24Sir in first one ...
H2O2 should react with NaHC2O4 left and NaC@O4 formed too naa ????
In third one I have found no .of moles of He=2*10-5
How to finish it ???
I mean my ans isnt matching...
can u do the rest of calculation part ??
Q1 Find out volume strength of H2O2 to eb added (V=1L) in an incomplete titration mixture of NaHC2O4 with NaOH.If mixture has ben prepared by mixing 500g of NaHC2O4 in 2L,0.5M NaOH.Assume H2O2 is st. reducing agent
Q2 Solid NH4I on rapid heating in a closed vessel at 357C develops a constant pressure at 275mm of Hg owing to the partial decomposition of NH4I into NH3 and HI but the pressure gradually increases further(when excess solid residually remains in vessel) owing to dissocaition of HI.Calculate final pressure developed under conditions
NH_4I\leftrightarrow NH_3+HI
2HI\leftrightarrow H_2+I_2 Kc=0.15 at 357C
Q3 Disintegration of radium takes place at an average rate of 2.24*1013 α particles per minute.Each α particle takes up 2 e from air and becomes neutral He atom.after 420 days the He gas was collected 0.5*10-3 L measured at 300K and 750mm Hg.
From above data calculate Avogadro's no.
-
UP 0 DOWN 0 0 4
4 Answers
1357For the first one
H2O2 is used to oxidize the remaining HC2O4- to CO2
So you to find the remaining amount of HC2O4-
Second one:
For first reaction Kp=(275/2)(275/2)
Further
NH4I→ NH3 + HI
275/2+y 275/2-x
2HI → H2 + I2
275/2-x x/2 x/2
So we have, (275/2)2=(275/2-x)(275/2+y)
and 0.15=(x/2)2/(275/2-x)2
1357For third one
no. of particles of He, x =2.24×1013×420×24×60
PV=nRT=xRT/N0
N0=xRT/PV
241357Yeah for the first one both of them will react with H2O2
For third no. of moles=2*10-5
N0=x/n=2.24×1013×420×24×60/2*10-5=6.77×1023