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oh ok thanks , so the ideal gas equation can actually be used for any gas ?
yes it can be..
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oh ok thanks , so the ideal gas equation can actually be used for any gas ?
9701_s09_qp_1.pdf
q 40 someone pleaseee explaainn!!
the bond will be broken from CO2 thats where the Na from NaOH will bond to oxygen.. it will not be 3 bcoz they have taken out the alkene completely so its wrong.. 2 is correct bcoz Na is bonded to O and the rest of the molecule is ok.. so definetly its going to be 1 also.. bcoz there is no option just for 2.. but u can still check it.. see if all the bonds are presnt or not and with OH which comes from NaOH
Thanks .. but can you explain it a bit more coz I still didn't understand why did the bond from CO2 break!
hmm :/
hmm :/
That question is f`d up.. totally went over my mind. Need help in this...
answer is B
amount of sulphite use = conc x vol
= 0.1 x (25/1000)
=2.5x10^-3mol
2 electrons are lost
amount of electrons lost = 2 x 2.5 x 10^-3 ==== 5 x 10^-3 mol
amount of electorns gained by metallic salt = amount of electrons lost by sulphite
= 5 x 10^-3 mol
amount of metallic salt used = 0.1 x (50/1000) ==== 5 x10^-3
amount of electrons gained PER MOLE of salt = (5 x 10^-3)/(5x10^-3) = 1 unit
hence oxidation state of metallic salt decreases by 1 unit meaning the oxidation satate becomes +2 from +3!!! this is for the question 9 , some other person posted it![]()
i got it looksee da above post
no no on the left side it has O and on the right it has H in B part. in C it has O on one side and Cl on the other side. that is what i meant by OPPOSITE poles. thus dipoles are formed.Hay there,
Thanks for helping out, but its still not solved
Option B has H and H on both sides.. (not H and Cl, if that would been, then it would be a sure answer, but its not like that)
Option C has Cl and Cl on both sides..
And I'll be grateful if you explain the question number 10 of the same paper. Why its answer is C, whats wrong in B ??\
Regards
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