is generated).
Do not get me wrong, I am just a high school student and I know my question might sound trivial, I am new to electrochemistry.

Anode and cathode are not the same electrode (or I don't understand the question). I took a liberty to correct slang and notations to more standardized forms. Also, please note that citing an actual textbook instead of saying "My textbook" is a good practice.

@andselisk I am not saying they are the same electrode, you do not understand my question, let me clarify with an example if oxidation potential of zinc electrode = -0.76V then it's reduction potential = 0.76V

Exactly, that's why only standard reduction potentials are tabulated. There must be another electrode to complete the circuit.

There exist two methods for teaching electrochemistry, and of course two schools of teachers. The two methods are equivalent, but they are exactly the opposite of one another. I will take an example to explain it correctly : Let us speak of the Daniell cell Zn/Cu, which uses reduction potentials equal to -$0.76 $ V for Zn and +$0.34$ V for Cu, if the ions concentrations are 1 molar. In this case, the overall voltage is $1.10$ V.
The Daniell cell is based on the two half-reactions :  $$\ce{Zn^{2+} + 2e^- -> Zn}$$ $$\ce{Cu^{2+} +2e^--> Cu}$$In the Zn/Cu cell, Zn is oxidized and acts as emitter of electron, which is in the opposite direction of the above equation, so the equation explaining what is going on in a Daniell cell contains the equation of Zn reversed. This becomes :$$\ce{Zn->  Zn^{2+} + 2e^-}$$ $$\ce{Cu^{2+} + 2e- ->Cu}$$When you want to describe the overall equation that goes in this cell, you add these two half-equations, remove the 2$\ e^-$ and this gives : $$\ce{Zn + Cu^{2+}->  Zn^{2+} + Cu}$$  We are here at the point where the two approaches differ. Strangely enough, one uses an addition, and the second a subtraction. Let me speak of the first method, which is used in your textbook.
First method. To obtain this overall equation, you had to add the two half-equations. To follow the same logic, you may simultaneously add the corresponding reduction potentials. But the reduction potentials of Zn and Cu are respectively $-0.76$V, and + $0.34$ V. And if you add them, you do not get $1.10$ V. So the chemists of this school have decided that when a half- equation is used in the opposite sense, with the electrons at the right-hand side, in an oxidation sense, like Zn here, the sign of the potential of this cell should also be opposite. $\ E_{anode} = - E_{cathode}$. Here : $E(Zn)_{anode} = - E(Zn)_{cathode)}$ = +$0.76$ V. With this assumption, you add $+ 0.76$ V plus $+0.34$ V and you obtain $1.10$ V, which is perfect.
Second method. Here you use the same half-reactions as described previously. But the values of the potential remains the same if the electrode works as a reducing agent or as an oxidant. It is always - $0.76$ V pour Zn and + $0.34$ V for Cu. You never change its sign. But remembering from the electricity course, that a measured voltage is a difference of potential, the voltage of the cell may be obtained by subtracting the redox potential of the cathode minus the potential of the anode, that is $E_{Cu} - E_{Zn}$. The result of this subtraction is: $+0.34$V -($0.76$ V) = $1.10$ V, which is perfect. This remains true even though you have obtained the overall equation with an addition of two half-reactions.
As you see, both approaches are equivalent. Have you understood my development ?


text:
Given the pH optimum of Angiotensin 1, how do I determine its charge?
I'm given the following information

I'm then asked to determine the charge of the substrate Angiotensin 1 at the pH-optimum. I have found the pH-optimum to be 6.9.
How do I go about answering this question only knowing the optimum pH and the protein chain that Angiotensin 1 is composed of?

The only charged groups are the N-terminus, the C-terminus and the arginine side chain. Depending on conditions, there is a slight chance for the histitidine side chains to be charged as well.


text:
Oxidation number of Si in SiH4
Wikipedia has the following reaction:

$\ce{Mg2Si + 4 HCl → 2 MgCl2 + SiH4}$

This seems to be a double displacement (metathesis) reaction, where $\ce{Si}$ has oxidation state $-4$. However, I thought that silane had silicon in the $+4$ oxidation state? For example, in this reaction:
$\ce{3 SiO2 + 6 H2 + 4 Al → 3 SiH4 + 2 Al2O3}$, silicon remains in the $+4$ oxidation state.
Why is there the discrepancy between the oxidation state of silicon in silane?

If one reacted SiO2 with two reducing agents, why you'd expected it to not get reduced?

The answer seems to be in a couple of paragraphs further away. Also, have a look at Negative oxidation states of Si.

Electronegativities: Mg < Al < Si < H < O. I would identify both reactions as redox, In the first, silicon and hydrogen change oxidation states. In the second, aluminum and hydrogen change oxidation states while silicon and oxygen don't.

@KarstenTheis No, in the second reaction, silicon, aluminium and hydrogen all experience a change in oxidation state. Only oxygen does not experience a change in oxidation state.

@Mithoron are you saying that Si in SiH4 goes from +4 to -4?

@TanYongBoon See https://chemistry.stackexchange.com/questions/37209/negative-oxidation-states-of-si/37214

The reaction given by DrPepper implies 13 substances to work simultaneously : one gas and two solids. It is highly improbable. This reaction must be the sum of two or more reactions, like $$\ce{6 H2O + 4 Al->2 Al2O3 + 6 H2}$$ $$\ce{SiO2 + 4 H2-> SiH4 + 2H2O}$$ And the H2O com ing from the last equation starts a new reaction  according to the first equation

Magnesium silicide is a somewhat strange beast, whose electronic structure has features that cannot be accounted for by just bonding between magnesium and silicon.
Basically, magnesium silicide has some properties of a salt and some properties of an intermetallic compound. As an intermetallic compound we can expect it to reduce hydrogen, just as an elemental metal would if it were as electropositive as magnesium or even silicon (which displaces hydrogen from steam when hot).
So hydrogen could reduced by magnesium silicide, and may thereby reach the -1 oxidation state (or zero, as magnesium silicide can also form elemental hydrogen from acid or water). With hydrogen thus reduced to the -1 state in silane, silicon goes from -4 to +4 which represents the oxidation of the silicide. The magnesium, already +2 in the silicide, remains there.


text:
What is the triple point pressure for Gallium?
I have seen various discussions about the triple point of Gallium determined to a very precise value, so precise that it is used as a reference for NIST scales and measurements.
However, these reference related documents mean 'temperature' only when talking about triple point. Triple-point is defined as a (temperature, pressure) pair, so I fail to understand why despite so many discussions on the triple point of Ga, the pressure value is not easy to find.
I did find a value of $10^{-38} \mathrm{atm}$ (if I'm not wrong) sometime in the past but I can't find the source. However I have two questions related to this value:

Does it even makes sense to talk about such a minuscule value of pressure? Wouldn't this mean one particle every cubic light-year or something? (edit: More like one particle every cubic kilometer, depending on temperature, but still).
Is it safe to say that Gallium melts at absolute zero pressure (perfect vacuum), and never sublimates or desorbs (outgassing of Ga atoms/clusters from solid Ga surface)?

In general, are there other metals that have such a small triple-point pressure that they are guaranteed to never exhibit the process of sublimation/desorption, even in a perfect vacuum? (i.e., they will always melt first before evaporating or boiling).

https://pubs.acs.org/doi/pdf/10.1021/ja01102a057 has the vapor pressure of gallium.

@JonCuster Thanks. Very helpful resource. I calculated the pressure using eq. 8 and temperature of $302.9166$ K. I get $6.77 \times 10^{-34}$ atm. Since this is pretty much as minuscule as $10^{-38}$ atm, posted questions are still relevant.

At the very least your pressure has to be consistent with a whole number of molecules or atoms in your vapor phase.  Suppose you have a pressure of $7×10^{-34}$ atmosphere or (roughly) $7×10^{-29}$ Pa at $300$ Kelvins.  The average volume per atom/molecule is computed by the molecular level version of the Ideal Gas Law, in which the gas constant $R$ per mole is replaced by Boltzmann's constant $k=1.38×20^{-23}$ J/K per unit entity.  Thus
$V=kT/P=(1.38×10^{-23})×300/(7×10^{-29})=6.9×10^7$
cubic meters per vaporized entity.
So unless you are equilibrating your gallium over that much vapor volume, with high probability you do not have in practice even one atom or molecule of vaporized gallium.  In realistic laboratory or even industrual settings the triple point content of gallium vapor is most probably zero.


text:
How can iodine oxidise hydrogen peroxide?
The following is mentioned in NCERT (government textbook in India used in all schools) under chemical properties of $\ce{H2O2}:$

(iv) Reducing action in basic medium
$$\ce{I2 + H2O2 + 2 OH- -> 2 I- + 2 H2O + O2}$$

I feel the reaction is not correct. Since $\ce{H2O2}$ is a moderate/strong oxidizing agent, I don't think it can be oxidized by iodine and the reverse reaction should be true since $\ce{O2}$ is a strong oxidising agent and $\ce{I-}$ is a reducing agent. Can someone please verify it?

Note the two hydroxyl radicals, which mans this reaction takes place in an alkaline solution. Indeed, adding iodine to NaOH in water produces colorless iodide (and iodate) ions. See https://www.jstage.jst.go.jp/article/jcej1968/7/5/7_5_360/_pdf

https://en.wikipedia.org/wiki/Bray%E2%80%93Liebhafsky_reaction

Harsh Jain, It may be a correct reaction, but it is a very complicated one.
Let us see what happens when iodine is added to a large volume of an aqueous base in the absence of $\ce{H2O2}.$ It forms iodide and iodate ion. If iodine is in excess, triiodide ion is also formed.
The complication arises from adding hydrogen peroxide because it can act as an oxidizing agent as well as a reducing agent. Similarly, iodate ion and iodide ion can react with each other as well as $\ce{H2O2}.$ For example, iodine can react with $\ce{H2O2}$ and form iodide ions.
On the other hand, iodide ion can be oxidized by $\ce{H2O2}$ to free iodine.
I would suggest that you read a paper by Bray and Liebhafsky [1].
Here is the summary by the authors in their table:

References

Bray, W. C.; Liebhafsky, H. A. Reactions Involving Hydrogen Peroxide, Iodine and Iodate Ion. I. Introduction. J. Am. Chem. Soc. 1931, 53 (1), 38–44. DOI: 10.1021/ja01352a006.

First, the slow action of iodine with water:
$$
\ce{I2 + H2O -> H+ + I- + HOI}
$$
In the presence of $\ce{OH-},$ the above equilibrium is moved to the right leaving $\ce{HOI}.$
Finally, a fast interaction with $\ce{H2O2}$ (which also occurs with $\ce{HOCl},$ (see [1, p. 1509]):
$$
\ce{H2O2 + HOI -> H2O + O2(g) + H+ + I-}
$$
References

Connick, R. E. The Interaction of Hydrogen Peroxide and Hypochlorous Acid in Acidic Solutions Containing Chloride Ion. J. Am. Chem. Soc. 1947, 69 (6), 1509–1514. DOI: 10.1021/ja01198a074.


text:
Black substances covering Aluminum Electrode
I have use two aluminum pieces as electrodes to electrolys salt (NaCl) water. After the experiment, I found that wherever electricity flows, the electrode will be covered in some blackish substances. One electrode has more dark substances than the other

Are they Al2O3? Or are they some other substances? Why is one electrode having more and the other having less?

Alloying in the aluminum, maybe?


text:
why is both types of oxides (Eo2 and Eo) of group 16 acidic?
the oxidation state of E in Eo2 and Eo3 is +4 and +6 respectively ...then how can both of them be acidic ?

All elements E appearing in the upper and right-hand part of the table are non-metals, have a high electronegativity, and their oxides are acidic. The doublet binding O and H in E-O-H is strongly attracted by the E atom. This favors the departure of the H+ ion.


text:
How does increasing the Hydrogen Concentration of CuSO4 (electrolyte) affect electroplating of copper
I'm trying to find an answer as to how lowering the pH of the CuSO4 solution during electroplating of copper affects the amount of copper deposited onto the cathode.
My very basic understanding is that if the concentration of hydrogen is too high, then there is greater competition and hydrogen will be reduced instead of the copper. (But why?)
Also the reason why CuSO4 is acidified is to prevent copper from hydrolysing from the solution as it prevents the oxidised Cu2+ from reacting with the OH- ions to form Cu(OH)2 precipitate.
However, I was wondering if this could be related to electrode potentials. That the increase of H+ concentration in the electrolyte affects the electrode potential, which is why increasing H+ increases the competition and makes it more likely for H+ to be reduced.
Im trying to explain why there is an optimal pH for the electroplating of copper and I need background information as to why too low pH or too high pH decreases the amount of copper deposition on the cathode.

In electrolysis, $\ce{Cu^{2+}}$ is reduced in two steps, first to $\ce{Cu^+}$ and later on to $\ce{ Cu}$. Of course $\ce{Cu^+}$ is automatically disproportioned into $\ce{Cu}$ and $\ce{Cu^{2+}}$ in acidic conditions. But in neutral solutions, $\ce{Cu^+}$ may react with $\ce{H_2O}$, forming an unwanted red precipitate of $\ce{Cu_2O}$. This will affect electroplating of copper.

If the available Cu ions in the electrolyte are decreased (ex. CuSO4 electrolyte isn’t replenished using an inert anode OR a copper anode passivates due to polarization) the excess H+ combines with the remaining Sulfate ions to produce H2SO4 which will react with any Cu2O and form more CuSO4.
During the temporary decrease of available Cu ions, the conductivity of the electrolyte drops, and in the event that the voltage is set higher than what it would take to electrolyze water (1.23v), Hydrogen will start bubbling at the cathode, thereby decreasing the amount of copper deposited on the cathode. This also can happen if the anode/cathode ratio is too low which will cause a higher current density at the anode causing more oxidation than can be reduced by the surface area of the cathode.
If I set my power supply to 31v/5A, if there are sufficient Cu ions available between the electrolyte and the anode, the voltage will never go above 0.6v, but if I have contamination in my electrolyte (nickel/iron sulfate, or excess chloride ions) or the anode/cathode ratio falls too much due to the anode being consumed, the voltage will spike and I’ll get uneven plating and Cu metal accumulation on the bottom of my cell. I’ve had this happen when a bubble in the middle of my anode opened up exposing a large surface area of un-oxidized copper causing all the current to flow quickly through that area releasing a ton of oxygen bubbles at once and my electrolyte to cloud completely black temporarily.
I think to answer your question, the optimal pH for electroplating is more in relation to the availability of Sulfate ions than the excess of H+. This can be adjusted by adding additional H2SO4 to the electrolyte which will be an additional source of sulfate ions and also lower the pH.


text:
Is organic chemistry mechanisms limited to organic chemistry?
I am curious to know the mechanism of each reaction. Organic chemistry is satisfying for me because there is a mechanism behind each reaction which make a sense but when I Google for the mechanism of inorganic reaction I do not find any good results. I applied an organic reaction mechanism on an inorganic reaction and it worked as shown in the picture. By summarizing "Can we apply organic reaction mechanisms to inorganic reactions?"

Is it a stupid question?...lemme know to delete it.

It would be nice to see your proposed mechanism (try to reduce image size or try VPN if your ISP is known for blocking stuff), but in general the question boils down to what makes sense and what doesn't. Mechanisms work fine for molecular substances, but are quirky when one tries to use them for network solid, which prevail in inorganic chemistry. BTW there is an entire book devoted to the mechanisms in inorganic chemistry: Ghosh and Berg's  Arrow pushing in inorganic chemistry.

If I were you, I would refrain from assuming your question is stupid. If you did the research and gave it a good thinking, but you still cannot come up with the solution or rationalization, it is stupid not to ask.

Related: Inorganic chemistry reaction mechanisms.

Actually I googled a lot and can not found anything related...so I thought maybe my question is soo easy that there is no need to answer it.

@andselisk uploading still not working...By the way it was the reaction of CO2 + H2O -----> H2CO3

Mechanism: H+ attacks on CO2 ...carbocation HCO2^+ formation occur....in the 2nd step OH^- attacks on carbocation and leads to the formation of H2CO3....Can you please input mechanism in my question by editing it?

My gut feeling says that we can apply organic reaction mechanisms to discrete molecules reaction in inorganic.

You should be able to it by yourself. If you still cannot sort out image uploading, just append the comments to your question by [edit]ing.

I approved the edit, but I think you should really check what's going on with your device or internet provider. Probably clear the cache and delete cookies for *.stackexchange.com site, then log back in.

I am using stack exchange app

Official SE app for Android hasn't been updated in 2.5 years and doesn't support many features, see e.g. Meta, so I would recommend not to use the app and stick with the web browser.

Mechanisms of hydrolysis, of certain common reactions and of organometallic compounds is known to me. But i mostly mug up the reactions

@andselisk bundle of thanks for  recommendation.


text:
Why do we need to add Al in excess for the reaction between a nitrate and a base to give ammonia?
I don't know if why we add Al for the reaction between Nitrate and a base so it would release NH3 gas.
So what I want to know is if what is the role of Al here?
And why do we have to add it in excess?
Can we do this with every nitrate? ( in my case I came across the reaction of Sodium Nitrate and Sodium hydroxide)
And thank you for your time!

This looks somewhat like a sub-question of the Why does Devarda's alloy include copper?

Pardon me I am not educated about it I will check that out. Thank you.

Hydroxide and nitrate alone do not react  with each other.
Their reaction with aluminium is the part of the classical quantitative determination of nitrate content by distillation of produced ammonia and by the back acid titration.
Nitrite + ammonium ions interfere, but the former can be eliminated due their high reactivity, the latter can be determined and/or eliminated by prior distillation without aluminium.
Aluminium acts as a reduction agents, reducing nitrates to ammonia, forming alkali aluminates.
$$\ce{3 NO3-(aq) + 8 Al(s) + 5 OH-(aq) + 18 H2O(l) -> 3 NH3(g) + 8 [Al(OH)4]-(aq)}$$
Generally, aluminium need not to be in an excess. But if there is the desire the all nitrate ions react quantitatively, aluminium and hydroxide must be in excess.
Usually, the Devarda's alloy is used, instead of pure aluminium, containing also copper and zinc.

Judging from the number of typos, someone started to celebrate the New Year early:)

Android typing does not require any celebration. I am sober, but Android is not. :-) I make typos all the time.

It is worth while mentioning that the reduction of nitrate to ammonia by aluminium corresponds to a huge change of oxidation number. Nitrogen oxidation number passes from +V to -III, which a change of 8, while Aluminium passes from 0 to +III. There are not so many known examples of a change of oxidation number as high as 8.

@Maurice One of few is the potassium perchlorate reduction to potassium  chloride in fireworks. ( +VII -> -I }, or the   sulphate reduction to sulphide ( +VI -> -II)

@Poutnik. I should have added that I was thinking of reactions in aqueous solution or near room temperature. The reductiosns you are mentioning happen at temperatures higher than 500°C.


text:
Is there any actually observed excited state of a "hypervalent" main-group molecule that actually involves d-orbitals?
It is well known that, even in periods 3 and higher, the bonding in so-called "hypervalent" main-group molecules nearly always do not involve d-orbitals, and I've heard that it's because the d-orbitals are just too high in energy.
I now have a question: if the energetic difference is the only problem, then can bonds involving d-orbitals exist ("exist" as in "actually be observed") in excited states of "hypervalent" molecules?
For a more concrete example- do the bondings in sulfonate groups in Brilliant Blue FCF and indigo carmine get "elevated" from S(-O)3 to S(=O)2(-O) (charges omitted and S=O bonds involving sulfur's 3d orbitals) as they absorb red-light (opposite of blue) photons, and, if they do, to what extent?

One candidate involves calcium, of all things. This paper by Krieck et al. [1] reports the synthesis of a dimeric calcium(I) complex $\ce{[(thf)3Ca\{μ-C6H3-1,3,5-Ph3\}Ca(thf)3]}$ in which the calcium atoms are bonded to each other and there are two unpaired electrons in antibonding orbitals of the aromatic ring system.
DFT calculations show that the orbitals containing the unpaired electrons have a significant contribution from calcium 3d orbitals which have the same symmetry as the ring antibonding orbitals.  A model of the compound is given below.

References

Krieck, S.; Görls, H.; Yu, L.; Reiher, M.; Westerhausen, M. Stable “Inverse” Sandwich Complex with Unprecedented Organocalcium(I): Crystal Structures of [(Thf)2Mg(Br)-C6H2-2,4,6-Ph3] and [(Thf)3Ca{μ-C6H3-1,3,5-Ph3}Ca(Thf)3]. J. Am. Chem. Soc. 2009, 131 (8), 2977–2985. DOI: 10.1021/ja808524y.

The preview says that "the compound is extremely sensitive to air and moisture". I can't get past the paywall so I can't read the full text but the molecule definitely looks unstable.

Unstable/sensitive, but still there.  Magnesium also forms organic complexes using the +1 oxidation state (with no d-orbital bonding), and these too are air- and moisture-sensitive.