The balanced chemical equation for above question is-

NaClO3 + H2O -> NaClO4 + H2

The oxidation state of Cl atom in NaClO3 is +5 and the oxidation state of Cl atom in NaClO4 is +7. So, the net change in oxidation state = 7-5 = +2.

This show that 2 faradays of charge is required to get 1 mole of NaClO4.

So if 1 faradays of charge is  passed through NaclO3  then (1/2) mole of NaClO4 obtained.

Then 3 faradays of charge is  passed through NaclO3  then (1/2)x3 mole of NaClO4 obtained that is 1.5 mole NaClO4 is obtained.

Hello candidate,

1. When 2-Bromo Propane undergoes dehydrohalogenation, action involves the removal of halogen and hydrogen group with the addition of a double bond which gives rise to the product as Pentene.

2. When Methyl Bromide reacts with potassium cyanide, then a simple substitution reaction happens which gives the product as methyl cyanide.

Hope this information was informational!

Hello aspirant,

Hope you are doing well

Elements of the lanthanide class have the most permanent oxidation state of +3, which is why lanthanides generally exhibit a +3 oxidation state. There are some lanthanide elements that exhibit +2 and +4 oxidation states, also known as incompatible oxidation states.

Hope this helps you

Feel free to ask any questions

All the best for your future

Hello,

Bragg's law-

In physics, the relation between the spacing of atomic planes in crystals and the angles of incidence at which these planes produce the most intense reflections of electromagnetic radiations, such as X rays and gamma rays, and particle waves, such as those associated with electrons and neutrons.

This is the main principle of Bragg's law.

All the best.

Hello Aspirant,

For JEE-Main 2021, complete chapter wise weightage for each subject is described below :-

@Physics

Modern Physics:- 20 marks

Heat and Thermodynamics :- 12 marks

Optics:- 12 marks

Current Electricity :- 12 marks

Electrostatics :- 12 marks

Magnetics :- 8 marks

Unit, Dimension and Vector:- 4 marks

Kinematics :- 4 marks

Laws of motion :- 4 marks

Work, Power and Energy :- 4 marks

Centre Of Mass, Impulse and Momentum :- 4 marks

Rotation :- 4 marks

Gravitation :- 4 marks

Simple Harmonic Motion :- 4 marks

Solids and Fluids :- 4 marks

Waves :- 4 marks

Electromagnetics Induction and AC :- 4 marks

@Chemistry

Transition Elements and Coordination Chemistry  :-  12 marks

Periodic table and Representative Elements :- 12 marks

Thermodynamics And Gaseous State:- 8 marks

Atomic Structure:- 8 marks

Chemical Bonding:- 8 marks

Chemical And Ionic Equilibrium:- 8 marks

Solid State And Surface Chemistry :- 4 marks

Nuclear Chemistry And Environment:- 8 marks

Mole Concept :- 4 marks

Redox Reaction :- 4 marks

Electrochemistry :- 4 marks

Chemical Kinetics:- 4 marks

Solution and Colligative Properties :- 4 marks

General Organic Chemistry:- 4 marks

Stereochemistry:- 4 marks

Hydrocarbon :- 4 marks

Alkyl Halides :- 4 marks

Carboxylic Acid and their Derivatives:- 4 marks

Carbohydrates, amino acid and Polymers :- 4 marks

Aromatic Compounds :- 4 marks

@Mathematics

Coordinate Geometry  :-  20 marks

Limits, Continuity and Differentiability :- 12 marks

Integral Calculus :- 12 marks

Complex numbers and Quadratic Equation :- 8 marks

Matrices and Determinants :- 8 marks

Statistics and Probability :- 8 marks

Vector Algebra :- 8 marks

Sets, Relation and Function:- 4 marks

Permutations and Combinations :- 4 marks

Binomial Theorem and Its Application:- 4 marks

Sequences and Series:- 4 marks

Trigonometry :- 4 marks

Mathematical Reasoning :- 4 marks

Differential Equation :- 4 marks

Statics and Dynamics :- 4 marks

Differential Calculus :-  4  marks

For complete details about important chapters and other weightage related information kindly check out the links given below :-

https://engineering.careers360.com/articles/jee-main-syllabus-weightage

You can also check out JEE-Main 2021 Learner package for your preparation :-

https://learn.careers360.com/jee-main-coaching/?utm_source=%20QnA

I hope this information helps you.

Good Luck!!

The anionic sites occupied by unpaired electrons are called F-centres (from the German word Farbenzenter for colour centre). Such defects also impart color to the crystals. The color results by excitation of these electrons when they absorb energy from the visible light falling on the crystals. For example, impart yellow color to the crystals of NaCl.

I hope my answer helps. All the very best for your future endeavors!

hello aspirant

structure of Propionic alcohol

CHOH

C - H-OH

Propanoic alcohol

It is a liquid with pungent and unpleasant  smell.

It is prepared from propanoic acid

The difference between 1- propanol and 2- propanol is position of OH group.1- propanol is chain structure while 2- propanol is branched structure.These are used in hand disinfectants

2nd excited state implies that n=3.

Now the inner most radius i.e., for n=1, r'= 5.3*10^-11 m

Now we know radius of nth orbit is given by r(n) = r' * n^2

There the radius of the second excited state i.e., r(n =3) = r' * 3^2 = 5.3*10^-11*9 = 4.77*10^-10 m.

I hope my answer helps. Have a great day!

Dear Student,

amino acids are essentially an organic compound which contains the compounds of amine and carboxyl functional groups along with a side chain specific to each amino acid.

Amino acids are amphoteric since they tend to have acid and basic tendencies. The carboxyl group of the amino acid is able to lose a proton and the amine group is able to accept a proton.