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29 Dec 2007 13:34:28 IST
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Hey Goiitians I guess the maths q bank is already running well Now here's the chem q bank Plz contribute ur questions..... CHEERS!!!!!
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29 Dec 2007 13:38:09 IST
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LET ME START
A first order reaction A-B, requires activation energy of 70 Kj/mol
. When a 20 % solution of A was kept at 25 degree celsius for 20 minutes, 25% decomposition took place. What will be the percentage decomposition in the same time in a 30 % solution maintained at 40 degree celsius?
Assume that activation energy remains constant in this range of temperature
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JEE 1993
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29 Dec 2007 13:45:12 IST
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OK one more 100 ml of a liquid contained in an insulated container at a presssure of 1 bar. The pressure is steeply increased to 100 bar. The volume of the liquid is decreased by 1ml at this constant pressure . Find the value of  H and U....... JEE 2004
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29 Dec 2007 13:48:08 IST
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OK some questions may come with soln but ill give u soln to my questions later
in this way u can solve the problem without thinking about the answer THERES ONE HAT FOR EACH QUESTION
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29 Dec 2007 13:48:53 IST
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Now the goiitians cmmon
lets post the questions so that others can solve it.....
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29 Dec 2007 13:54:02 IST
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OK now an easy one
The molecule having linear structure is
A. CO2
B. NO2
C. SO2
D. SiO2
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29 Dec 2007 13:54:43 IST
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Simple:
Here apply
K = (2.303/t) Log (a/a-x)
Therefore T = 250C
So T = 273+25 = 298K.
a = 100
a-x = 100 - (25% of 100)
a-x = 100 - 25
a-x = 75.
t = 20 min.
K = 2.303 Log ( 100 )
20 75
K = 0.014 /min
Now we say at 400C it is K'
T = 273+40 = 313K.
Now as per Arrenhius Equation:
Log K' = E (1/T - 1/T')
K (2.303)R
So we put hthe values and we have:
Log K' = 70000 (1/298 - 1/313)
0.014 2.3*8.314
On solving for above expression:
we will get:
Log K' = 0.014 * 0.59
And
K' = 0.055 / min.
Now for Percentage of Decomposition at T' = 400C
lets say it is x'
a = 100
a-x = 100-x'
t = 20 mins
K' = 0.055
K' = 2.303 Log (100/100-x')
20
0.055*20 = Log (100/100-x')
2.303
On solving above we will get:
x' = 67.173
And so percentage = 67.17%
Hope it is useful.
Cheers !!!!!!!!!!!!!!! 
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29 Dec 2007 13:57:04 IST
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A sample of gas is compressed at a pressure of 0.5 atm from a volume of 400 cm3. During the process 8.0 J of heat flows to the surroundings. Calculate the change in the internal energy of the system.
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29 Dec 2007 14:00:37 IST
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Now cmmon some more.....
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29 Dec 2007 14:05:00 IST
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On the basis of ground state electronic configuration arrange the following molecules in increasing O-O bond length order
KO2, O2, O2[AsF6]
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29 Dec 2007 14:10:15 IST
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Two students use same stock solution of ZnSO4 and CuSO4
. The emf of one cell is 0.03 V higher than the other. The conc of CuSO4 in the cell with higher emf value is 0.5 M. Find the cocn of CuSO4 in the other cell.
2.303 RT/F= 0.06
JEE 2003
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29 Dec 2007 14:11:07 IST
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For JEE-04:
From the Question its clear that the process is Adiabatic.
For the first process:
P = 1 bar , V = 100 ml  P = 100 bar , V = 100 ml
Since  V = 0
.'.W = 0
dU = Q + W
dU = W ..........( for Adiabatic process )
dU = 0
For Second process:
P = 100 bar , V = 100 ml  P = 100 bar , V = 99 ml.
W = -P(V)
W = -100(99-100)
W = +100 atm ml
dU = 100 atm ml
as Adiabatic
Therefore, Net
dU = 0 + 100 = 100 ml atm = 0.1 litre atm
For H:
H1 = U + P(V) + (PV)
= 0 + 0 + 100(99)
= 9900 ml atm.
H2 = U + P(V) + PV
= 100 + 100(-1) + 0
= 0.
H = H1 + H2
= 9900 ml atm
= 9.9 Litre atm
Hope it is useful
Cheers !!!!!!!!!!!!!!!!
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Always available for help !
But Remember Don't hesitate to ask a good Question but
Be damn serious for Questioning a weak one.
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29 Dec 2007 14:13:56 IST
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A gas of identical H like atom has some atoms in the ground energy level a and some in excited level b and there are no other atoms in any other energy level. The atoms of the gas make transition to a higher energy level by absorbing a monochromatic light of 2.7eV and subsequently the atoms emit radiations of 6 different photon energies, some of which have energy 2.7eV some have gtreater than that, and some have less than that. (i) Find the prinicipal quantum number of initially excited level b. (ii) Find the ionisation energy of the gas atoms. (iii)Find the maximum and minimum energies of the emitted photons.
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"Do not pity the dead Harry, pity the living, and above all, those who live without love." ~Albus Dumbledore, Half Blood Prince. |
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29 Dec 2007 14:28:48 IST
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Here is one from ma side:
Q)A 3 gm sample containing Fe3O4,Fe2O3 and an inert impure substance is treated with excess of KI sloution in presence of Dilute H2SO4.The entire iron is converted into Fe2+ along with liberation of Iodine.The Resulting sol. is diluted to 100 mL of the diluted solution which requires 11 ml of 0.5M Na2S2O3 solution to reduce Iodine present.A 50 ml of the diluted solution , after complete extraction of the iodine requires 12.8 ml of 0.25M KmnO4 solution in dilute H2SO4 medium for the oxidation of Fe2+ . Calculate the precentages of Fe2O3 and Fe3O4 in the original sample
Ans:49.33% , 34.8 %
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Always available for help !
But Remember Don't hesitate to ask a good Question but
Be damn serious for Questioning a weak one.
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29 Dec 2007 14:33:03 IST
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GR88888 keep it going guys and gals........
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29 Dec 2007 14:41:24 IST
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Ok from inorganic
Element A burns in nitrogen to give an ionic compound B.
Compound B reacts with water to give C and D.
A solution of C becomes milky on bubbling CO2. Identify A, B , C and D
JEE 1997
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29 Dec 2007 15:10:10 IST
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5 g of a substance with molecular wt. 200 is dissolved in 50g of a solvent with molecular wt. 60 and vapour pressure 40cm Hg at 270C. Calculate the vapour pressure of the soln. at this temperature.
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29 Dec 2007 15:14:06 IST
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Thionyl chloride can be synthesized by chlorinating SO2 using Pcl3
. Thionyl chloride is used to prepare anhydrous ferric chloride starting from its hexa hydrated salt.
Alernatively the anhydrous ferric chloride can also be prepared from its hexahydrated salt by treating with 2, 2 dimethoxy propane
Discuss all this using balanced chem eqn.....
JEE 1998
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29 Dec 2007 15:35:04 IST
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A direct current diposits 29.25g of potassium(at. wt. 39) in one minute. How many grams of aluminium (at. wt. 27) will be diposited by the same current in the same time?
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29 Dec 2007 15:54:13 IST
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qs asked by apurvIITJEE Two students use same stock solution of ZnSO4 and CuSO4
. The emf of one cell is 0.03 V higher than the other. The conc of CuSO4 in the cell with higher emf value is 0.5 M. Find the cocn of CuSO4 in the other cell.
2.303 RT/F= 0.06
we can represent the cells as
Zn | Zn2+ || Cu2+ | Cu , its EMF = E1 (LET)
here conc of Zn2+ is = c , we have to find conc is CuSO4
other cell
Zn | Zn2+ || Cu2+ | Cu , its EMF = E2
in this
conc of Zn2+ = c , and of Cu2+ = 0.5M
the total cell rxn =
Zn + Cu2+ -----> Zn2+ + Cu
so , E1 = Eo - (2.303 RT / 2F) log [c/Cu2+]
E2 = Eo - (2.303 RT/2F) log [c/0.5]
E2 - E1 = 0.03 (its given)
also , we have given 2.303RT/F = 0.6
jst solve this
u will get answer as 0.05M = conc of Cu2+
HOPE THIS HELPS
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