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for ur Q x^2+y^2 ,is it from 1st 100 nos?????????
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here A=4.5*10^7 * e^(2.2*10^--------)
take log of e.now multiply it by its power.now take antilog.
u can then multiply the ans with 4.5*10^7.
remember wat we do for squaring.
for eg: to cal a*a
v take log of a and multiply it with 2,for sq root v divide it by 2.
same way can b used for this.
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first oxidise the CHO grp to COOH with PCC.C=O will not b affected.
now reduce C=O grp using H2/Pd to CHOH.COOH will be unreactive to H2/Pd.convert COOH to COCl and do rosunmund's reduction to get back CHO grp.
plz tell me if there is anything wrong in this method
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i think it hsud be NH4+ not -.its hybridisation is sp3.
shape:tetrahedral.
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sorry , i agree with harsha.dint see the - sign.
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but thats not the right ans .
me too did the same way.in this method v get a cube term.but in the solution there is no such term.
the ans is Vr= r^2-r^2/2 +r/2
hows that possible.
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Let r V denote the sum of the first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r-1) . Let Tr= Vr+1 -Vr -2 andQr= Tr+1- Tr for r = 1, 2,?. find Vr!!!!!!!!!!!!!!!!!! plz reply soon rates assured !!!!!!!!!!!!!!!!!!
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Br will be replaced by CN.3-cyano nitro benzene will be formed.
correct me if m wrong.
but i think this is what wil happen
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thta is the case when the 2 eqns of the lines r given,but here combined eqn is given ,how did uu split it up.
xplain!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!plz
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ans is 2/3 which is the fractional part of the integral.
@ ashish_banga i think he is asking for the fractional part.,otherwise ur ans is right.
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thats right but how did u find the eqn of the bisectors????pl xplain
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its easy!!!
in the 1st one 3x-2y-5=0
=>3x-5=2y
multiply by 2, 6x-10=4y--------------(1)
4y-3z+7=0
=>4y=3z-7-----------(2)
from 1 and 2,
eqn: 6x-10=4y=3z-7
=> (x-5/3)/1/6 = y/(1/4) = (z-7/3)/1/3
u can try the same for the other eqn
hope this helps
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area of triangle formed by the angle bisectors of the pair of lines x2 -y2+2y-1=0 and the line x+y=3 is: a)1 b)2 c)3 d)4 my doubt is how to find the eqn of the bisectors??????? plz reply soon rates assured. !!!!!!!!!!!!!!!!!!!!
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yup chandini is right.
arrehenius eqn is K=Ae^-E/RT
=A*1/e^E/RT
so as E increases,the denominator becomeseven more smaller , the value on RHS bcomz larger and so K increases.
so rate inc with inc in energy gap.
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FACIAL:it means smane type of atoms or grps are present on the corners of a triangular face.
MERIDIONAL:it means atoms or grps r present along the meridian of the square plane.
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m.wt of water content=249.17-159.6=89.57g
so there is 89.57 of water in 249.17g of hydrate.
amt of water x is in 3g of hydrate
x=89.57*3/249.17=1.07g
so wt of anhyd shud be 3-1.07=1.93g
but the wt is 2.00g
so error%=1.93*100/2 =96.5%
hope this helps
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thats right ,it cannot be a polymer, it serves as the monomer for many other polymers.
polymers are those which r composed of long chains of same units.
sucrose on hydrolysis gives only 1 molecule of glucose and fructose,so its not a polymer.
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a,b,c are non-coplanar vectors and b1=b-a(b.a/|a|^2 ) b2=b + a(b.a/|a|^2 ) c1=c-a(c.a/|a|^2 ) + b1(b.c/|c|^2) c2=c-a(c.a/|a|^2 ) - b1(b.c/|b1|^2) c3=c-a(c.a/|c|^2 ) + b1(b.c/|c|^2) c4=c-a(c.a/|c|^2 ) - b1(b.c/|b1|^2) then the set of orthogonal vectors is 1) (a,b1,c3) 2) (a,b1,c2) 3) (a,b2,c2) 4) (a,b1,c1) ans is 2).plz xplain how. rates assured !!!!!!!!!!!!!
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