brilliant MANSI !!!!!!!!!!!!!!!!!!!!!!!! , WAISE AAP PADHTI KAB HAI( JUST KIDDING)

 

GOOD JOB.............

DONT EVER STOP WRITING.......................

U R GOOD AT IT..................

ONCE AGAIN

BRILLIANT..............( BOTH OF THEM............................)

AAP SHAYAR HI BANENGI ??????????

KYUN...........

_^g2HEY NUPUR , DONT WORRY , MAIN HOON NAA................

(KIDDING).....

NOW ,look...........

use conservation of  energy ........

initial energy= -GM_{^{1 m}}/(d/2) -GM₂  m/(d/2) + 1/2 m v²

⁽  as mass m is d/2 apart from moon as well as earth , therefore , gravitational energy is there , and mass is given a velocity v , therefore kinetic energy is also present.....RIGHT)

NOW ,

FINAL ENERGY= -GM₁  m/$ - GM₂  m/$     ( $--- REPRESENTS INFINITY)

(THERE IS NO KINETIC ENERGY BCOZ AT THAT POINT V=0)

NOW EQUATING , WE GET

  GM₁m/ (d/2) + GM₂ m/(d/2) = 1/2 m V²

A- 1(ankita}
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H- (1) Himanshu
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J- (1) Jyothi
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N-1(nupur)
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U-2 Uday 
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GOOD TRY DUSE , BUT I HAVE TO SAY THIS WANNA NOT WORK .................I HAVE ASKED MY SIR & HE SAID ANS WILL BE IN COMPLEX SORRY 4 DISTURBING U................

GOOD TRY( THUMBS UP)

IN ETHERS ,

CH₃ - O - CH₃

NOMENCLATURE: 

ALKOXY ALKANE

TEREFORE 4 THIS ANS:  METHOXY METHANE

UNDERSTOOD

NOW FOR

ANHYDRIDE:

NOW   4 this ........

 let us take example:

 1)  CH₃ - CO  - O  - CO -  C₂H₅

ETHANOIC ( 4 C2H5)  METHANOIC (4 CH3) ANHYDRIDE .

note: ethanoic is named first bcoz " E" comes first than " M"

2) CH3 - CO -O- CO- CH3

THEN

ONLY      METHANOIC ANHYDRIDE      (NOT METHANOIC METHANOIC ANHYDRIDE)

3)  C2H5 - C0 - O - CO - C3H7

  then

I HOPE U HAVE UNDERSTOOD...........

THANKU............

CONGRATTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS..................................

 

 

 

BUT U KNOW

 

 

 

 

 

I M FEELING JEALOUS..............

 

KOI BAAT NAHIN ...............

 

 

U ROCK MAN...................

KEEP UP.........

all r not absolutely right .........................

now look,

1) inert pair effect  is reluctance of ELECTRONS of s orbital to move out of the shell ,

this is because of 2 options :

   a) s orbitals is very close to nucleus and therefore due to higher coulombic charge , they r very difficult to remove

b) higher SCREENING EFFECT CAUSED BY Selectrons than others

( understood)

 

2)

 N(SiH3)3 is planar  because SI IS PRESENT IN SECOND PERIOD , THEREFORE IT HAS D ORBITALS LEFT AND THEREFORE

LONE PAIR OF N2 r transferred to Si,

therefore planar

but in N(ch3)3  u know CARBON HAS NO D-ORBITALS LEFT...........................THERFORE PYRAMIDAL ...............

NOW U SHOULD HAVE UNDERSTOOD ..

THANKU

 

DIPOLE MOMENT IS A VECTOR QUANTITY ,

= Q * L

( DIRECTED FROM + TO - CHARGE)

IT IS A COMBINATION OF +Q CHARGE & -Q CHARGE PLACED L DISTANCE APART .....................

HOPE U HAVE UNDERSTOOD.............

THANKU

HEY , LOOK .............in these type of compounds

1) find OS of nickel ( it is 0)

2)Ni=28=4s² 3d⁸  and CO IS STRONG LIGAND( NOTE : CO & CN- ARE STRONG LIGAND) , THEREFORE PAIRING OCCURS................        AND ALL HE ELECTRONS COME IN A D ORBITAL 

AND S ORBITAL LEFT BEHIND

AS WE HAVE TO FILL 4 ATOMS OF CO ,

NOW WE HAVE 1 S & 3 P OUTER (N=5 TYPE) ORBITALS

THEREFORE HYB = SP³

ITS DIFFICULT TO UNDERSTAND THE CONCEPT HERE BUT

HOPE I HAVE MAKE U UNDERSTAND ( SORRRRRRRRY OF NOT UNDERSTAND)

BUT IT IS RIGHT

NOTE :

Z=PV/RT....................ONLY 4 ideal gases........

and Z>PV/RT  (FOR H2 & He ) or Z<PV/RT (FOR OTHER REAL GASES)

 SO EQ 2 IS NOT VALID 4 REAL GASES SO NO POINT OF PUTTING RT in EQ 2

HOPE U HAVE UNDERSTOOD ......

anurag has done it nicely,  BUT the confusion i think TRISHA has IS that

as converting

         Ag⁺  +  e-   = Ag        ,

 E⁰ = +.799V  ( Reduction half cell) and note

while converting

      2Ag⁺  +  2e-   = 2 Ag  

E⁰ is ALSO EQUAL TO  +.799V 

RIGHT THEREFORE     IN UR

ANSWER  2* 0.799 V WAS COMING AS u were also multiplying it,

but actually , it remains +0.799V .

understood............

right

 

HEY vishal , look

learn : length of internal bisector is from A is ( 2bc/b+c) cos A/2

THEREFORE , b option is correct .

first u draw the diagram .

now , u take triangle AED ,   

use cosA/2= AD/AE =  ( 2BC/B+C) cosA/2  /AE

cancelling cosA/2 , WE GET ,

AE= ( 2 bc /b+c)

hence option (A ) IS ALSO correct......

now ,

again take triangle AED,   HERE  DE= (2BC/B+C) sin A/2 (USING SINA/2)

AND TAKE TRIANGLE ADF, DF= ( 2bc/b+c)sin A/2

EF= DE+DF =4 bc / b+c  sin A/2

hence C IS CORRECT

now DE=DF ( THERFORE TRIANGLE IS ISOCELES)

OPTION D IS ALSO CORRECT..........

ANS   ----  A, B , C, D,

look ,

electon has higher mobility than hole because when electro moves from VALENCE BAND to CONDUCTION BAND , then it create HOLES .............( WHICH R VERY CLOSE TO NUCLEUS AND experience COULOMBIC FORCE ) , there fore holes have their mobility less than electrons.

SECONDLY , in valence band, CHARGE density is larger , therefore holes have difficulty to move ........but in CONDUCTION BAND where electrons have gone , has charge density very less.............therefore

now u have understood electrons have higher mobilty than holes

hope im right ,

please give some grades........