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i am in from 24th . then only mcqs will be done
yep thnx a tonMoles Of KOH= 0.025*0.01 = 0.00025 moles
H2SO4:KOH
2:1
Moles of H2SO4 = 0.000125
H2SO4:CaSO4
1:1
0.00125 moles of CaSO4 .. volume was 50cm^3
M= 0.000125/0.05
M=0.0025 mol/dm^3 answer so A would be the answer.. is that correct?
yep thnx a ton
actually I got it alreadyLol.. What do I say! I logged in to see your post and after finishing the question in just 3mins, my internet connection was gone. Really sorry for replying late
The way I did it is like this:
50cm3 of H2O and calcium sulphate
1Ca ion is replaced by 2H ion
so, the reaction of KOH with the H2SO4 is:
H2SO4 + 2KOH --------> K2SO4 + 2H2O
no. of moles of KOH = (25 x 1.0x10^-2) / 1000 = 2.5x10^-4 mol
Stoichiometric ratio is 1:2
Therefore, no. of moles of sample solution is (2.5x10^-4)/2 = 1.25x10^-4
So,
1.25x10^-4 = (50x concentration)/1000
therefore, concentration = 2.5 x 10^-3 mol/dm3
So A is the answer
actually I got it already
whateverLol! I saw it after I posted mine
Mass of 1cm3 of H2O = 1g
no. of moles of H20 =( 1/18 ) = 1.33mol
Now, solve this using unitary method:
Volume of steam at 298K = 1.33
Volume of steam at 1K = (1.33) / 298
Therefore, Volume of steam at 596K = (1.33 x 596 )/ 298 = 2.6666... round off to 2.67. So answer is C
If you follow the syllabus, there is a better way to find this.
By using the following equation. PV = nRT
where pressure is in Pascals, Volume is in m^3 , n = Moles, R is a constant which is 8.31 and Temperature in Kelvins (K)
Here water is 1 g/cm
So the number of moles will be 1 / mr = where mr of water (H2O) is 18, So 1/18 is the number of moles of water.
So to get the answer is simple.
PV = nRT
101000 x V = 1/18 x 8.31 x 596
V = ( 1/18 x 8.31 x 596) / 101000
V = 0.0027242904290429 m^3
Multiply the answer by 1000 to change m^3 to dm^3 this gives you = 2.72 which only shows similarity to option C ... Good luck.! the answer is a bit different because i used the R value as 8.31 , it might be different in the data booklet but 8.31 is also the closest value which has always worked for me.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
how to do question no 3 and 36?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
how to do question no 3 and 36?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
how to do 15, 17 , 20, 26 , 28....
15. The oxidation number of Ag in AgCl is 1+; in the [Ag(S2O3)]3- ion, it's 1+ too. Therefore B, C & D are ruled out.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf
how to do 15, 17 , 20, 26 , 28....
. okayWish I could explain it to you now.. Got an exam tomorrow.. Nai kar sakta today.. I did this paper long time ago. Need to study it again :L
Thanks do u know the other questions i mentioned?15. The oxidation number of Ag in AgCl is 1+; in the [Ag(S2O3)]3- ion, it's 1+ too. Therefore B, C & D are ruled out.
For Q.17 is actually very simple. (A) can occur because NH3 can be a base and SO3 can be an acid. (B) is also correct because in this reaction, ammonium salt is formed where ammonium itself involved dative covalent bond. (C) is correct because NH4+ and SO42- to give (NH4+)2SO4 which can be considered as Ionic bond formation.Wish I could explain it to you now.. Got an exam tomorrow.. Nai kar sakta today.. I did this paper long time ago. Need to study it again :L
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