Dear Aspirant,

the repeated questions for NEET examinations are generally in most cases low, as the level of competition is very high. In most cases similar kinds of questions are asked instead of repeating the same question. However, there are a few times where a question or two are repeated from the previous years. Hence to get to know of some of these repeated questions or similar questions that are asked through out the year in NEET, you may solve the previous years question papers. They are very beneficial as it not only gives you an idea of the similar questions but also helps you get familiar with question paper pattern.

To view the previous years question papers, kindly visit the link provided below.

https://medicine.careers360.com/articles/neet-question-paper

Hi,

The process of formation of soap is referred to as saponification.

Basically, it is a  conversion of fat, oil, or lipid , into soap  and alcohol by the action of heat in the presence of aqueous alkali. Soaps are basically salts of fatty acids that have long carbon chains like Sodium palmitate. These reactions are irreversible in nature and are exothermic.

I hope this helped you.

Dear Student,

Please study from books your NCERT books or you can search your textbook related questions on different platforms like toppr and doubtnet because here I can not teach you and explain you all these.You have to remember the formulas to solve this kind of questions and should know basics of balancing the chemical reactions.

Thanks

Hello Aspirant,

DISSOCIATION CONSTANT AND IONIC PRODUCT OF WATER - DEFINITION

H2OH++OH−

This ionization is known as self-ionization of water.

Dissociation constant=K=[H2O][H+][OH−]

Ionic product of water may be defined as the product of molar concentration of H+ ions and OH− ions.

Kw=[H+][OH−]=10−14

K=55.5510−14=1.8×10−16

IONISATION OF WATER :-

The self-ionization of water (also autoionization of water, and autodissociation of water) is an ionization reaction in pure water or an aqueous solution, in which a water molecule, H2O, deprotonates (loses the nucleus of one of its hydrogen atoms) to become a hydroxide ion, OH−.
H2O→H++OH−

I hope this will help you.

Feel free to ask any query.

All the best for your bright future ahead.

Hello,

Aluminium has vacant d-orbitals. So, it can expand its coordination number from 4 to 6. Thus, it forms octahedral [AlF6]3- ion in which Al undergoes sp3d2 hybridisation.

On the other hand, Boron does not have d-orbitals. Therefore, it can have a maximum coordination number of 4. So, boron can not form [BF6]3- ion.

Hope it helps!

Good luck!

12. Assertion and Reasons are both true and it is the correct explanation.

Si has d-orbitals to accept electrons donated by water but C does not.
Therefore, SiCl4SiCl4 reacts with water but CCl4CCl4 does not.

13. Assertion is not correct but Reason is correct

Resonance hybrids are always more stable than any of the canonical structures would be, it they existed. The delocalisation of the electrons lowers the orbital energies, imparting stability. The gain in stability of the resonance hybrid over the most stable of the canonical structure is called resonance energy.
A canonical structure that is lower in energy makes a relating greater contribution to resonance hybrid. Thus, the correct assertion will be energy of resonance hybrid is equal to the sum of energies of all canonical forms in proportion of their contribution towards the resonance hybrid.

Hello,

I will share with you a link from where you can download JEE Mains related Ebooks available on this platforn:

https://engineering.careers360.com/download/jee-main-ebooks?page=1

Also you can download previous year question papers and sample papers which you can practice during your preparation time. Below I have shared a link where you can get sample question papers.:

https://engineering.careers360.com/articles/jee-main-question-papers

v https://engineering.careers360.com/articles/jee-advanced-mock-test

I will also share a link from where you will get to know the best study materials for JEE Mains exams:

https://engineering.careers360.com/articles/best-study-material-for-jee-main

Good Luck

Dear student,

Equilibrium Constant ( k ) = [products]^x / [reactants]^y  { where x and y are stoichiometric coefficients of products and reactants respectively}.

Now, coming onto the question :-

K1 = ([A]/[B])^3

So, [B] = [A]/(K1)^1/3

K2 = ([B]/[C])^2

So, [C] = [B]/(K2)^1/2 = [A]/((k1)^1/3 (k2)^1/2)

Now, K3 = [D]/[C]

So, [D] = [C]k3 = [A] K3/ ((k1)^1/3 (k2)^1/2)

Now, as D is in equilibrium with A ,

So, K = [D]/[A]

K= [A] K3/ [A] ((k1)^1/3 (k2)^1/2)

So, the equilibrium constant for this reaction is:-

K = K3/ ((k1)^1/3 (k2)^1/2)