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thanksOh, thats right, i apologised.
Option A is actually C-F bond broken.
For option C, it is the C-Cl bond broken.
So its is option C, but using the explanation I mentioned in post #9124
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thanksOh, thats right, i apologised.
Option A is actually C-F bond broken.
For option C, it is the C-Cl bond broken.
So its is option C, but using the explanation I mentioned in post #9124
2003can i know the year please?
may june!2003 what session may or oct?
ques 27 see itit's not there!!!!!!!
D should be the correct answer!
becoz faster the reaction, lower the Ea and lower the hump and since reaction 2 is faster it should have a lower hump!
hahahaSORRY
Qn 9 :
But 1 mole of so3 2- gives 2 electrons so 2 moles will give 4 ?Qn 9 :
moles of SO3 2- : moles of electrons : moles of metal
0.0025 : : 0. 005
1 : : 2
1 : 2 : 2
2 mol of metal gained 2 mol of electrons
1 mol of metal gained 1 mol of electron (oxidation state will thus decrease by 1)
Original oxidation state of metal = +3
Final oxidation state = +3 - 1 = +2
Q32. Info is implying that CO2 is more stable than CO
A. Is correct as CO2 (+4) and CO (+2)
B. CO2 is more stable than CO, so formation of CO2 is more exothermic than CO.
C. Eqm lies strongly to left as we expect a high amt of CO to be converted to CO2. Kc is thus high.
Yes, based on the half equation, your statement that 2 mol of SO3 2- loses 4 mol of e is true, just that it's not relevant to this particular question.But 1 mole of so3 2- gives 2 electrons so 2 moles will give 4 ?
For elimination, the Cl is removed together with H from a "next door" carbon.
For elimination, the Cl is removed together with H from a "next door" carbon.
Note that there are no H on the "next door" carbon for option D.
Yes, that is a valid general guideline.Thanks a lot! The thing that got me confused is that chemguide.com mentions that Tertiary Halogenoalkanes mainly undergo Elimination, whereas Primary undergo substitution.
Thanks , what about q37 in the same paper do u know how to solve it ?Yes, based on the half equation, your statement that 2 mol of SO3 2- loses 4 mol of e is true, just that it's not relevant to this particular question.
If my previous presentation is not clear, I'll put the reasoning in step by step statement form.
0.025 mol of sulfite reacts with 0.05 mol of metal
1 mol of sulfite reacts with 2 mol of metal
1 mol of sulfite loses 2 mol of electrons, which are gained by 2 mol of metal
Therefore, 1 mol of electron is gained by 1 mol of metal
The key to solving such redox mol calculations is usually to find out the ratio of
Mol of oxidized substance: mol of electron transfered: mol of reduced substance.
Hope it makes sense.
but y kc is half how we square root 1 to get 0.5 i didnt ge t this partThe Kc expression for reaction I is
Kc(1)= [X2Y]^2 / [X2]^2 + [Y2]
And for reaction two
Kc(2)= [X2] + [Y2]^1/2 / [X2Y]
Now we should rearrange the first expression in order to get a form similar to second expression.
( [X2]^2 + [Y2]) Kc(1) = [X2Y]^2
( [X2]^2 + [Y2]) Kc(1) / [X2Y]^2 = 1
[X2]^2 + [Y2] / [X2Y]^2 = 1/Kc(1)
now square root all terms to get
[X2] + [Y2]^1/2 / [X2Y] = 1/(Kc(1))^1/2 (see the power of half is a square root)
we can see that
Kc(2) = 1/[Kc(1)]^1/2
= 1/(2)^1/2
Q6... first of all we will find the moles of ice => mass=density x volume = 1x1= 1ghttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf q6. ,19 and 28 can someone please help me with them
Q28 ?Q6... first of all we will find the moles of ice => mass=density x volume = 1x1= 1g
the mole of ice = 1/18 =0.06
assumng ice as steam:- at room temp the volume will be 0.06 x 24 = 1.32
and then by the ratio we will find the volume of steam produced at 596 K
V1/T1 = V2/T2
1.32/298 = V2/ 596
V2 = (1.32 x 596) / 298 = 2.64 cm3
Q19:- when Ca(OH)2 reacts with SO2 so initially CaSO3 is formed and then CaSO4 will form... so the ans is C
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