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These institutes promise many things but ultimately what matters is the self study. They will only give give you proper direction but it does not mean that if you join a reputated institute then you will surely clear JEE. Understand the concepts and practice problems, show dedication towards studies and manage time properly. If you want , join any institute and chapterwise keep pace with them. If you will work hard throughout I assure you that you will enter the door of your dream institute.
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We use a mole concept to bring together the concepts of counting numbers and atomic weights of elements.The mole is derived from the following information.Atomic weights are an average of the relative masses of all of the isotopes of the given element.The number of C-12 atoms in exactly 12.00 g of C-12 is 6.02 X 1023.This is called Avogadro?s number.
An amount of a substance that contains Avogadro?s number of atoms, ions, molecules, or any other chemical unit is called a mole.A mole of C-12 atoms is defined as having a mass of exactly 12.00 g, a mass that is equal to its atomic weight.
1 mol 12C atoms = 6.02 x 1012 atoms
1 mol H2O molecules = 6.02 x 1023 molecules
1 mole NO3- ions = 6.02 x 1023 ions.
For atom or compound : no. of moles = (mass)/(molecular mass)
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SNi or Substitution Nucleophilic internal stands for a specific but not often encountered nucleophilic aliphatic substitution reaction mechanism. A typical representative organic reaction displaying this mechanism is the chlorination of alcohols with thionyl chloride and the main feature is retention of stereochemical configuration. Thionyl chloride first reacts with the alcohol to form an alkyl chloro sulfite. In the second step the sulfite group is lost and just like in an SN1 reaction an alkyl carbocation is created. The actual nucleophile in the third reaction step is the chlorine atom attached to the sulfite group which recombines with this carbocation. The crucial difference with the standard SN1 mechanism is that since the nucleophile resides at the same side as the original leaving group i.d. the hydroxyl group, the stereochemistry is retention of configuration and not racemization.
The E1cB elimination reaction is a special type of elimination reaction in organic chemistry. This reaction mechanism explains the formation of alkenes from (mostly) alkyl halides through a carbanion intermediate given specified reaction conditions and specified substrates. The abbreviation stands for Elimination conjugate Base.The reaction takes place around a sp3 - sp3 carbon to carbon covalent bond with an ?-acidic hydrogen atom substituent and a ?-leaving group.This leaving group can be a halide or a sulfonic acid ester such as a tosyl group. A strong base abstracts the ? proton generating a carbanion.The electron pair then expels the leaving group and the double bond is formed. When the first step to the carbanion is slow and the second step fast the reaction is irreversible and named (E1cB) i. When the first step is fast and the deprotonation reversible then the reaction mechanism is(E1cB) r. In the (E1cB) anion variation the carbanion is especially stable with a rapid first step and a slow second step.
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Let the degree of dissociation be x.
At equilibrium PCl5 remaining is 5(1-x) mole and PCl3 formed is 5x mole and Cl2 formes is 5x mole.
Hence total no. of moles = 4 + 5(1-x) + 5x + 5x = 9 + 5x
Now apply PV=nRT
(4.678)(110) = (9+5x)().0821)(523)
x  0.6
Mole fraction of PCl5 = 5(1-x)/(5+9x) = 0.17
Partial pressure of PCl5 = (0.17)(4.678)
Mole fraction of PCl3 = 5x/(9+5x) = 0.25
Partial pressure of PCl3 = (0.25)(4.678)
Mole fraction of Cl2 = 5x/(9+5x) = 0.25
Partial pressure of Cl2 = (0.25)(4.678)
Hence Kp = (0.25)(4.678)(0.25)(4.678)/{(0.17)(4.678)}
= 1.75 atm
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Faculty is good in UIIT.
Lat year following organizations came for recruitments :
Quark, Mohali
Infosys, Bangalore
Teleatlas, New Delhi
Tech Mahindra
IBM,Gurgaon
Wipro Spectra, Bangalore
Shoghi Communcations, Shimla
Nucleus Software, Noida
HCL Technologies, Noida
Perot Systems, Noida
Impetus, Noida
Oracle, Banglore
Infotech, Noida
Cellebrum, Parwannu
RMS, Chandigarh
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Start with basics.
Physics :Refer H C Verma and understand the basic concepts.
Chemistry : Refer NCERT and use P Bahadur for numerical problems.
Maths : Refer R D Sharma initially and then go for Arihant for higher level problems.
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In a screw gauge, the pitch is defined as the distance travelled in one rotation.
Distance travelled in 5 rotations = 5,4 mm
Hence distance travelled in 1 rotation = 5.4/5 = 1.08 mm
Hence the pitch = 1.08 mm
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Start with basics.
Physics :Refer H C Verma and understand the basic concepts.
Chemistry : Refer NCERT and use P Bahadur for numerical problems.
Maths : Refer R D Sharma initially and then go for Arihant for higher level problems.
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Since it is an objective type question, put n=3.
y = 4x3+3x2+2x+1
dy/dx = 12x2+6x+2 > 0 for any x since its discriminant is less than 0.
Hence y is an increasing function and hence it will cut the x-axis only once.
Hence no. of real roots is 1.
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I will explain with the help of an example.
Force of attraction between two bodies is given by :
F = Gm1m2/r2 , to fond the dimension of G.
G = Fr2/m1m2
As force is (mass)x(acceleration0 its dimension is [M][a] = (M)(LT-2)
[G] = [Fr2/m1m2] = [F][r2]/[m1][m2]
[G] = (MLT--2)(L2)/{(M)(M)}
[G] = M-1L3T-2
Write down the dimensions of the quantities in terms of base quantities and simply do multiplication and division of them.
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Cold KMnO4 is a mild oxidizing agent and it will cause the hydroxylation of of alkene forming CH3-CH2(OH)-CH2(OH) whereas hot KMnO4 will cause oxidative cleavage of alkene thus forming CH3COOH , CO2 and H2O.
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It can be written as :
x =  (8+2x)
x2 - 2x - 8 = 0
(x-4)(x+2) = 0
Hence x = 4 , x = -2
But x is not negative hence x = 4
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Yes the shortest distance between the parabolas is along the common normal to both of them. Find the equation of common normal and then its intersection with parabolas. Then apply distance formula to find the shortest distance.
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Due to strong field they force the electrons to get paired thus violating the Hund's rule.
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Velocity of centre of mass = (mv1+mv2)/(m+m) = (v1+v2)/2
From COM frame velocity of 1st mass = v1 - (v1+v2)/2
From COM frame velocity of 2nd mass = v2 - (v1+v2)/2
Hence KE = (m/2){v1 - (v1+v2)/2}2 + (m/2){v2 - (v1+v2)/2}2
= m(v2-v1)2 = mv2
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