Autogamous pollination takes place in
(A) Geitonogamy
(B) Xenogamy
(C) Cleistogamy
(D) Chasmogamy
Correct Answer - A `(phi_(1))/(phi_(2)) = (lambda_(2))/(lambda_(1))`
2 Answers 3 viewsCorrect Answer - B No reaction takes place
2 Answers 1 viewsCorrect Answer - A `Ae^(-E_(a)//RT)=[(Ae^(-E_(a_(1))//RT)xxAe^(-E_(a_(2))//RT))/(Ae^(-E_(a_(3))//RT))]^(2//3)` `=[Ae^((-E_(a_(1))-E_(a_(2))+E_(a_(3)))//RT)]^(2//3)` `E_(a)=(2)/(3)[E_(a_(1))+E_(a_(2))-E_(a_(3))]` `=(2)/(3)[180+80-50]=140 ` kJ/mol
2 Answers 1 viewsCorrect Answer - C (c) The first phase (proliferative phase) of menstrual cycle ends on `14^(th)` day, the ovarian follocles rupture and ovulation occurs.
2 Answers 1 views(i) Socialization (ii) Internet
2 Answers 1 views4000 crore, 3400 crore
2 Answers 2 views`t_(1)=(AB)/(v_(b)+v_(R)) . T_(2)=(AB)/(v_(b)-v_(R))` or `v_(b)+v_(R)=(AB)/(t_(1))` and `v_(b)-v_(R)=(AB)/(t_(2))rArr 2v_(b)=(AB)/(t_(1)+(AB)/(t_(2)))=AB((t_(1)+t_(2))/(t_(1)t_(2)))` or `((2t_(1)t_(2))/(t_(1)+t_(2)))=(AB)/(v_(b))`= time taken by the boat to cover AB
2 Answers 1 viewsCorrect Answer - C `DeltaH_("reaction")=DeltaH_(f(C_(6)H_(6)))^(@)-3DeltaH_(f(C_(2)H_(2)))^(@)` `=85-3(230)` `=-605kJ" "mol^(-1)`.
2 Answers 1 views`E_(cell)^(ɵ)=E_(Cd^(2+)//Cd)^(@)-E_(Cr^(3-)//Cr)^(@)` `=-0.40-(-0.74)` `=-0.40+0.74=+0.34V` the half cell reaction are `2Cr(s)to2Cr^(3+)+6e^(-)` `underline(3Cd^(2+)+6e^(-)to3Cd(s))` `underline(2Cr(s)+3Cd^(2+)to2Cr^(3+)+3Cd(s))` `therefore` No. of electrons `n=6` `triangle_(r)G^(ɵ)=nFE^(ɵ)` `=-6xx96500xx0.34` `=-196.86kJmol^(-1)`
2 Answers 1 views(a) Geitonogamy : This is transfer of pollen grains from anther to the stigma of another flower of the same plant. (ii) It is genetically smilar to autogamy since pollengrains...
2 Answers 1 views