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We can think in the reverse way .
If we remove the halogen atoms then carbocations are formed
Now since 1 deg carbocations are least stable thus is most stable
alkyl halide and similarly the 2 deg and 3deg are more stable then their corresp halides will be unstable.
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Sorry I wrote this at the wrong place
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Moderator Sir ---you must do something about the new shameless user He is not fit to be here I think.
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I have noted that the electronegativities can be used to decide these matters in case of cells at least
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Zero because the charge will directly flow to earth (and because in the sphere the charges will not be static)
At least I think so.
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Capacitor can be defined as a device that stores electrical energy or electric field .
Saying STORES CHARGE WOULD BE WRONG AS NET CHARGE IS ALWAYS ZERO
It can also be used to provide a potential difference though it will vary with time as charge decreases.This is used in giving electric shocks.
You can find one in you house in the ceiling fan.It is cylindrical in shape there.
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What rubbish Anbareesh ......Check out your book
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If you could apply the brackets a bit more properly
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y=sin(-1)x +cos(-1)x
=pi / 2
thus it is a constant function
dy/dx =0 ANS
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nth term is a+(n-1)d
2n+1th term is a+2nd
a+(a+2d)+(a+4d)............a+2nd
=(n+1)(a+a+2nd)/2
=(n+1)(a+nd)
Second Part--
(a+d)+(a+3d)+..............(a+2nd-1)
=n(a+d+a+2nd-1)/2
Sum = (n+1)(a+nd) - n(2a+2nd+d-1)/2 ANS
(YOU CAN SIMPLIFY IT)
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The substance whose oxidation no decreases is reduced i.e it is the oxidizing agent
The substance whose Oxidation Nomber increases is the reducing agent as it is itself oxidized
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I am getting the answer probably it is
a0.t0 / 2 Ans
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LHS=sinA/cosecA +cotA
=sinA(cosecA-cotA)
=1-cosA
RHS=2-sinA(cosecA+cotA)
=1-cosA
Thus LHS=RHS
Hence Proved
It is hence clear that the question has a misprint
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Because carbon is also a small atom and due to the three hydrogen there exists high e- density over the C-atom of CH3
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The answer is only that to cancel the away field of point charge at the centre of cube the cube also exerts the same field towards charge q as the field inside will be zero
Thus ans is q/4piED^2
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The Kirchoff's Laws are the major tactics to solve the problems....
They give a definite surity to solve the questions
Master them --you will never face problem
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Hope you write the rest of HCV soon
Just Joking
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Dont you have RCM it contains the theory
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Himanshu what you wrote is useless until you try and understand what I am confused with----------One cant change the law like that
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The best way to get the limiting reagent ---which is the reactant that controls the amount of products that will be formed is to calculate the produce due to each reactant and then choose the one that gives the least amt --it is the limiting reagent
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