text:
Removal of electrons from orbitals doubt
I recently encountered a question asking whether [PtCl4]2- has square planar geometry. I know that Pt has an electronic configuration of [Xe] 5d9 6s1, in the given compound Pt2+ is there, so the E.C will be [Xe] 5d8 . But I'm confused whether Pt2+ will have four pairs of electrons or three pairs of electrons and two unpaired electrons in the 5d orbital. It seems like the latter should be true because paired electrons may be harder to take out. If that is true, is it always true as a general rule that unpaired electrons are always taken out instead of paired ones if both types are there (during the formation of the oxidised form of the element)?

If you're trying to draw a Lewis structure, you're going to have a bad time. You need a different model to be able to deal with transition metal complexes.

No I'm using valence bond theory to model this structure. If the electronic configuration of Pt2+ was such that there were 4 pairs of electrons in the 5d orbitals. I know that there would be dsp2 hybridization and square planar geometry. What I'm not sure about is   how the electronic configuration of Pt2+ will be (to use VBT).

How would you know what the hybridization is? Generally, it's not a great idea to use hybridization with transition metals.

Ok, but this is like a method we've been taught in school- its probably not what's actually used while studying transition complexes. So to answer questions on coordination complexes we use VBT - and that uses orbital hybridization as one of the main concepts to explain bonding (https://courses.lumenlearning.com/boundless-chemistry/chapter/bonding-in-coordination-compounds-valence-bond-theory/ ). How I found the hybridization was by considering the empty orbitals left (one 5d, one 6s and two 6p) for the ligands to give their four lone pairs of electrons. What I want to know is how to getthe EC

Your intuition that the electrons are all paired is correct, but the explanation is somewhat beyond what is provided by VBT. In VBT terms, the best way to explain it is that the one d orbital that is involved in bonding results in a low energy bonding orbital that is filled by electrons from Cl- and a high energy  antibonding orbital (as happens with all bonding interactions). This antibonding orbital is higher in energy than the remaining four d orbitals, so the lowest energy arrangement is to put pairs in those 4 and leave the antibonding orbital empty.

So VBT cannot explain why one d orbital is involved in bonding at all, instead of just the s and p orbitals being used in bonding leaving two unpaired electrons (in the d subshell) and so no available d orbital for bonding?

Hybridisation is useless for predicting geometries of transition metal complexes, and it also "predicts" some scientifically unjustifiable rubbish, so it's worse than just being useless - it's actually misleading. To predict the geometry or the electronic configuration you need more accurate theories.

@orthocresol I found your post here : https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-for-transition-metal-complex/76740 , on how hybridisation theory is misleading- very helpful, thanks.


text:
Why does ionisation energy to remove electrons from the same subshell increases?

Electrons in the same subshell possesses the same energy.
So why does ionisation energy to remove electrons from the same subshell increases?
For example, $\mathrm{2s^2}$ has two electrons. Since both of them are in the same subshell shouldn't ionisation energy to remove each of them stay the same?

"Electrons in the same subshell possesses the same energy." That's not true between different atoms.

Beside the comment above, which is related, you already see the same with a classical distribution of charges. .The remaining electron is more bound after removal of its companion.

By related, we mean the following: an atom without an electron has the same electronic structure as the atom with atomic number $Z - 1$.

The ionisation energies would be same if there was no shielding effect but it exists. The electrons due to mutual repulsion  prevent the outer electrons from the nuclear pull. Thus the force with which the nucleus attracts the electrons decreases as we move away from the nucleus both because of the increasing distance and also due to the increasing shielding effect of the inner electrons . Now in the same shell when one electron is removed the shielding effect decreases . The shielding effect due to the electrons of the same shell is the greatest. Thus the successive ionisation energies of any atom increases due to decreasing shielding effect.

source: Stack Exchange

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text:
How dependent is the molal depression constant on the nature of solute?
I've been reading about the molal depression constant (freezing point depression constant) for binary solutions in class.
I encountered a problem (see below) in which the constant $K_f$ was to be evaluated first for a sucrose-water solution ($\Delta T$ and concentration of the solution was given).
Then this value of $K_f$ was used to determine the depression $\Delta T$ of the freezing point of a glucose-water solution when given it's concentration.
This is the problem:

A $5\%$ solution (by mass) of cane sugar in water has freezing point of $271.00\  K$. Calculate the freezing point of a $5\%$ glucose water if freezing point of pure water is $273.15\ K$.

My confusion: Is $K_f$ solute dependent? If no, then why not?
I have this confusion because I'm used to solving problems in which if the solute is changed, then most of the constants related to various properties of the solution also changes.

You're right to be skeptical. This is another case where  "ideal behavior" is assumed. More complicated mathematical models can be used to obtain more consistent results. See Wikipedia article Freezing-point depression

Some solutes form nearly ideal solutions up to moderate molalities, examples being glucose and sucrose in water. Such solutions allow fitting of freezing point temperature data to the following equation:
$$\mathrm{log}(1-x_s)=\frac{\Delta_{fus} H_m}{R}\left(\frac{1}{T_{m}}-\frac{1}{T}\right)$$
Taking various approximations which includes assuming that the heat of fusion is constant from $T$ to $T_m$ (the melting point of the pure solvent), that $x_s<<1$, and that $T_m\approx T$ you obtain the expression for the freezing point depression, $\Delta T = T_m - T$, in terms of the cryoscopic constant $K_f$ and solute molality $m_s$:
$$\Delta T = K_f m_s$$
where
$$K_f = \frac{M_wRT_m^2}{\Delta_{fus}H_m}$$
Since $K_f$ contains parameters that depend only  on the solvent (not on the particular solute) the equation can (to within the limitations imposed by the above approximations) be applied to any solutes with which the solvent forms ideal solutions. Which is why you can determine $K_f$ with one solute only to use that same constant to later determine the concentration of another solute.

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text:
2,4 dihydroxy benzoate formation
When resorcinol and potassium hydrogen carbonate are added to each other in acidic conditions together with water, 2,4-dihydroxybenzoate is formed.
What kind of reaction is this? and what is a possible mechanism for it?

Are you sure about the acidic conditions? That will destroy the hydrogen carbonate. Now in basic conditions it can work.

HCl is also used so I assumed it is simply for acidifying

HCl would be used at the end of the reaction to form the acid from the potassium salt for precipitation or extraction.

This reaction has been studied in basic conditions by Barbarossa et al. [1] using $\ce{KOH}$, $\ce{KHCO3}$ and $\ce{K2CO3}$ under $\ce{CO2}$.
Two mechanisms are offered in the paper, the first one looks to be more relevant for the example the OP quoted.

References

Barbarossa, V.; Barzagli, F.; Mani, F.; Lai, S.; Vanga, G. The Chemistry of Resorcinol Carboxylation and Its Possible Application to the CO₂ Removal from Exhaust Gases. Journal of CO₂ Utilization 2015, 10, 50–59. https://doi.org/10.1016/j.jcou.2015.04.004.


text:
Why 1 mole of H2 occupied the same volume occupied by 1 mole of O2?
I'm asking a question about the volume occupied by gasses in standard temperature and pressure.
My textbook said that a mole of any gas occupies $\pu{22.4 L}$ at standard temperature and pressure.
But $\ce{O2}$ is greater than $\ce{H2}$ in size

I asked my teacher why the volume occupied by 1 mole of $\ce{H_2}$ is equal to the volume occupied by $\ce{O_2}$ at standard temperature and pressure, but he seemed to not know the answer, and I remain confused.
Please answer in simple words.

The basic assumption is ideal gas behavior. // PS -- STP changed in 1982. An ideal gas has a volume of 22.7 liters at STP.

Think of tennis and football balls in a very big hangar to see that the frequency of collision is basically the same, unless you pack the hangar with a lot more balls. What you refer to is the proper volume, which is not the volume of a gas.  Reread what your book say about ideal gas, if the level of your class includes it. Else use my pictorial example.

The volume occupied by a gas has no relationship to the size of the molecules of the gas
This is one of the principles of the gas laws in chemistry. It is based on observation: in the early days of chemistry it is what chemists observed to be true. And it is a very important chemical law.  But, of course, that doesn't explain why it is true.
There are several ways to gain some appreciation of this without too much theory. consider water as a liquid compared to water as a gas (I'm simplifying a bit but not by much). One mole of liquid water occupies about 18mL of space and this space is largely dominated by the size of the water molecule (again reality is a little more complex as the specific way the molecules interact makes a small difference to the volume occupied). As water vapour (the same substance but as a gas) 1 mole of water occupies over 22L of space (ignoring the fact that water is not a gas at STP, but bear with me as this is the space it would occupy if it were). That is well about 1200 times more space. This implies that very little of the space occupied has anything to do with how big the water molecule is.
The same broad idea applies to all gases.
So why does the molecular size not matter? A detailed understanding requires a fair few calculations about how gas molecules behave. But, the bottom line involves thinking about the molecular cause of pressure and the meaning of temperature. Pressure is caused by gas molecules bouncing off the walls of a vessel. Temperature is the average energy of those molecules which is a function of their speed and their mass. Doing the calculations shows the relationship between volume, pressure and temperature. This calculation explains the properties of gases but the volume of the molecules has no part.
Strictly speaking the ideal gas laws assume that the size of molecules don't matter. But this is a good approximation when the gases are far away from the point where they become liquids. The simplified theory matches actual observations well. And the fact that, in typical gases, the volume occupied by the molecules is less than 0.1% of the volume occupied by the gas shows why the molecular volume is usually irrelevant.

Let's take a look at the mean free path in a gas, which is defined by Wikipedia as

the average distance travelled by a moving particle (such as an atom, a molecule, a photon) between successive impacts (collisions)

On the same page, one can find a value of $\pu{68 nm}$ ($\pu{6.8 \cdot 10^{-8} m}$) for the mean free path at room temperature and a pressure of about 1 atmosphere. This is to be compared with the bond lengths of $\ce{H2}$ and $\ce{O2}$, which are about $\pu{74 pm ( 0.74 \cdot 10^{-10} m)}$ and $\pu{121 pm ( 1.21 \cdot 10^{-10} m)}$ respectively. Even if we generously assume and effective molecular diameter of three times the bond length, we can conclude that molecular size has little impact given that there are two orders of magnitude difference between size and mean free path.
This is true as long as we deal with a gas that is reasonably close to ideal. In reality, the size does matter indirectly. Larger and heavier molecules typically have more polarizable electron clouds, leading to stronger intermolecular forces. This is the reason why at low densities, gases with otherwise rather different boiling points (that are largely dependent on the strength of the intermolecular forces) still behave rather similar.

I down voted this answer. The OP is obviously an absolute novice. I think you need to first explain the basic notions of ideal gas behavior.

I'd restructure that last paragraph since "This is the reason..." doesn't follow from the preceding sentence (it follows from the first sentence). Also, nonideality is probably unimportant except for charged particles, polarization being important only when you get to much higher densities compared to a gas at STP (maybe a dipole moment will have some effect).


text:
What happens electrochemically when bird droppings corrode cars?
I think the metal of the car first goes through an acid corrosion, followed by an electrochemical corrosion. First, the uric acid in bird poop corrodes with iron: 2Fe+C5H4N4O3 -> C5N4O3Fe2+2H2. Then, after the acid is used up, the moisture in the poop participates in an electrochemical corrosion(?). Can someone explain what exactly happens electrochemically?
And please correct me if any of my conclusions were wrong, as I mainly used my own limited knowledge and guessed. I am confident that there is some electrochemistry involved though.

Yes, corrosion is an electrochemical reaction. Exactly what is happening I'm not going to speculate. I will point out that it is much more complicated than it might seem. If you have a big sheet of steel it isn't just one big homogeneous plate. Rather there are small cells within the structure that have grain boundaries. Thus there is an unevenness to the composition  of the steel when the plate is considered on a microscopic scale.

@MaxW Thank you. I read that there is an acid corrosion theory, an electrochemical theory, and some others. In this case, is it correct that there is first an acid corrosion, and then an electrochemical one? Or are all corrosions electrochemical? Also, in this case, what are the reactants in the electrochemical reaction?

@MaxW Also, the unevenness that you mentioned, does it mean that each cell is going through a reaction, and it ends up having thousands of reactions on one sheet of steel? Does its microscopic composition affect the overall corrosive reaction?


text:
Effects of Ka on the Half-Equivalence Point
I'm new to studying chemistry at a university level and am currently reviewing acid-base titrations. I am wondering about the relationship between your ionization constant for a weak acid ($K_\mathrm{a}$) and both the volume of strong base needed to reach the half equivalence point and the $\mathrm{pH}$ at the half-equivalence point.
Assuming these two weak acids are of the same concentration, if you had a higher $K_\mathrm{a}$ value for one of them, my reasoning is that your $\mathrm{pH}$ at the half equivalence would be lower than that of the weak acid with a lower $K_\mathrm{a}$ value because given the equation
$$\mathrm{pH} = -\log[\ce{H3O+}],$$
a greater hydronium ion concentration (from being present in larger amounts as a product given the larger $K_\mathrm{a}$ value) would result in a lower $\mathrm{pH}$ overall.
However, I am not entirely sure if this is correct, and I am also unsure as to how the volume of strong base titrant needed to reach the half equivalence point would be affected.

The half equivalence point is not really a "thing" in chemistry because it would depend on some much else. That said, with a strong base titrant, the acid that had the smaller pKa would also have the lower pH at its half equivalence point.

If we consider $\ce{HA}$ as a weak acid, then at the half equivalence point, $$\mathrm{p}H = \mathrm{p}K_\mathrm{a}$$ As $$\mathrm{p}H = \mathrm{p}K_\mathrm{a} + \mathrm{p[\ce{HA}]} -\mathrm{p[\ce{A-}]}$$ and for the half equivalence point,
$$\mathrm{p[\ce{HA}]} =\mathrm{p[\ce{A-}]}$$
So the higher the $\mathrm{p}K_\mathrm{a}$ is, the higher is $\mathrm{p}H$ at the half equivalence.


text:
GAMESS Gaussian qst2 method alternative
I've heard about the qst2 Gaussian method but I don't have any licence so I use the free GAMESS-US. Nevertheless, I didn't find any Gamess method which allows a TS finding like qst2. Have you heard about an alternative ?
Thanks

I don't tend to use the QST2 method in gaussian but perhaps the Nudged Elastic Band approach that has been implemented in ORCA 4.1 may be of some use. It too takes the structure of both the reactant and product as inputs. ORCA is free for academic use unless the licencing has changed recently. https://sites.google.com/site/orcainputlibrary/geometry-optimizations/tutorial-neb-calculations

Thanks, I'll check it out, looks like a new and good alternative !

I'd also recommend the growing string method. We've recently implemented it into python https://github.com/ZimmermanGroup/pyGSM. It doesn't have Gaussian hooked up right now but it's fairly easy to do that. Orca works with it.


text:
Low-cost nontoxic corrosion inhibitor for steel exposed to hydrochloric acid
It seems that many commercial descaling solutions designed to remove scale (calcium carbonates, etc that form on water heaters) are a simple combination of an acid and an acid corrosion inhibitor for metals.
The acid renders the scale soluble in water, while (some) inhibitors coat metal surfaces preventing the acid from corroding the metal.
The problem is that these descaling solutions, despite their simplicity, are often \$30-\$100 per gallon (at just 5% HCl concentration!). Furthermore, most corrosion inhibitors are toxic.
A smart cost-sensitive consumer would obtain hydrochloric acid at a low cost (\$10 for 1 gallon of 35% HCl), and obtain or make a non-toxic, low-cost corrosion inhibitor.
Is there a non-toxic, low-cost, and reasonably well-established corrosion inhibitor that can be reliably obtained by a consumer for use as a descaling agent in combination with HCl?

I assume you mean for a hot water boiler system, not the hot water heater that would supply domestic hot water.

Hot water heater, as in your shower etc. Meaning preferably food-grade corrosion inhibitors and acids. Would also work for descaling boilers for heat.

An extremely wide variety of plant extracts have been shown to serve as corrosion inhibitors, including black tea, chamomile, orange peels, black pepper, mustard seed, neem, the list goes on an on.
I have skimmed almost 100 papers so far and handfuls of review papers, and it is not clear to me that any of these "results" can be trusted. Most of these papers (all?) are from "PhD Factories" in developing countries, and zero of them give actionable insights. However, apparently almost every plant extract any author has ever tested offers protection against corrosion by a factor of roughly 5x in grams-per-liter concentrations.
A plant extract is generally made by taking about 10 grams of dried plant, and heating it in 0.5-1 liter of water (or some other solvent), and removing all but ~20ml/20g of liquid, leaving a very concentrated solution of plant compounds ("plant tea"). Or you can mix the dried material directly with the hydrochloric acid if you can then filter it.
Thus a food-grade hydrochloric-acid / "plant-tea" mixture might be an effective and cheap alternative to extremely expensive commercial products. 1 gallon of muriatic acid is \$10, producing 5 gallons of plant-extract/HCl descaling solution at a cost of \$2/gal. It may have comparable performance to commercial descaling solutions at more than \$50/gal.
However, these extracts are poorly understood and are extremely complicated bio-chemical solutions. My search continues for a single well-understood compound with low-toxicity and good corrosion inhibiting properties for steel in acids like HCl.

Much more practical to flush the water heater occasionally as recommended by the manufacturers ( in the US). Keep in mine a normal water heater also has a magnesium /zinc/aluminum anode that will dissolve in acids. However citric, formic and acetic acids are used to cleans steels with minimal damage. PS : Refineries will pay you a fortune for your HCl inhibitor when you find it, they have been looking for almost 100 years. PS #2; I flush a white product out of my water heaters ( city water) that is not acid soluble.

not only will the anode dissolve in acid, in many cases it will generate hydrogen gas :) HCl-steel corrosion inhibitors exist and they are plentiful. Their job is to slow corrosion, not stop it. They are used extensively in boiler descaling / pickling, oil and gas activities, etc. The problem is finding one that is non-toxic.

Properly operated boilers ( industrial and power) generally do not have scale problems ; water chemistry  ,  inhibiters ,chealates and blow-down control scale. Calcium Sulfate scale is a problem in oil field heaters but is not removable with HCl.


text:
Do all polymers contain either carbon or silicon?
Is either carbon or silicon present in every known polymer?
If not, is there a small group of elements that, between them, are present in every known polymer, or can polymers be based on a lot of different elements?

Not all and answer may depend on what you'd call a polymer.

No, neither C nor Si have to be present in a polymer.
IUPAC defines polymer simply as

A substance composed of macromolecules.

In turn, there is no limitations on the elements the macromolecules may consist of.
In fact, there is an extensive class of inorganic polymers free of carbon and silicon.


text:
What is the concentration of H2 for the equation H2+I2=2HI if the equilibrium constant is 64 and the initial concentrations are both 1.9
So for this, I set up the equation =
$$64= \frac{(x)^2}{(1.9-x)^2}$$
I took the sqrt of both sides and got
$$8=\frac{x}{(1.9-x)}$$
So then I did $$8(1.9-x)=x$$
And simplified I got $$15.2=9x$$
So $$x=1.688$$
So to find concentration it’s Initial- Change
$$1.9-1.688=.212$$
But it’s telling me this answer is wrong

$x$ moles of reactant consumed will yield $2x$ moles of product, so the first equation needs to be modified.

So you get 0.212 sausages per square meter, or what is your unit?

let the amount of $\ce{H2}$ reacted = $\pu{x~mol}$
Use the molar ratio to detemine the amount of $\ce{I2}$ reacted and the amount of $\ce{HI}$ formed:
$$\begin{align}
&\text{The amount of }\ce{H2}~\text{reacts to reach equilibrium }&=\pu{x~mol}\\
&\text{The amount of }\ce{I2}~\text{reacts to reach equilibrium}&=\pu{x~mol}\\
&\text{The amount of }\ce{HI}~\text{formed at equilibrium }&=\pu{2x~mol}\\
\end{align}$$
Use the following table to determine the amount and the concentration
of each species at equilibrium :
$$\begin{align}
\ce{&H2 &&+ &&O2 &<=> &&2HI\\
I~~~~  &1.9   &&&&1.9      &&&0 \\
C ~~~  &-x    &&&&-x       &&&2x \\
E ~~~  &(1.9-x) &&&&(1.9-x) &&&2x}
\end{align}$$
Th concentrations at equilibrium :
$$[\ce{H2}]_\mathrm{e}=(1.9-x)  , [\ce{I2}]_\mathrm{e}=(1.9-x) ,  [\ce{HI}]_\mathrm{e}= \pu{2x~M} $$
Sustitute the concentrations in the following formula:
$$\begin{align}
K_\mathrm{C} = \frac{ [\ce{HI}]_\mathrm{e}   }  {[\ce{H2}]_\mathrm{e}[\ce{I2}]_\mathrm{e} }\\
64= \frac{(2x)^2}{(1.9-x)(1.9-x)}
\end{align}$$
Take the square root of both sides :
$$\sqrt{64}=\sqrt{\frac{(2x)^2}{(1.9-x)^2}}$$

$$8=\frac{2x}{(1.9-x)}$$
- Solve for $x : x=1.52$

Calculate the concentrations at equilibrium :
$$\begin{align}
[\ce{H2}]_\mathrm{e}&=(1.9-x)&=(1.9-1.52)&=\pu{0.38~M}  \\
[\ce{I2}]_\mathrm{e}&=(1.9-x)&= (1.9-1.52)&=\pu{0.38~M} \\
 [\ce{HI}]_\mathrm{e}&= \pu{2\times{x}}&= 2\times{1.52}&=\pu{3.04~M}
\end{align} $$

Why is it 2X^2, instead of just X^2

When $\pu{x~mol}$ of $\ce{H2}$ reacted ,$\pu{ 2x ~mol}$ of $\ce{HI}$ formed ,so $[\ce{HI}]\mathrm{e}=\pu{2x~M}$ and $[\ce{HI}]^2\mathrm{e}=(2x)^2$ .


text:
What is wrong with storing deionized water in glass containers?
I have heard that in general storing deionized water in glass bottles is ill advised and that it is better to use plastic. Additionally this is especially important for water used in ICP. I have noticed that ICP standards (which use 1-2% nitric acid) all come in polyethylene bottles as well though the nitric acid used for rinsing (10%) is stored in a glass reservoir. Is there some reason that glass is unsuitable for ICP applications or is it another just economical to use plastic?

It is very difficult to make a container that does not introduce impurities to a liquid you place into it. As mentioned in the comments to the question, storing in a glass bottle will result in ions in your formerly pure water.
This is thought to proceed by a mechanism called ion exchange (source):
$$\ce{ Glass.Na+ + H3O+ -> Glass.H3O+ + Na+}$$
The Wikipedia article on dealkalisation states:

Routine tests for surface dealkalization in the glass container industry all generally aim to evaluate the amount of alkali extracted from the glass when it is rinsed with or exposed to purified water. For example, dealkalization can be quickly checked by introducing a small volume of distilled water to a freshly made bottle and rolling the bottle gently to pass the water completely over its inside surface. The pH of the rinse water is then measured; untreated containers will tend to yield a slightly alkaline pH in the 8-9 range due to extracted alkali, while dealkalized containers tend to yield a pH that remains approximately neutral.

The article also mentions a technique called secondary ion mass spectrometry that can probe the composition of solid glass up to 1-2 nm depth below the surface. Leaching has been studied quantitatively in the context of storing nuclear waste in the glass state, see for example this PhD thesis (abstract in English, main text in French).
So storing in a glass bottle is not ideal because of contamination with ions. If you store it in a plastic bottle, on the other hand, you will find organic substance in your formerly pure water. Depending what kind of analysis you want to use your water for, different types of impurities will be more relevant (i.e. give spurious signals in the analysis). If you are analyzing samples with a GC/MS, glass bottles would be better (well, you would use an organic solvent, not water). If you are analyzing samples with atomic emission or absorption spectroscopy, plastic bottles would be better.
A first pointer would be the materials used in sample handling by the instrument itself.


text:
How to keep raspberry ketone from forming crystals again after dissolving in alcohol?
I dissolved raspberry ketone 700 mg in 10 ml of 99% isopropyl alcohol. The solubility according to pub chem is 1 g/ml. After dissolving, I added room temperature water to bring the final volume to 70 ml but suddenly crystals form again. Why is that? What am I doing wrong? I have to heat it above 90 °C to dissolve it but I'm afraid the melting point is 83 °C.
P.S : I'm not consuming this solution, I'm using it for other applications.

Just why wouldn't that be?

Why would the Crystals form again if it's already dissolved ?

So what? It is quite plausible that your compound is well-soluble in alcohol, but only poorly soluble in water-alcohol mixture.

The solubility data are for the isopropanol, not for the total amount of the solvent in abundance of water.
Check the solubility in water.
Edit:
The important thing is the compound solubility in the final solvent mixture, not in the initial solvent.
A very rough (as it is generally non-linear) solubility estimation could be done from the molar composition of solvent. The ketone solubility in the used mixture is much closer to its solubility in pure water than to its solubility in pure isopropanol.
If you e.g. make a cupric sulfate water solution and if you add ethanol, the sulfate immediately precipitates as well.
Suggestion:

make a test for various ratios isopropanol: water,
note the edge ratio where the ketone seems to stay dissolved yet.
user as final solvent a mixture with some excess of isopropanol to be on the safe side.
If it solves perfectly in e.g. 30:70 mixture, use rather 40:60 one.


text:
Why are higher oxides of non-metals more acidic?
I have studied that if the electronegativity difference between O- and element M is more then it is a basic oxide and if the difference is less then it is an acidic oxide. However, it is a fact that higher oxides of Non metals are more acidic than lower oxides. If the theory concerns polarisation, is it not contradictory?

Getting idea about cutting out "acidic oxides" out of every textbook and forcible teaching the authors the Lux-Flood theory, after seeing this...

It is expected oxides with more polar $\ce{M-O}$ bonds are stronger Lewis bases than those with less polar bonds.
Furthermore, for the same element, the polarization gets weaker/Lewis alkality weaker when more $\ce{O}$ atoms are attached.
When hydrated to acids, another phenomena plays a role in acidity: a charge delocalization.
It is well observable that acids like $\ce{HNO3}$, $\ce{HClO3}$, $\ce{HClO4}$, $\ce{H2SO4}$, $\ce{HBF4}$ are strong, as the delocalization  has a good sandbox to play in.
It also plays role in solubility of salts in organic polar solvents, where there is a large  organic cation with one of the above small anions.


text:
Are the C-C bond lengths different in 2-methyl propene
When we were first taught resonance it was with the example of benzene and I myself justified it by thinking that the double bond oscillates because no two carbon atoms are special and hence there is no reason there will be a double bond between any two and not any other two. I now know that's not true. But just to avoid any latent confusion, can you verify my reasoning is wrong by answering this question? By my method, all three bonds will have same length and bond order 4/3

For the purpose of this question, it might be good to assume for a while that what you thought earlier about benzene is true. 2-methylpropene, however, is not like that. Its carbons are special. Some have 3 hydrogens, some only have 2.

You're correct. I didn't see that. Guess I'll have to find another example to attempt to disprove my theory. Thanks a bunch!

It is not clear " what isn't true and now you know it" but independent of the question resonance is not a kind of oscillation.

@Alchimista I understand that it is not an oscillation, just that's the terms all us amateurs(students around me) use without thinking about it and I too just went with the flow

The cases of benzene and 2-methyl propene are different.
In benzene, all carbon atoms are equivalent. One can also imagine the electron resonance.
In case of $\ce{CH3-C(CH3)=CH2}$ , the electrons have no place to move to create a delocalization cloud.
Additionally,  it would have to switch a hydrogen atoms between 3 atoms,
or create highly unstable $\ce{C=CH3}$
Such molecule would be highly reactive.

I completely ignored that the hydrogen would have to be displaced for it to happen. Thanks!


text:
How does conversion of a free radical to a carbonyl compound take place? Unsure about mechanism
Came across a question that talked about fragmentation of peroxyesters, about the peroxide linkage of course.
After the fragmentation of the peroxyester was done, it gave out three products (2 being free radicals, and the other being carbon dioxide)
One of the products(i.e. free radical), went under a reaction (with itself) and gave two products, one being another free radical and the other being a carbonyl compound. I am unsure about the mechanism.

Here's what I thought:
The single bond between the C1 and C2 will cleave in a homolytic manner. The (now) unpaired electron on C1 will combine with the unpaired electron of oxygen to form a $\pi$ bond. Also, the unpaired electron will remain on C2.
Why am I not sure about my reasoning? Because why would a homolytic cleavage take place between C1 and C2 after all?
Any hints or tips would be highly appreciated.

In what situation it's supposed to be happening?

You seem unsure about the homolysis between C1 and C2, but I speculate to consider the product's stability in this suggested reaction:

For one, two previously unpaired electrons form a $\pi$ bond in acetone.  For the other, you form a benzyl radical which may be described in more than one mesomeric form, for example for a radical bromination reaction of toluene.  Plus the neuter tropyl radical, the seven membered ring structure on the bottom right, result of a rearrangement.
Of course, the benzyl radical won't be persistent, eventually either looses a hydrogen, or combines with an other radical to form a closed shell molecule.


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A theoretical framework for the formalization of chemical interaction and configuration based properties
Is there an attempt in any area of chemistry to formalize the way a molecule's properties change based on its configuration and the resulting changes in interaction with other molecules. For example, exchanging an ethyl for methyl group and vice versa.
What I am thinking of is analogous to the pursuit of a ToE(theory of everything) in theoretical physics. If any research like this is occurring or even being conjectured on, what is its current state?

Technically, one could argue that the entirety of chemistry is trying to formalize how properties of molecules change under a given perturbation. From a computational perspective, some work is being done in the context of machine learning towards developing a continuous mathematical representation of molecules, essentially mapping out chemical space. In this way, you could determine how properties change based on how close different configurations of a compound are in this chemical space.
For details/an example, see 1.
For more on this topic, search "chemical space" or, for some of the commonly used mathematical tools used, "principal component analysis" and "variational auto encoder".
Overall, I think a chemical Theory of Everything would have to necessarily be much more qualitative than that of Physics, just due to the emergent complexity that comes from the vast number of combinatorial possibilities of different compounds.

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


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electron donation affecting carbonyl bond strength
I have been told that electron donation decreases carbonyl bond strength and thus decreases the bond stretch frequency, but electron withdrawal increases it.
This does not seem to make any sense as I presume that increasing electron density between two atoms should increase the attraction of their nuclei to those electrons and thus to each other. Is this reasoning flawed? or does this just apply to carbonyls?

Decreasing bond strength decreases the stretching frequency, first of all. It is not clear to which bonds your sentences apply.

I took the liberty to edit the question (in accordance with the suggestions of another reviewer). Please check!

@Alchimista i'm referring to the c=o bond in the carbonyl. Now that you mention it i'm not sure if the strengthening of the bond is from the sigma or the pi bond of the cabonyl.

Your thinking isn't flawed.  What is flawed is where/what are the effect you mention.  The point is that electron withdrawing exerted by a group attached to the carbonyl reflects to the other side, but not subtracting electron density, opposite decreasing the weight of -O---C+. Bond order increases and thus frequency increases. I will have a calm look at the answer and if necessary comment or answer back.

Let's take the above molecules, acetyl chloride and ethyl acetate, as examples of acyl compounds with electron-withdrawing and -donating groups, respectively.
We know that Cl has a higher electronegativity than O as a halogen, so the electron density from the carbonyl carbon should be pulled toward the chlorine in acetyl chloride and pushed toward the carbonyl oxygen in the case of the ester. This might not be immediately obvious, as the two oxygens should have the same electronegativity, so I like to relate this latter point to electric force laws:
$$ F=\frac{kq^2}{r^2} $$
Since C=O has a shorter bond distance than C-O, the C=O will exert a greater force on surrounding electrons than the C-O (this assumes electrons to be point charges, so it is only useful as a trend).
Therefore, we draw the following picture:

Meaning that there is more electron density on the carbonyl carbon for ethyl acetate than for acetyl chloride.
Now, that's somewhat of a problem for the carbonyl oxygen of ethyl acetate! It already has a lot of negative charge as an oxygen, so it will experience repulsive forces due to the increased electron density on its adjacent carbon. Remember that oxygen does not like having more than a charge of -2 formal charge, which is already satisfied by the double bond it has to the carbonyl carbon; a similar relation can be made for actual charge density.
These repulsive forces, well, repel the oxygen away from the carbon a bit, increasing the C=O bond length. And we also know from physics that increasing the bond length decreases the frequency of vibration. This should make intuitive sense: if the bond distance increases, then the atoms have more distance to travel to complete one cycle of vibration, thus decreasing the number of cycles that can be completed per unit time.
Now onto our example electron-withdrawing group:
Acyl chlorides pull electron density away from the carbonyl carbon, as mentioned earlier. Following the same logic, we can conclude that the C=O bond distance decreases(... to a point*). It follows that the C=O vibrational frequency should increase.
Well, that was a lot of words. But let's approach this from another, much simpler, way. In an acyl chloride, the chlorine acts as an electron-withdrawing group, destabilizing the already electrophilic carbonyl carbon. While this decreases the stability of the molecule, that does not mean that all bond strengths are diminished. In acyl chlorides, the C-Cl bond is the weak, labile bond, not the C=O bond. In fact, the C=O bond strength increases for the reasons outlined above. In our example ethyl acetate, the C-O donates electron density to the electrophilic carbon, thus stabilizing the compound. This does not mean that all bonds are strengthened. If anything, it might indicate a greater C-O bond strength than would otherwise be expected. The C=O bond however,  is a different bond that is being affected inductively, so it must be considered differently.
Personally, I am a fan of the first explanation, as that provides an actual understanding of what's going on, whereas the second explanation leads to a more memorization-focused approach to chemistry. I hope this helps!

*Extra information--Let's consult the following graph (ignore the specific numbers on it):

What this shows is the distance between two atoms and the potential energy of their "bond" (this is one of those areas where it should be noted that bonds are really just constructs to help us make sense of molecules... but this is not important for now). To the right of the minimum, attractive forces (nucleus --> electrons) dominate. To the left, repulsive forces (nucleus <--> nucleus) dominate. As two positive charges get further away from each other, their energy goes to zero (this can similarly be rationalized by simple physics II-type equations). As their distance approaches 0 (where a distance of 0 indicates superposition, or direct overlap), the energy between the two atoms goes to infinity. This is all just to say that there's a limit on how close the atoms can get to each other. You can see that the slope to the left of the minimum is much steeper and increases at a much faster rate than the slope the the right. It follows that there should be less difference in the C=O bond distance as compared with some fair reference molecule such as acetaldehyde than between the ester and that reference. Part of why I'm bothering with this at all is because of how useful this graph is in just about everything I have studied in some shape or form, so it is very useful to familiarize yourself with it.

I don't think that your analysis involving the forces of electrostatic interactions is sound. Could you cite a reference for the theory you have presented?

@TanYongBoon I cannot, that is just a nice way to rationalize the electron-donating effect of the ester's C-O bond that I came up with in undergrad. It is, however qualitatively accurate, which is all that is needed to answer this question. What exactly do you think is unsound about it?

After stating Coulomb's Law, you write: "Since C=O has a shorter bond distance than C-O, the C=O will exert a greater force on surrounding electrons than the C-O". This is not physically incorrect. The electrostatic force would be experienced by the oxygen and carbon nuclei. So yes, it is certainly correct to say that the force experienced by each nuclei is greater. But you seem to be trying to say that the force experienced by surrounding electrons is also greater?

I personally do not think it is wise to consider these electrostatic interactions using such physical laws in chemistry. I agree that it definitely can be done. In fact, high-level computational programmes do these sort of analysis, considering all the different electrostatic and quantum chemical interaction that occur. But for the student, this is beyond our abilities. And it can also be dangerous to do so as it is likely that you probably neglected some other important consideration since so many interactions are at play.

@TanYongBoon What I am saying with that statement is that the O on the C=O pulls electrons off the carbon more effectively than the O on the C-O, and the closeness of the C=O bond helps the oxygen suck in electrons around it; precise reasons are unnecessary. There are other effects to consider, but this is not the place to discuss that - it's just a simple way to qualitatively rationalize an effect. Physics is inseparable from chemistry in both concepts and in history, and everything presented in my answer is literally all stuff learned early on in an undergraduate chemistry degree.


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Which of the following show hyperconjugation

None of the given compounds have any alpha hydrogen to show hyperconjugation, but the given answer is c.

I pick a black bear in a coal mone. Please draw the structures in rather than relying on screenshots. You shoukd have software that can do so.

I guess you've only considered one resonance form of compound C. If you consider the other major resonances, there are alpha C-H bonds that can participate in hyperconjugative effects:


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Freezing point lowering for a mixture of many liquids
I have been taught that if I have a solute of molality $m$ mixed with a liquid, the freezing point in lowered by an amount $K_\mathrm{f}\cdot m$ where $K_\mathrm{f}$ is the molal freezing depression constant.
But if I have a solution containing three liquids, and if all the $K_\mathrm{f}$ values are known, what would be the lowering of freezing point of such a solution?

At some point one of the liquids will reach its own freezing point lowered by presence of two other liquids and whatever else you have there. Then it will freeze.

How do we calculate the freezing point?

Precisely as you described, via molality and Kf.

Molality is moles per kilogram of solvent right? If there are 3 liquid components do we take the one with highest contribution to the solution as the solvent?

Sounds about right.

Under the Calphad (Calculation of Phase Diagrams) framework, besides the binary terms representing the interactions of components A, B, and C (AB, BC, and AC), you may need a ternary term as well to represent ABC interactions in the Gibbs free energy.

@JonCuster So the conventional Kf*m  method would not hold ?

It is, not surprisingly, more complicated than that. The Kf*m model is an approximation to begin with, since it presume that Kf does not depend on concentration which is a poor approximation for many real liquids. But, an extension to a ternary would seem quite possible (and ternary interaction terms are not always needed). Depends on the liquids.

It seems suspicious.What is methanol molality in 1:1 molar mixture with water ? Or 1:1:1 methanol:water: acetonitrile ? Is methanol molality in the former the same or different to methanol molality in the latter ?

IMHO, molarity relates to total volume and molality to total mass.

Or to total mass of major solvent components. Wikipedia mentions "mixed solvent mass".


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Why does the degree of polymerisation remain the same throughout chain-growth polymerisation?
It is known that the degree of polymerisation, which is the average number of monomer units in each polymer chain formed, increases as the reaction proceeds for step-growth polymerisation as longer and longer chains are formed throughout the reaction. However, the degree of polymerisation supposedly remains approximately the same through the reaction for chain-growth polymerisation. Why is this the case? Shouldn't the degree of polymerisation also increase throughout the reaction?
More interestingly, it is also mentioned in Introduction to Polymer Chemistry (3rd ed.) that:

Also, unlike the stepwise process, average chain length decreases with increase in reaction temperature.

If this is the case, wouldn't the degree of polymerisation decrease as reaction proceeds?
I was thinking, could it be because in the chain-growth process, polymer chains of roughly the same length are formed as the reaction proceeds. For example, at the start, a chain is initiated and the chain propagates until the number of monomer units it contains is $\ce {100}$. Once it has reached this length, the chain terminates. As the reaction proceeds, more chains are initiated but all of which terminate when the chain length is about $\ce {100}$ monomers. Is this what actually happens?
Reference
Carraher Jr., C. E. (2003). Introduction to Polymer Chemistry (3rd ed.). CRC Press

Mhm, pretty much this.


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Types of concentration units
I would like to ask about a concentration unit.
What does it exactly mean when stock concentration of something is,
for example, 5X?
Thank you in advance!

To me, this has nothing to do with concentration or units in general. "5X" is probably a dilution factor or just poor typography. If you have the label scanned, feel free to post the image here to make the question more clear.

The “X” factor simply indicates that the solution is in a concentrated form that must
usually be diluted to a “1X” concentration for use. For example, a 5X concentrated solution must
be diluted 5-fold, while a 100X concentrated solution must be diluted 100-fold. The dilutions
are usually done using water.
See more  here under part 3:
http://csmbio.csm.jmu.edu/biology/courses/bio480_580/mblab/SolutionPrepReview.html

Often, it is followed by the abbreviation of a solution commonly used in a biochemistry lab, such as PBS (phosphate buffered saline) or TE (tris EDTA) or sample buffer (for SDS gel samples). Of course, it means nothing unless you know the composition of these solutions.


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Formamide and Hoffmann Bromamide Reaction
I encountered a problem to identify the compound which does not give Hoffmann bromamide degradation reaction. One of the options was formamide, and it was given as the answer.
Why can't formamide give Hoffmann bromamide reaction? I feel that the answer is wrong (it should produce ammonia).

I guess hydride isn't sufficiently nucleophilic to do the 1,2-migration?

If hydride were to migrate, isocyanic acid from hydride migration would be deprotonated under the basic conditions of the Hofmann rearrangement.

I have spent a lot of hours to find the evidence to support OP's idea of formamide may undergoes Hoffmann degradation reaction, but I can't find any. However, I find at least one publication, which may suggest formamide may decompose to formic acid and ammonia (ironically) in hot alkaline medium, before exchanging amide $\ce{N-H}$ to $\ce{N-Br}$ (please keep in mind that $\mathrm{p}K_\mathrm{a}$ of $\ce{HC(O)NH2}$ is about 24 at $\pu{25 ^{\circ}C}$ in DMSO, Evan's pKa Table):

Ref.1 has described the rates of decomposition rates of acetamide and formamide, which have been studied by individually decomposing them in pressurized hot water. According to the studies, while apparent decomposition rates for both amides were more represented by the second order rather than the first order reaction kinetics, the rate constant for formamide was more than 10 times higher than that for acetamide (Ref.1).

Even though, the work has been done in high temperature and high pressure, the authors observed the decomposition without using acids or bases, and throughout wide range of concentrations. They also found that rate increases with increasing concentrations. Thus, one can expect that in highly alkaline media used in Hoffmann degradation, some amide may hydrolyzed before rearrangement. According to a mechanistic study of the rearrangement, hydrolysis of even aromatic amides under alkaline conditions has been observed:

It has been found that in the presence of excess alkali at $\pu{30 ^{\circ}C}$, p-nitrobromobenzamide gives approximately as much p-nitrobenzoic acid by hydrolysis as p-nitroaniline by rearrangement; at $\pu{90\!-\! 100 ^{\circ}C}$, however, a $90\%$ yield of p-nitroaniline is obtained.

All of these evidence directed to the theory of hydrolysis of formamide (which is aliphatic and most susceptible to hydrolyze under the conditions due to its size) before it converted to N-bromformamide. Further more, the Ref.3 described a theoretical study of the Curtius rearrangement, which also believed to be going through same mechanism as Hoffmann degradation. Although authors have discussed various pathways of rearrangement including going via oxazirene ion, and identified the most stable intermediate is isocyanic acid during the rearrangement, their conclusion wasn't clear on the hydride ion shift.

References:

M. Okazaki, T. Funazukuri, “Decomposition of acetamide and formamide in pressurized hot water,” Journal of Materials Science 2006, 41(5), 1517–1521 (DOI: 10.1007/s10853-006-4616-1).
C. R. Hauser, W. B. Renfrow, Jr., “The Removal of $\ce{HX}$ from Organic Compounds by Means of Bases. III. The Rates of Removal of Hydrogen Bromide from Substituted N-Bromobenzamides and their Relative Ease of Rearrangement in the Presence of Alkali. The Hofmann Rearrangement,” J. Am. Chem. Soc. 1937, 59(1), 121–125 (DOI: 10.1021/ja01280a029).
A. Rauk, P. F. Alewood, “A theoretical study of the Curtius rearrangement. The electronic structures and interconversions of the $\ce{CHNO}$ species,” Canadian Journal of Chemistry 1977, 55(9), 1498–1510 (https://doi.org/10.1139/v77-209).

So can we conclude that if in place of an alkyl group we have  H or some electron withdrawing group then, hydrolysis would take place before rearrangement because of increase in electrophilicity of the carbonyl carbon?

Can't make it universal rule on everything. My argument is pure speculating, but have some evidence in literature. Also, if you look at reactivity of aldehyde vs ketone, my argument make more sense. Keep in mind also that, $|ce{H}$ is not electron withdrawing (compare electronegativity of $|ce{H}$ vs $|ce{C}$). I believe it's more of its size.


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Which ions determine ionic conductivity of a salt solution?
Let's take $\ce{ZnSO4}$ salt solution in water for example. So the ions inside my solution will be $\ce{H+}$, $\ce{OH-}$, $\ce{Zn^2+}$ and $\ce{SO4^2-}$.
My questions are:

Is the ionic conductivity measured for the solution gives an idea of mobility of zinc ions and sulfate, or is it a result of all four ions combined?
When we change the salt concentration of zinc sulfate, do we change the ionic conductivity of zinc and sulfate ions only, or all four ions?

Also, any undergrad university level source to understand this concept will be helpful.

Why is this being flagged-to-close as homework?!?

Di water H2O has very poor conductivity. Because H+ and OH- only driven by the ionic product.
2 H2O <=> H3O+ +OH-
If any other ions will be add the condictivity rises . So that can be for example by acids (H+), alcaline  solutions (OH-) and salts (Cations and anions). The conductivity of the salts  >>>>>> as of the water. So mainly in your case zinc and sulphate drive it. But to calculate correctly all 4 ions participate.


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How are t-butyl substituents named according to IUPAC nomenclature?
What is the IUPAC name of this structure?

I've actually came up with two names, the first one is "1-tert-butyl-5-heptyl-2-propylcyclononane" and the other one is "1-(1,1-dimethylethyl)-5-heptyl-2-propyl-cyclononane". I'm confused which one is right or if both are correct.

@KarstenTheis I've actually came up with two names, the first one is "1-tert-butyl-5-heptyl-2-propylcyclononane" and the other one is "1-(1,1-dimethylethyl)-5-heptyl-2-propyl-cyclononane". I'm confused which one is right or if both are correct.

So your question is not about how to name it, but rather whether the tBu group should be written as "tert-butyl" or "1,1-dimethylethyl".

Note also that there should not be a hyphen directly before cyclononane; one of your proposed names is correct in this regard and the other isn't.

According to Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book) (P-57.1.2), the preferred prefix for $\ce{-C(CH3)3}$ is "tert-butyl".
As for where it comes in the name:

P-14.5 Alphanumerical order
  [...] Alphanumerical order is applied as follows in organic nomenclature. Nonitalic Roman letters are considered first, unless used as locants or part of a compound or composite locant [...]

so the prefix tert is not considered (except in more complicated cases, cf. P-14.5.3). Therefore, the preferred IUPAC name for the molecule in question is 1-tert-butyl-5-heptyl-2-propylcyclononane. Note that there are no parentheses around tert-butyl.
Be warned: tert-butyl has received special treatment from IUPAC on this front. Other non-systematic prefixes have fallen from favour, most notably isopropyl, sec-butyl, and isobutyl. The preferred prefixes are respectively propan-2-yl, butan-2-yl, and 2-methylpropyl.


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Why do we begin an experiment involving solutions of different concentration from the least concentrated solution?
Is there a particular benefit to start any kind of measurement with the least concentrated solution and then work my way up to the more concentrated solution?
In one of my experiment, we measure the UV absorbance value of a solution beginning with lowest concentration. Weirdly, the demonstrator did not clean the apparatus and simply rinse it with higher concentrated solution that is to be measured.

Can you be more specific with your question? For example, what type of experiment are you running? If this is an experiment using the same measurer, you might think that less concentrated solutions might become more concentrated if more concentrated solutions are measured first. However, it is always good practice to clean measurers after a measurement. Also, the effect of the bad practice depends on the volume of solution.

In one of my experiment, we measure the UV absorbance value of a solution beginning with lowest concentration. Weirdly, the demonstrator did not clean the apparatus and simply rinse it with higher concentrated solution that is to be measured.

That is a valid experimental procedure as well. I believe your instructor measured absorbance starting with the lowest concentration solution is because of the technique used (rinsing apparatus with next solution) as it would be highly problematic to follow this procedure if we measure starting from the highest concentration solution. This procedure should work just as well as cleaning the apparatus if  the solutions do not vary in chemical composition. Thus, choosing which procedure to follow often comes down to personal preference.


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How to standardise NaOH?
My aim is to "To determine the concentration of citric acid in lemon juice using titration technique." and as following i need to standardise the NaOH in order to titrate it with validity. Is there a procedure and step to standardising NaOH. Thank you

You should find it in any analytic chemistry textbook, considering volumetric methods.
We did so in the high school.
In short, a solution of known molar concentration of well defined acidic substance is made, that get titrated by the $\ce{NaOH}$ solution.
Typically oxalic acid $\ce{(COOH)_2 . 2 H2O}$, weighted at good scales, preparing solution $\pu{0.05 mol / l}$.
When you Google "acidimetric titration sodium hydroxide oxalic acid" , you get tons of references, from illustrating videos to PDF documents.

You might have used oxalic acid in high school, but potassium hydrogen phthalate (KHP) is the go to primary standard.  You dry it in an oven before use. // I can't imagine a hydrate being used for serious work.

It frequently  used in universities and lab praxis  as well.Potassium hydrogen phthalate is primary standard for pH. Hydrates are used often, if they are well defined. Another example of a hydrate standard  is the Mohr salt $\ce{(NH4)_2Fe(SO4)2.6H20}$ for oxidimetry as a stable source of $\ce{Fe^{II}}$

I mentioned oxalic acid as even since high school ( chemical  industry oriented), not only in high school. The acid is used since a dawn of volumetric. In fact, I do not remember, if we ever used KHP for that purpose.

Additionally, KHP is less suitable for the particular $\ce{NaOH}$ case, due its rather high $\mathrm{p}K_{a2} = 5.51$, compared to the $\ce{(COOH)2}$  $\mathrm{p}K{a2} = 1.27$  and $\mathrm{p}K_{a2} = 4.27$.
As it would interefere with $\ce{H2O + CO2 <=> H+ + HCO3^-}$ ( of $\ce{Na2CO3}$ origin ) with the apparent  $\mathrm{p}K_{app} = 6.35$

Using KHP and phenolphthalein works fine to standardize NaOH. For both KHP and oxalic acid you'd need to boiled distilled water to remove dissolved CO2 from atmosphere.

Using KHP with FF converts undefined amount of Na2CO3 to NaHCO3.

I don't understand. What is "FF"?

Phenolphthalein ( I am broking my fingers typing it ) . I apologize, it is a common abbreviation among Czech chemists, as we write it with F. I have forgotten that and it could confuse you.

So what indicator would you use with oxalic acid when NaOH is the titrant?

In 80s, we titrated NaOH by HOx on methyl orange (3.1-4.4) I guess. When yellow showed hints of red, we stopped and heated the solution to strip it from CO2 from Na2CO3. Then we cooled it down. It turned back to yellow, but the red titration end was already close.

Hm, When I think about it , it would match rather HCl. HOx would be probably reversed, on the PhPht as well, as you suggested.

I've asked the question about oxalic acid -- https://chemistry.stackexchange.com/questions/111646/is-oxalic-acid-dihydrate-considered-a-primary-acid-standard-in-analytical-chemis


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1000 mg/L alkalinity as CaCO3 standard preparation
How do I prepare a 1000 mg/L alkalinity as CaCO3 standard?  I found a method that says to dissolve 1.06 g of sodium carbonate (Na2CO3) in 1L of water but I'm not sure why I can't use 1g instead?  Thanks!

Do you suppose both substances have  the same molar mass ?

The problem is the soloubility of calcium carbonate is very poor. It is 14 mg/l. So it is not possible to prepare a solution with 1 g/l, but instead you can use sodium carbonate, the soloubility 217 g/l. To get the same molarity you have to dissolve 1,06 g of it.


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Find out of the friend is getting a good deal
I have a strange chemistry question that I've never came across before, and I have no idea about how to solve it.

Your friend comes across a good deal to purchase a gold ring. She asks you for advice for you to test the ring. The ring has a mass of 4.54 g. When the ring is heated with 94.8 J of energy, temperature rises from 23.0 °C to 47.5 °C. The gold specific heat capacity is 0.1291 J/g. Is she getting a good deal?

So I did the equation to solve for Q which is 14.35 J, and I have all the numbers the (considering they are given) but I have no idea how to actually calculate how much of this ring is pure gold.
I came here as a last resort.

You cannot. All you know is that it's not pure gold. There's not that much they can mix into it to increase its value, so...

I actually just figured it out. You solve for C with the given numbers, you get a number that is 0.8xxx which is far off from the heat capacity of gold, therefor it is not a good deal. That's what I was looking for :)

@RobertOchinski You are allowed and encouraged to write and accept an answer post to your own question. This will help future readers who come here with the same question, while your comment might get deleted in future.

The question, as posed, isn't answerable. We can determine that the ring isn't pure gold but whether or not it's a good deal depends on the price being asked for. A dollar for an impure gold ring is still a good deal; a hundred might not be; a thousand probably isn't.

As written its underdefined, but its clear that "Is she getting a good deal?" is meant to be taken as "is the ring actually pure gold?"

The jewelry gold usually comes historically rated in grades 14 or 18 karats, where 24 means pure gold. The rest is mostly silver.
Find out specific heat capacity for the complementary metals like silver or copper, so you could roughly estimate via interpolation.
Another method would be pyknometry, determining the ring density by differential weighting the mass of displaced water - old good method of Archimedes. Gold has almost double density, compared to other expected metals.
Edit:  list of densities:

Gold 19.3.
Silver 10.5.

If linearly interpolated by mass fraction:.

14 karat gold  15.63.
18 karat gold  17.1

I agree with Robert Ochinski (see above comment) that this question, as posed, is not answerable, because the price of the ring was not given. Yet, we can assume what amount of gold in the ring by given data. The specific heat capacity of ring ($c_{p(\text{ring})}$) can be calculated using the equation, $\Delta Q = mc_p\Delta T$. Thus, substituting given values:
$$c_{p(\text{ring})}= \frac{\pu{94.8 J}}{\pu{4.54 g} \times \left(47.5-23.0 \right)\pu{^{\circ}C}}= \pu{0.852 Jg^{-1}{^{\circ}C^{-1}}}$$
This indicate that the ring is not in pure gold, and the specific heat capacity of the mixing metal should be greater than $\pu{0.852 Jg^{-1}{^{\circ}C^{-1}}}$. This makes only mixable metal be aluminum with $c_{p(\ce{Al})}$ of $\pu{0.91 Jg^{-1}{^{\circ}C^{-1}}}$.  Other disregarded metals and metalloids are $\ce{Li}$ ($c_{p(\ce{Li})}=\pu{3.57 Jg^{-1}{^{\circ}C^{-1}}}$), $\ce{Be}$ ($c_{p(\ce{Be})}=\pu{1.83 Jg^{-1}{^{\circ}C^{-1}}}$), $\ce{Na}$ ($c_{p(\ce{Na})}=\pu{1.21 Jg^{-1}{^{\circ}C^{-1}}}$), and $\ce{Mg}$ ($c_{p(\ce{Mg})}=\pu{1.05 Jg^{-1}{^{\circ}C^{-1}}}$), all of which are unstable under ambient conditions (Data are from: Engineering Toolbox). Keep in mind that well know metals used to make alloys with gold such as $\ce{Ag}$ and $\ce{Cu}$ have specific heat capacities less than $\pu{0.50 Jg^{-1}{^{\circ}C^{-1}}}$. Accordingly, we can set the two equations:
$$m_\ce{Au} + m_\ce{Al} = 4.54  \enspace \enspace \enspace \enspace (1)$$
$$ 0.129 \left(\frac{m_\ce{Au}}{m_\ce{Au}+m_\ce{Al}}\right) + 0.91 \left(\frac{m_\ce{Al}}{m_\ce{Au}+m_\ce{Al}}\right)   = 0.852 \enspace \enspace \enspace \enspace (2)$$
$$ 0.129 \left(\frac{m_\ce{Au}}{4.54}\right) + 0.91 \left(\frac{m_\ce{Al}}{4.54}\right)   = 0.852 \enspace \enspace \enspace \enspace \enspace (3)$$
Where $m_\ce{Au}$ is mass of $\ce{Au}$ in the ring and $m_\ce{Al}$ is mass of $\ce{Al}$ contributed to the ring (all in $\pu{g}$). The equation (2) is derived from assuming heat did not loose when heating the mixture and each metal increase the temperature relevant to their specific heat capacities (Thermtest).
When eq (3) is simplified, it gives:
$$ 129 m_\ce{Au} + 910 m_\ce{Al} = 3868 \enspace \enspace \enspace \enspace \enspace (4)$$
Using eq (1) and (4), solve for $m_\ce{Au}$:
$$ (910-129) m_\ce{Au}  = 4.54\times910-3868 $$
$$ \therefore m_\ce{Au}  = \frac{4.54\times910-3868}{910-129} = 0.337$$
Thus, the ring contains only $\pu{0.337 g}$ of gold. The rest is believed to be aluminum. Today's gold price is USD 42.13/g (Gold Price). Accordingly, the ring contains USD 14.20 worth of gold. I'd say if your friend spent more than $40/- for the ring today (giving some space for labor), she's not having a good deal! :-)


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Trend in ionisation potential in period 2
Carbon has valence electronic configuration of $2s^2$ $2p^2$. Beryllium has configuration $2s^2$.So carbon should have less ionisation potential than Beryllium as it is easier to remove electrons from $2p$ subshell than fully filled $2s$ subshell.
But I found that in JD Lee it is mentioned that carbon has more I.P. than beryllium. How can it be explained?

Emperically speaking, Be is on the "metallic" side of the table. Lower IE than non-metals is obviously expected.
A quick glance at the $IE_1$ trends for period II tells us that like the OP noted, Be certainly has an abnormally high $IE_1$, deviating significantly from the expected plot, due to the fully filled configuration.
However, Carbon (Z=6) has a higher nuclear charge(=more protons) than Be (Z=4), a smaller radius and hence a stronger hold on its valence electrons.
Even Nitrogen displays an exception, due to half-filled $2p$ subshell, as evident in the plot.
Just because of a tiny bump in the plot, expecting N and Be to have higher IE than all the elements of the periodic table is clearly an exaggeration.

These periodic trends in 1st ionisation energy (I.E.) are always the result of more than one factor changing. The most important factors are:

The effective nuclear charge (Zeff) that the valence electron being removed experiences (roughly equal, for s and p blocks, to the number of valence electrons)
The potential energy of the sublevel (1s, 2s, 2p etc) that the electron is being removed from
Any electron-electron repulsion that the electron being removed experiences in its orbital

As we move left to right across the period, factor 1 increases and this dominates the overall shape shown in the graph. A drop occurs from Be to B as we move to the higher potential energy 2p orbital (so the electron is easier to remove) and this is more important than the increasing Zeff from 2+ to 3+. As we move to C, the Zeff increases to 4+, so it has a higher I.E. than B (same sublevel, 2p).

The reason that C's I.E. is higher than Be also is because the change in Zeff from Be to C is +2 to +4 (it roughly doubles) and this is more important than the difference in potential energy between the 2s and 2p sublevels, so factor 1 takes over again, over factor 2.
I haven't discussed factor 3. Anytime an electron is removed from a paired orbital, it is a little easier (lower I.E.) than exactly the same situation (same Zeff, same sublevel potential energy) from an unpaired orbital, due to electron-electron repulsion within the paired orbital. It is a small factor between Be and C (Be is a paired 2s, C is an unpaired 2p) but it is very minor compared to factors 1 and 2 changing between these species, so C is still harder to remove.
It is tempting to say that a filled sublevel, or orbital, is "stable" = higher I.E. This statement comes not from the fact that the sublevel is full (these paired electrons are actually easier to remove than you would expect, due to electron-electron repulsion) but because filled sublevels always correspond to the highest Zeff for that sublevel.
You can employ these arguments to explain all the trends in the simple (s and p block) 2nd and 3rd periods. Things get much more complicated in other parts of the table!


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Counting basis sets in quantum chemistry
I have to use STO-nG (a "minimal basis set" - meaning that only one basis function is used for each atomic orbital in the atoms of which the molecule is made from). Let's take the example of a water molecule.
The water molecule has two H atoms and one O atom. Thus, we have a total of 7 orbitals (two 1s of H, one 1s of O, one 2s of O and three 2p of O). So when using STO-nG, would that then mean that three 1s type basis functions, one 2s type basis function, and 3 p-type basis functions are used, totalling at 7 basis functions, each being made up from a linear combination (LC) of n simple gaussians?

Possible duplicate of How many basis functions used in STO-3G and 6-31+G** for the water molecule?

@Tyberius Yes. But if you see the answers, this has not been answered.

Sorry I missed that they never explicitly answered that. But you are correct, STO-nG for water will have 7 basis functions, each of which is formed from n-Gaussians. As an aside, you could always check this using a free electronic structure program like Psi4 or ORCA. @hhsomething69

Yes, you have counted the number of basis functions correctly:
    #1s    #2s    #2px    #2py    #2pz    #Total
H1   1                                      1
H2   1                                      1
O    1      1       1       1       1       5

Which as you also counted is 7 basis functions.
As you also correctly state, each of the basis functions will be a linear combination of $n$ primitive Gaussians:
$$\phi(r)=\sum_{i=1}^nc_i\exp\left(-\alpha_i\left(r-r_0\right)^2\right)$$
As you have observed that is what the $n$ in STO-$n$G denotes.


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How to draw protonated benzene (horseshoe) carbocation structure with ChemDraw?
Is there a way to draw the aromatic "horseshoe" (forgot the official name) structure of a protonated benzene cation (arenium) with ChemDraw, and not the cation with double bonds?

I've added an image produced with chemfig to illustrate your question — hopefully it reflects your point. If not, feel free to add your own image instead:)

It's called Arenium ion. https://en.wikipedia.org/wiki/Arenium_ion

Draw cyclohexane. Open Drawing Elements. Select 270o circle. Orient with lasso and adjust size. Insert in cyclohexane. Add H's and +.
ADDENDUM: For a centered structure. Draw cyclohexane, arc and + inside each other. Select them. Go to Object>align. Select L/R centered then T/B centered. Add hydrogens. See also Comments. PS: Clearly, ChemDraw does not recognize the structure as protonated benzene. Can't have everything ;).

Interesting, this is exactly what I had in mind. What prevented me from writing an answer is that this way the charge is treated by ChemDraw as the stray one, and the entire molecule is recognized as cyclohexane if one would try to generate its name (Alt+Ctrl+N). Also, what bugs me is that vertical alignment tool doesn't work nicely for arcs, so it cannot be perfectly centered within the ring. Are you aware of any means to overcome all those issues?

CD13 or 17 cannot handle "Structure to Name". But that wasn't the question. On an Apple laptop I had no problem (CD13) with orienting the arc or reducing its size. I hope I understood you correctly.

@andselisk: About centering.. Draw ring, arc and + inside each other. Select them. Go to Object>align. Select L/R centered then T/B centered.

CD18 on Windows can generate a name, but, as I said, it's cyclohexane and is wrong. As for alignment, that's exactly what I did, but it looks like the cut edge is used for alignment of arcs, not the circumference, which drives my internal perfectionist nuts:) (my screenshot) But you are right, that's nitpicking and not what OP asked, and the proposed method helps drawing something that is close enough, so +1.

@andselisk: I was able to make a centered structure that now reads cyclohexane. How do you put a screen shot in comments?

I upload the image in the answer field, copy the URL, paste it in the comments normally via [screenshot](URL), and discard the answer :) But that's precisely what I was talking about, CD doesn't treat this structure as arenium, which is a problem IMO. As for alignment, maybe I messed something up, I'm not a CD guru by any means (the vertical alignment on your second screenshot is also wrong — the arc is shifted to the top a bit).

Let us continue this discussion in chat.

That is a terrible way of drawing that intermediate. You should draw a cyclohexane, then right-click on the four bonds where the delocalization takes place and choose double -> tautomeric.

Welcome to ChemSE. @andselisk was asking for a way to draw the structure he provided. He did not ask for the "best" way to do it. We strive to be a kind and caring community.  " Here is an alternative way of drawing that intermediate" would have been a more civil statement. You may wish to consider an edit. https://chemistry.stackexchange.com/conduct

@user55119 I didn't ask the question, Cloud Chem did:) I only added an illustration to what is commonly shown in the literature as a "horseshoe benzene". Despite the main part of the post being critics, I think it is still a valid alternative answer.

@andselisk: My mistake. I still stand by the rest of my Comment.


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Why isn't my galvanic cell making enough voltage?
I'm in kind of a pickle here. I made a galvanic cell of Lead and aluminium for a chemistry internal assessment but unfortunately, it didn't turn out so well and neither me nor my teacher has any clue why.
The E0 value of the cell is 1.53v. I got a maximum reading of 0.29v
If you don't have time to read here is what I think the sources of error could be:
1. I used filter paper as a salt bridge and I don't think it was thick enough to allow ions to flow through it.
2. I used around a 1M concentration of KCl to dip the salt bridge in.
3. I suspect either of the metals was not correct but this seems improbable as I got a voltage just not the correct one.
4. The solutions of AlNO3 and PbNO3 were kept in round-bottomed flasks with a glass cork in them for a few days before I decided to do the experiment.
5. I got a negative reading for the voltage which was weird.
6. I got a higher voltage reading only when I touched the aluminium rod to the salt bridge in the solution
I used the following procedure:

I made 0.01M, 0.1M, 0.25M and 0.5M solutions of aluminium nitrate and a 0.1M solution of lead nitrate (both well dissolved) from lab equipment.
I poured each of them into separate 50ml beakers. I then placed an aluminium rod (that was 50% submerged) into the AlNO3 beaker and a thin piece of lead (that was well sanded before each reading) that was fully submerged in PbNO3.
I connected each of the metals to a multimeter via alligator clips
I then prepared a salt bridge using 5x25cm strips of filter paper which were dipped in a solution of KCL of about 1M concentration. The filter strips were folded and stapled.
I then placed the salt bridge in both the beakers and conducted the experiments.

I'm not in school right now and I don't have time to conduct the experiment again as the assessment is due very soon.

You do know that SEM isn't the same as E0 and depends on concentrations, time, and other stuff?

Another question is possible passivation of lead and mainly aluminium by oxidized layers acting against electrode kinetics. Potentials are thermodynamic quantities, they do not say, how fast are the reactions ongoing. If the reaction is very slow, then the cell may act externally as a power source with huge internal resistance. Additionally, the potentials in open state may not be established. Salt bridges are used to have at least one common ion with electrolyte, so KNO3 Is preferred. The forming of PbCl at solution/paper contact can be an issue as well.

Just keep in mind that Eo is determined under highly controlled conditions and concentrations. I sense two key problems here. The major problem is already stated above that the resistance of a wet filter paper must be high, and this is giving you lower voltage. Evaporation of water from the filter paper would make the situation worse. Since you are not repeating the experiment anyway, make sure to identify the causes of lower voltage. Secondly, both lead and aluminum nitrate tend to form hydroxides in water if you don't add little bit of acid in them. Did you notice that lead solutions were cloudy or they were clear and transparent?
As a side note, negative voltage reading of the battery just means that your connections of the lead are the other way round on the voltmeter. Test your voltmeter with a battery keeping the signs on the battery in your mind. Once you get a positive reading of a simple AAA cell you have correctly made the connections.

+1. All good calls and the lead cathode should not have been totally submerged with the alligator clip and its attached wire. I wish the filter paper/paper towel salt bridge would go away. Also, the suggestion by @Poutnik (to use potassium nitrate rather than KCl in the salt bridge) is good: no need to risk clogging the ersatz salt bridge with lead chloride.

Eliminate the salt bridge. Test Al and Pb in a) Al(NO3)3 solution and b) (separately) Pb(NO3)2 solution. The salt bridge is for purity of thought and separation of the electrochemical reactions. A third possibility is to put an Al and a Pb electrode in HNO3 solution: any output voltage?
Just try to get a cell going. Aluminum does not dissolve easily in HNO3, and may require some catalyst, like fluoride or chloride. In other words, your aluminum may be passivated. If the lead surface is oxidized (most lead I see is dull, oxidized), it may require some activation to dissolve off the oxide before it is really a good electrode.

I think, there is still something wrong with the salt bridge. It works like a resistor I guess. You loose voltage according the law of Ohm. U = R*I.  Did you try for a short time to use a wire between your two cells. If yes how is the voltage reading. Other possibilities are oxidising of the lead and aluminium as well AND KCl means chloride ions , which can form precipitation of PbCl2. Blocks the saltbridge.
Did you immerse both eletrodes in KCl solution for test? Exchange the saltbridge to potassium nitrate. KNO3.

Using a wire would create a serie of 2 cells. Al wire||wire Pb.


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Can I boil off chlorine? Does it evaporate quickly at high temperatures?
Can I boil off chlorine?
Does it evaporate quickly at high temperatures?
I am asking because I want to remove it from drinking water, and I don't want to wait 24 hours for it to evaporate naturally.

Isn't it usually chloramine, rather than elemental chlorine, in drinking water?

@DavidRicherby Depends on the municipality. Some stick to the traditional molecular chlorine as a disinfectant, but others have indeed switched over to using chloramines, as they're more persistent. (And thus you don't lose disinfectant capability at the edges of your water distribution network.) -- If JMN hasn't already, they should check with their local water works about the type of disinfectant they use. (Most will happily provide information on this and other water quality metrics.)

Yes, solubility of chlorine decreases drastically as the temperature rises, and it's almost insoluble in boiling water.
That's also the reason why in the areas where tap water is chlorinated, it advised to boil it before drinking.
Data from [1, p. 8]:

Figure 5. Solubility of chlorine in water, hydrochloric acid (two concentrations), and sodium chloride solutions (three concentrations) All percentages are weight percents.
In aqueous solutions, chlorine is partially hydrolyzed, and the solubility depends on the pH of the solution. Below 10 °C chlorine forms hydrates, which can be separated as greenish-yellow crystals. Chlorine hydrate is a clathrate, and there is no definite chlorine: water ratio.

References

Chlorine: Principles and Industrial Practice, 1st ed.; Schmittinger, P., Ed.; Wiley-VCH: Weinheim ; New York, 2000. ISBN 978-3-527-29851-8.

Can you comment at what level of water chlorination "they" suggest you boil the water before drinking?

@costrom I cannot speak for the entire world, but in Russia there should be between $\pu{0.3 mg L-1}$ and $\pu{0.5 mg L-1}$ of residual chlorine according to the state sanitary and epidemiological standards. In general I guess it won't hurt to boil any tap water before drinking as it also reduces probability of infection and reduces water hardness. As they say, to make holy water, boil the hell out of it.

Hmmm; boiling water also a way to make untreated water safe to drink.


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How does dehydration of an alcohol occur in alkaline medium?
During aldol condensation reaction the aldol formed reacts further to give alkene (which is probably called enone)

I am really struggling with this part of the reaction as my textbook says nothing about the mechanism of this particular step
the medium during aldol condensation reaction is alkaline but the end product is a dehydrated product and the conversion of alcohol into alkene is dehydration. Do we need to make the medium acidic at later stage of reaction, or does some mechanism causes dehydration in alkaline medium?
Also please suggest if there is some other reaction, or examples depicting similar phenomenon that I could go through for better understanding
(I am really new to chemistry as well as to asking questions. thank you)

Google “e1cb mechanism”.

The commonly accepted mechanism for the base-catalysed aldol example here is that the alpha hydrogen (shown in blue on your scheme) is removed by OH- present in the reaction medium, and the negative charge on the enolate then forms the alkene by loss of OH-. This mechanism falls under the general heading of E1cB, further information on this mechanism may be found here and here 3.
Thanks to @orthocresol for the note about this

Usually aldol products are dehydrated under acidic conditions or at elevated temperatures . But it can also be done employing strong bases like alkoxide or hydroxide ion , and it follows E1CB mechanism . The mechanism involves the formation of an enolate ion which subsequently eliminated the hydroxide ion

Here is the mechanism of Aldol .
Next is the Dehydration Step , shown as below

More Detailed Base Catalysed Dehydration :


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Steps of crystallization

An excess of copper(II) oxide is added to dilute sulfuric acid to make crystals of hydrated copper(II) sulfate.
The processes listed may be used to obtain crystals of hydrated copper(II) sulfate.

concentrate the resulting solution
filter
heat the crystals
wash the crystals

Which processes are needed and in which order?
A 1, 2, 3 and 4
B 1,2, 4 and 3
C 2, 1, 2 and 3
D 2, 1, 2 and 4

I know the correct answer is D. I understand that the solution is first filtered to remove excess copper(II) oxide. Then copper(II) sulfate is saturated by adding excess of it. Then filtered again to remove excess. The last step is wash the crystal. Isn't it needed to heat the solution to get crystal? Also is my understanding so far correct?

There is no mandatory requirement to heat the solution. Then again, how exactly were you going to concentrate it?

@IvanNeretin
Been trying to understand the answer for quite some time now. And these are just some assumption I made

Yes, your understanding is correct. As for the heating, see my previous comment.

Step 2)
You would filter the solution as you would have added excess copper oxide to ensure all acid has reacted.
Step 1)
Then you would concentrate/saturate (saturated solution is a solution containing the maximum concentration of a solute dissolved in a solvent at a specific temperature) the solution by heating,
Step 2)
You would leave the solution to cool and crystallise (form crystals), and so there would be crystals at the bottom and so you would need to filter to collect the crystals
Step 4)
You wash the crystals with cold distilled water to remove impurities. distilled water is purified (no salts, minerals etc) water and lacks ions. This ensures that it will not solubilize the crystals and plus you are only using a tiny amount and not submerging the crystals in distilled water, right after that, you dry the crystals in between filter paper to remove any of the little remaining distilled water


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What reaction takes place here?

Upon reading the given question, I realized that:
The first reaction is an aldol condensation reaction between the given compound and formaldehyde.
So the product after 1st step is:

Now, the second part has thrown me off. The current ketone has no $\alpha$ hydrogen left. Condensation reactions are out of the question.
Cannizzaro reaction can not take place as this is a ketone (and not an aldehyde) also the medium is now acidic.
However, I saw one possibility of formation of hemi-acetals.
(I did not think about full acetals, as such a structure was not present in the given options to this question)
Which gave me the following product:

However, the answer provided seems to be not this.
The given answer is this:

Can someone give a suggestion that where I want wrong and/or how could I reach the answer and also what was wrong in my thinking as well?
This question is taken from the JEE(A)-2016 [Paper - 2]

If that ketone part would've been an alcohol, an easy intramolecular reaction would take place - giving me a full acetal as well as the answer...if only.

@Waylander, thanks for your input. Amazing. However, when my hemi-acetal is formed, I am already on step (ii). At step (ii), an acidic medium is provided instead of a strongly basic one that is needed for Cannizzaro reaction.

You have misidentified the intermediate at the end of step i. The formaldehyde/NaOH conditions will deliver the diol (intermediate 9 in the scheme in the linked answer) through aldol followed by crossed Cannizzaro. The conditions for step ii are for forming the acetal only.

@Waylander, you're correct. Thanks for the explanation. Everything seems to fall in place now.

I think a cross Cannizzaro with the Ketone and Formaldehyde takes place. The ketone give the alcohol and Formaldehyde formiate. In a second step a full acetal will be created with formaldehyd and now the Diole.
Sorry I cannot add comments here. So use answer function.

Isn't it that Cannizzaro reactions happen only between two aldehydes? Sure that there are no $\alpha$ hydrogen present, but they aren't aldehydes either.

It is a "special case" Cannizaro. See the answer linked in the comments for the mechanism.


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Why does water expands when we add salt to it?
When we add salt to water the volume of the water expands. Why is that

When you add salt to water, there is no longer such thing as volume of water. Instead, there is volume of (salt+water).

If one adds potatoes to a cooking pot with soup, the volume of soup expands as well. Any surprise?

Said in easy words: The salt will dissolve into ions, and in this particular case these need space between the water molecules, so the volume will expand.
However this is not always the case! As @JohnCuster points out:

There is no requirement that the volume expands. Things that hydrate (MgSO4 is one example) may actually reduce the volume under appropriate conditions.

There is no requirement that the volume expands. Things that hydrate (MgSO4 is one example) may actually reduce the volume under appropriate conditions.


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Gibbs Free energy of a reversible reaction
From what I understand, it sounds like the Gibbs Free Energy change of a reversible reaction at equilibrium is zero. However, since I know that Gibbs Free energy change depends on temperature, does this not imply that equilibrium can only ever be reached at one very specific temperature? This doesn't sound right, as I know that equilibrium can be established at lots of different reaction temperatures, although the position of equilibrium changes with temperature of course. I asked three chemistry teachers at my school, who all seem just as baffled as I am about this.

At each specific temperature, there is a relationship between the concentrations of the products and reactants at equilibrium such that the Gibbs Free Energy change is zero when the concentrations satisfy this relationship.  The relationship involves the equilibrium constant for the reaction, which is a function of temperature.

A change in temperature may put the system in a state that is not equilibrium. At this point, the reversible system will readjust to be in equilibrium at which point, the free energy change between reactants and products will be zero again.

@Zhe in order to get the free energy change between reactants and products to become zero again, either the temperature, the enthalpy change of the reaction, or the entropy change of the reaction must change, since ΔG = ΔH - TΔS. Which of these changes when the system readjusts to be in equilibrium again?

$\Delta G^0$ is a function of temperature, as are $\Delta H^0$ and $\Delta S^0$.

@ChetMiller My chemistry teacher told me that ΔH and ΔS only negligibly vary with temperature - hence why it's possible to answer questions such as "At what temperature does this reaction become feasible?" using very simple linear algebra.

Well, that statement by your teacher does not address just how much they may change with temperature. For a degree of two, not much. However, remember that in G one has a T*S term...

I'm so confused! @JonCuster I know that ΔG depends on temperature, but what I'm saying is that my teacher stated that ΔH and ΔS only negligibly vary with temperature.

Well, they really can't have it both ways. if G depends on temperature, and G = H-TS, well, there is temperature variance in H and S (generally speaking). What is 'negligible' is open to interpretation.

@JonCuster I don't understand what you mean. If there is temperature variance in H and S, this would imply both can be written in terms of temperature and some other variables. In which case, surely the ΔG equation could be written in terms of only temperature and those other variables? (so you could eliminate ΔH and ΔS from the equation?)

Your teacher is incorrect.  Both $\Delta S^0$ and $\Delta H^0$ depend on temperature.  The correct equation for the temperature dependence of $\Delta G^0$ is $d\Delta G^0=-\Delta S^0dT$.  The superscript 0 designating the standard state means that the pressures of the pure reactants and products are all taken to be 1 bar at all the temperatures considered.  And, of course, $\Delta S^0$ must be a function of temperature because the heat capacities of the reactants and products are not zero.

Thanks, @ChetMiller. I'm still confused because I thought that the little superscript thing implies standard conditions, which means the temperature is 298K. But then that means surely ΔH0 cannot depend on temperature, since by definition, it is measured at 298K? Or do I have the wrong understanding?

It doesn't mean 298 K.  It only implies pressures of pure reactants in the initial state and pressures of pure products in the final state are all 1 bar.  However, because we know how to determine the temperature dependence of $\Delta G^0$, we only need to tabulate $\Delta G^0$ at 298 K.  But the symbol $\Delta G^0$ means $\Delta G^0(T)$.

It's fair to assume in most cases that $\Delta H^\circ$ and $\Delta S^\circ$ do not vary enough with temperature over a range of 100 K or so that you have to factor that temperature dependence into your calculations, and this assumption is made on most school tests. Because of the T in the equation for $\Delta G^\circ$, however, its temperature dependence must be considered. If you have a reaction at equilibrium and you change the temp, $\Delta G$ becomes non-zero and you get net reaction in the forward or reverse direction until $\Delta G$ again becomes $0$ with new concentrations.

it sounds like the Gibbs Free Energy change of a reversible reaction at equilibrium is zero

Yes, that is true. The Gibbs free energy is zero when the reaction has reached an equilibrium, i.e. the reaction quotient Q is equal to the equilibrium constant K. You can express the Gibbs free energy in terms of the standard Gibbs free energy and the reaction quotient:
$$\Delta G = \Delta G^\circ + R T \ln(Q) = 0$$
No matter what the value of $\Delta G^\circ$, there is always a matching $Q$ that will result in a $\Delta G$ of zero. This is why at any temperature, the reaction will be able to reach equilibrium.

However, since I know that Gibbs Free energy change depends on temperature, does this not imply that equilibrium can only ever be reached at one very specific temperature?

As explained above, as long as $Q$ is able to change, there is a state of equilibrium for any value of $\Delta G^\circ$, i.e. at any temperature. However, if all of the reactants and products are pure liquids or solids, the expression for $Q$ is simply 1, and so $\Delta G$ does not change when the reaction proceeds forward or backward. In those cases, $\Delta G^\circ$ has to be zero in order to attain equilibrium, and that only happens at a specific temperature.
There are some electrochemical reactions where all reactant and product species are solids. This is great for making a battery because the voltage won't drop as you discharge the battery. Another more familiar example is the process of ice melting:
$$\ce{H2O(s) <=> H2O(l)}$$
The equilibrium constant expression and the reaction quotient for this process is simply 1. There is only one temperature at normal pressure where ice and liquid water exist side by side, and this temperature is called the normal melting point of water. At a temperature higher than that, water is all liquid, and at a temperature lower than that, it is all ice.

Thanks. Do ΔH and ΔS depend on Q, then?

$\Delta S$ does.

Thanks, @KarstenTheis . So, I think I now understand: as the reaction progresses, the relative concentrations of reactants and products changes, so ΔS changes which causes ΔG to change, until ΔG becomes zero at which point equilibrium is established. In terms of the standard entropy change, then, does that assume that the concentrations of all reactants and all products are equal (since under standard conditions, concentrations are all 1M?) And is that why the standard entropy change has only one value even though actually entropy change is affected by Q?

Yes, that’s it exactly.

Let us continue this discussion in chat.

In general, any expression for $G$ will by necessity have a dependence on $T$, if only through the explicit incorporation of temperature in $G = H - TS$.
Looking at the Calphad (Calculation of Phase Diagram) community, they go further. Referring to A.T. Dinsdale, "SGTE Data for Pure Elements", CALPHAD 15(4) 317-425 (1991), they state, for the purposes of defining $G$ for the elements:

The Gibbs energy is represented as a power series in terms of temperature $T$ in the form:
  $ G = a + bT + cT \ln(T) + \Sigma d_{n}T^{n}$

From the definition of $G$, $S$, and $H$ one can then get:
$S = -b-c-c\ln(T)-\Sigma nd_{n}T^{n-1}$ and
$H = a-cT-\Sigma (n-1)d_{n}T^{n}$
So, one sees, given the chosen representation for $G$, that both $H$ and $S$ should depend on $T$ unless $G$ can be represented in a very simple form (only $a$ and $b$ being non-zero). Since Dinsdale uses a more complex power series, one can rest assured that $G$, even for the elements, tends to be rich in temperature dependence.

In which case, does this mean that equilibrium can only happen at one very specific temperature? Since the only thing that affects G is the temperature, but ΔG is zero at equilibrium? And therefore the power series of temperature must be zero at equilibrium?

@Purpledragonunicorn - no, it just means the equilibrium point (see comments to your question) will shift. For example, the solubility of salt in water is a function of temperature. That comes from the free energies changing with temperature. There is still an equilibrium point, it just changes.

Surely @JonCuster if you change the temperature, that changes ΔG such that ΔG is no longer zero. Now that ΔG is not zero, it will be negative for the forwards reaction and positive for the reverse reaction (or the other way around) meaning that only the forwards reaction is feasible, the backwards reaction is infeasible. And therefore, surely only the forward reaction will proceed, and not the backwards reaction, so it is therefore no longer in equilibrium? Or does something change after that to put it back into equilibrium?


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Molecular weight of NiMnGaFe compund problem
I have a big problems calculating molar/molecular etc masses, because in my case I have a monocrystal, co I cannot imagine it as a molecule and my compound is made of Ni-Mn-Ga-Fe. I know the mass and the atomic and the mass percents and I need to get molecular weight.
In the atomic percents I have: 50% of Nickel, 25% of Manganese, 17,5% of Gallium and 7,5% of Iron. So the ratio (of number of atoms) is 2:1:0,7:0,3 and I can find the atomic masses of all these elements but what is going to be the molecular mass?
I am sorry for probably trivial question (I am really dumb in this area unfortunately and I was searching on the internet for answers but every time the structure of molecule was known and this is a crystal)

:/ It's an alloy not a compound, so there's no molecular or even molar mass.

I was thinking that but I needed to calculated the Bohr magneton f. u. from emu, so I need to search for different equation, thank you

At least now I know for sure.

Seems it's not even a single phase https://www.sciencedirect.com/science/article/abs/pii/S0925838818321406

This alloy has too much of iron so after Laue diffraction it looked as a polycrystal but my samples with less of iron are single phase and succesfuly oriented. Making monocrystal is just harder so there are plenty of articles about polycrystals. The magnetization measurement however were succesful for all.

https://www.researchgate.net/post/How_do_you_convert_emu_g_to_Ub_mewB_per_TM_atom I found the aswer here.