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assumin HI to be in excess i.e at least 2 equivalents

yaar actually there are two ans to this question

if you have C6H5-O-C6H5 WITH HI it depends on the condition in which HI is used.

if you use dil HI it will form C6H5OH+C6H5I

if you use hot conc HI then 2C6H5OH

please correct me if i am wrong or nudge me if you have any doubts

yaar another reputation problem. guys you are preparing for IITJEE one of the toughest exams in the world. you know dude prioritise what you wan't~~~~~~~~~~a topper in boards or clearin JEE through a good rank. obviously the ans is the second one. there it is you have ans your own question. PLEASE MAKE IT CLEAR I AM IN KNOW WAY ASKIN YOU TO NEGLET YOUR BOARDS YOUR BOARDS ARE ALSO IMP. you know today there is not much value difference between a 85percent and 95percent. none of them fetch you something special. so basically a guy clearin JEE with a good rank and gettin odd 85 percent in his boards is in a better position than a person who is landing with 95 percent in boards but no way near clearin JEE. and you know that in cbse gettin an 85 percent is not tough. so basically what i wanna say is the only soln to this problem is give time to both the things. now you can afford to give 25 percent of your study time to boards and rest to JEE [assumin your study time is atleast 6-8 hrs excludin your school. maybe you can try do most of your school stuff at school itself so that you get time for JEE other necessary activities. but never neglect your boards. you can even increase the time of your board preps if need be at later stages maybe a few 1,2 months before boards as required. one more thing when you are preparin for JEE your mind should be completely in JEE.forget the rest of the world. coz if you get worried abt anything else nothin will be done properly. same thing applies foe boards. distribute your time well and don't get worried, stay calm and as some one has said you can even overcome a storm..................cheers!

wass the prob?

please could some one provide the detailed soln

sorry made a mistake

2-D
13-B

ans to 4th one comes to 297.62K or 24.62C

4 pretty same asw the third one

use the equation k=Ae^-E/RT

so for the process occurin at different temperature energy of activation will remain the same

k1=Ae^-E/RT1

k2=Ae^-E/RT2

so k1/k2=e^(E/R)(1/T2-1/T1)

so let k1=4.5*10^3

T1=283K

k2=1.5*10^4

T2=?

E=60kJ/mol=60000J/mol

solve and calculate the ans

it is a little lengthy calculation but if you have any doubts then please feel free to ask

for the second one. the decay constant of the reaction is ln2/T
where T is the half life of the reaction = 5730 years
so letAo be the initial conc of carbon in the sample and At be the final so At = 0.8Ao and so usin that formula
ln(Ao/0.8Ao)=(ln2/5730)t
and so t comes out to be round abt 1910 years
please correct me if i am wrong or nudge me if you have any doubts

good yaar

WELL I THINK @yugu901234 is correct

DO YOU REALLY THINK THAT THIS IS THE CASE.
cmon people the equation Eq=Kx just gives us the condition for equilibrium condition. when initially the electric field force is more than the spring's backward pull it gains some velocity as there is some net acceleration acting at the equilibrium position it has a maximum velocity and it goes further due to that velocity it still has until it is stopped by the spring force. the point where the equilibrium is attained is actually the mean position of the SHM and the distance it goes further now from this mean position is the amplitude.
well this is the case when the electric field has not been switched off. and no where in the prob is it mentioned that the electric field is switched off
please correct me if i am wrong

ok the following reactions take place as follows
P4+3O2 - P4O6
lets say x grams was used of P4 in this reaction so the no of moles consumed will be x/124 and no of O2 moles consumed is 3x/124
the other reaction which takes place is
P4+5O2-P4O10
let 2-x grams of P4 take part in this reaction and hence the moles consumed are (2-x)/124 and total number of moles of O2 consumed are 5(2-x)/124
so total no of moles of O2 consumed = 2/32 = 1/16
= ]5(2-x)/124]+[3x/124]
and when you solve it x comes as 1.125
and so 1,125 grams of P4 was used up to produce P4O6 and 0.875 grams was used to produce P4O10
so you know the moles of P4 that reacted in each reaction and hence you can get to know the amt of P4O6 and P4O10 produced
please correct me if i am wrong or nudge me if you have any doubts

i think @rahul1993 method is correct