Messages posted by mudit.jain

Plz do not ask more than one que in one post. We r given instructions to answer only one.

So here is ans to ur first que.

Conductometric titration is used to estimate the concentration of a solution.

Difference in the conductivity is used as the basic principle for this.

->Suppose u want to find out concentration of NaOH by conductometric titration with HCl

The conductivity of OH- is more than Cl- .

Take NaOH in the beaker and HCl in the burrette.

Initially there were only Na+ and OH- ions in the beaker , when HCl is added then H20 is formed which dissociates very very less?..and now the conductivity is due to Na+ and Cl- ion, cond of Cl- is less than OH-,so the net conductivity decrease till a certain point . And after that it starts to increase when the whole NaOH is consumed.

see attached file for graph

Now , I hope u got the basic concept.

Now u can urself find out graphs for other cases.

ya harsha is correct.

Nitration of aniline gives more than 40% m-nitro aniline.

This is due to protonation of - NH₂ to give ⁺NH₃, which is meta directing and deactivating.

{Because nitration is carried out in the presence of conc HNO3 and concH2SO4, so a lot of H+ is present.}

i'm just adding my views to an already answered query

bahut accha socha baalak!....

So you r thinking that if u pass ray parallel to the surface, it will pass straightway without going into glass slab. You r partially correct.

When we say that the ray is grazing on the surface, the situation actually is little complex.

when the ray is incident exactly at critical angle then the refracted ray is just 90deg to normal. Pay ur attention to 'exactly', i mean to say that as soon as the angle is just more than critical angle(even .0000001deg) then the ray would undergo Total Internal reflection. And when it is just less than critical angle(even 0.0000000001 deg) then the ray is not 90deg to the surface.

By heisenberg uncertainty principle there will always b some error, i.e. one can not make rays perfectly grazing the surface.

Practically a perfect angle of 90.000000000000000000000000000000 deg cant b achieved. So the almost 90 deg ray will always find some point of incidence.

But let us assume somehow one succeeds in making incident rays grazing the interface, then we want only one point of incidence on the glass slab. But suppose glass surface is perfectly smooth and plane such that the ray is just above the surface. Then we will need only one point of glass just one point above the surface so that the ray gets a surface to strike. And then it will trace the same path.

So practically the principle of optical reversibility is always true.

Its BLACK BALL POINT pen....

its more clear in scanning...

"hamen jankar laga accha

jo tum khush hue baccha"

i was busy wid my studies...now xams r over

I am not administrator. But you have asked this question in the expert zone. So my advice is-- you better contact narayana by email or by posting a letter or by phone...only they can give u a satisfactory answer.

f(x)=f(x+pie) but not equal to f(x+pie/2)
so the period of f(x) is a=pie.

similarly g(x)=g(x+pie).
so b=pie

now see the func p(x)=f(x).g(x)

p(x)= 1*(modsinx-modcosx) for all 2n(Pie/2)<=x<(2n+1)pie/2

=> p(X)= p(x+pie/2)

so c=pie/2;

so a-b+c=pie/2....ans

so did u get it?

yes Sreedharan this forum is not for jobs,, We are sorry we cant help u in this

Kane(ofcourse its not his name) has given a perfect solution...u just have to visualise how the hanged bridge looks like.

I hope the solution of Kane is quite satisfactory.

if u have filled the form online...then u can create a new login...and fill the application form again.

The mistake in email id will not b a prob at all ( they r not going to send u an email).

But the address must have been complete.....u can contact JEE office by phone or by mail...they will provide u the solution.

Nivedh has written a nice brief desc of po box.

But i will suggest you to do it practically in ur phy lab.

or atleast read the full procedure of experiment in detail from a lab manual of class 12.(It will b a hectic task to type everything here).

Please prepare this topic and other experiments well, questions will come.

tarin, u can get quizzes on fiitjee(only if u r enrolled in fiitjee) and there r many other webs providing online quizzes on each topic.

U plz search this on web.

Even The ques of cbse are also gud, not only for boards but also for jee.

And i think if you do the ques frm book u will get a thorough practice.

hi my kid,

I think ur friends and relatives r frm cbse or icse. That's why u r worried about ssc.

Harsha and deedee r perfectly right, After all the concept r the same.

But just a friendly suggestion, try to go for cbse in 11th and 12th.......-------------LIFE is very peaceful in CBSE.............
No burden at all, so there is more time for preparation of JEE. And there is one more advantage in CBSE that anyone can get very high %ge in CBSE.....without much effort.

But those guys r right.....the syllabus is almost the same in all the boards..
..if preparing for JEE, it doesnt really depends on which board u r studying frm...

Gud u asked question.

But this is basically IIT-JEE,AIEEE Forum. And the topics u r asking are not in this syllabus.

ur question seems to be incomplete.....

Pervoskite structure is of type ABO₃ (MgTiO₃, CaTiO₃)

->One cation A is bivalent and other cation B is tetravalent

->oxygen occupies centre of the faces

->bivalent cation, A at the corners

-> Tetravalent at the body centre

Now I hope u can do the question easily.

If the atoms of one face diagonal are removed then, there are 2.5 O ,0 .75 Mg, 1 Ti in a unit cell.

Total mass of a unit cell= (2.5*16+0.75*24+1*48)/N_A

Volume of unit cell= 2*Radius of Mg ( because the corner atoms are touching)

Density=mass/volume;

Just calculate?..OK

P4 is insoluble in water(ie P does not react with water), so the water added will form bromine water with bromine(with some bromine in +1 OS).

This Br+ will oxidise P4 to H3PO4 and itself will get reduce to Br-.

Now dont ask why P4 will oxidise to +5 not to +3.

JUST remember the thumb rule->

-> P4, H3PO2, H3PO3 generally oxidise to give H3PO4 and
-> PH3 oxidise to give H3PO3

harsha has written all the equations correctly(though unbalanced)

But harsha the compound is not CrCl3 but is K2Cr2O7 or Na2Cr2O7 ...ramya is right here.

And man ..."coffeebyte"(nice name yaar!!)...the two reactions are just given to confirm what ever u have guessed in the problem statement. There might have been few changes and yet all the reactions might look same...so it is always good to confirm ur guesses in such ques(only if u have time in the exam).

In the first rxn the initial white ppt may be due to formation of AgCl.

As per my knowledge u can take as many courses as u want. You will not find any difficulty in ordering for YG files.

For further queries u can send an email to brilliant tutorial. They are quick in replying.

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