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Potential at the centre of 1st ring = kQ1/R + kQ2 / R(-/2)
Potential at the centre of 2ndt ring = kQ2/R + kQ1 / R(-/2)
p.d = k(Q2-Q1){1-1/ -/2 ) / R
=k(Q2-Q1)(-/2 -1) / -/2 R
(B) ans
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Answer is (b) also because since we know that bases are good leaving groups we can have the basicity order
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This is a sense less question without some more data.
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I would myself prefer answer (d) .
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Reading Morrison is very interesting -----its very rude to say its lengthy ---I have not enjoyed any other book more---even if you read FOR FUN YOU WILL LEARN A LOT and that too faster.
Have you read Kekule's Dream of Benzene ring..........excellent....
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How obvious ....
The horse is pushed forward by the grounds reaction force --implies---horse pushes in backward direction---implies friction acts in forward direction.
Now when the wheel rotates the wheels move forward --implies friction is backward---(its a matter that the friction will oppose the motion.
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Actually the answer given to us by our teacher is also c but I am confused that Raoults Law does not state anything about i .It is a matter of vant hoff's correction.
We do not write raoults law in the way as (c).
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Gas eudiometry deals with mainly the computation of amt of given gases in the reaction mixture which gives product mixture
It requires a knowledge like oxygen is absorbed by alkaline pyragallol ,CO2 by lime water.
The theory is long you should refer a good book.I studied it in by Brilliant's Corresp. Course
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According to the raoult's law the rel. lowering of vapour pressure is directly proportional to- a-moles of solvent b-mole fraction of the solvent c-i times the mol fraction of solute ,i -->vant hoff's factor d-none of these
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Thats too confusing at the moment for me
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Question not clear
Explain the specific arrangements
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Sorry the formatting went wrong
--------- Ist Group------2nd Group--------3rd Group
1I.P------low--------------low------------Moderately High
2I.P-----very high------- low---------------- High
3I.P.---very very high--very high-----------High
Prefer---ionic----------- ionic-------------- covalent bonding
Ex.------ Na--------------- Mg---------------- Al
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Ist Group 2nd Group 3rd Group
1I.P low low Moderately High
2I.P very high low High
3I.P. very very high very high High
Prefer ionic ionic covalent bonding
Ex. Na Mg Al
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We know that the atomic size decreases across the period due to increase in the attractions with charge
Hence the given result but you are wrong to say that
Al < Mg ---Mg comes second in the period 3 and Al comes later
I hope you understand now
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Solder is an alloy of tin and lead
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Have the soln but can't write it here
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(a)
Since when the alpha particle as it approaches the nucleus the P.E. increases and K.E. decreases i.e K.E. changes into P.E.
0.5mv^2 = k (q1.q2)/ r = X/r since kq1q2 = X = constant
r=2X/mv^2
When the velocity becomes thrice then the closest distance of approach becomes 1/9 times
Thus ,
x/9 is the answer.
Second Question--
Since V= E.d
Pot on surface = kq/R
Pot at 3R = kq/3R
thus acc to you V=2kq/3R
Thus V at 3R =V/2
Since V= E.d
E = V/6R (d) Ans
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No book better than Morrison And Boyd for reaction mechanisms
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Zn + CuSO4 ------------> Cu + ZnSO4
0.4 1mol 0 0
0.4-x 1-x x x
1-x = 0.8
x= 0.2
Thus Zn = 0.2 but ZnSO4 is insoluble in water thus I think the M =0
else if you don't seek the solubility in water then 0.2
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