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Just draw the mirror diagram of the organic compoundhttp://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_2.pdf
4 (b) (ii)
someone PLEASE help!
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Just draw the mirror diagram of the organic compoundhttp://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_2.pdf
4 (b) (ii)
someone PLEASE help!
Not getting it.Just draw the mirror diagram of the organic compound
okay now question 26 .. hope I'm not disturbing you
See the data booklet for electrode reactions.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_42.pdf
Please explain Q2 b....i will be very thankful!
How A is Oxidised ? can you write the equation ... !?A and C get oxidised with nothing getting reduced. D no reaction. In B, the organic reagent gets oxidised and Ag gets reduced, thus answer is B.
i asked for part b, thanks for the reply thoughSee the data booklet for electrode reactions.
Use Cr+3 + e=Cr+2 value is -0.41
Cr2O7+14H+6e.............
Rest just write the reaction.balance by multiplying eq1by 6.Cr+3 on right will add up.
any A2 candidates???
E is CH3COO-http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s09_qp_2.pdf
4 (b) (ii)
someone PLEASE help!
pH = pKa + lg([salt]/[acid]) (or pH = pKa - lg([acid]/[salt]) )i asked for part b, thanks for the reply though
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf
Q3
31
Question 36 ( why 2 is wrong ?)
please help !
Loads of ThanksQ3: C
A Charge on the cation is ruled out. Mg is 2+ and Na is 1+ yet they both have the same lattice.
B Ratio of the ionic charges is ruled out. The ratio is the same for all three lattices.
C NaCl = (102/181 = 0.56) MgO = ( 72/140 = 0.51) CsCl = (167/181=0.92). So definitely this.
D Ruled out. Sum of ionic charges is 0 in all cases.
Q31: D in a heartbeat.
2 is ruled out the second you see it. P4O10 will react with the ammonia produced. This alone eliminates A B C. 3 is wrong anyway and even an O Level kid will tell you that!
36: D again.
1- Increased surface area = more catalyst present = reduced NOx conc.
2- More exhaust gases same surface area. Catalyst being used completely. Some gas escapes without making contact with the catalyst. NOx goes up
3- Higher temperature = much more NOx produced.
Plz help me with these questions
Octobernovember 2009 paper42 q5 part A.
Octobernovember 2009 paper41 q3 bi and bii
October November 2008 paper4 q2 part b.
THANKS!!
Abby can you answer the question of paper5 posted ab ahmed abdullahQ3: C
A Charge on the cation is ruled out. Mg is 2+ and Na is 1+ yet they both have the same lattice.
B Ratio of the ionic charges is ruled out. The ratio is the same for all three lattices.
C NaCl = (102/181 = 0.56) MgO = ( 72/140 = 0.51) CsCl = (167/181=0.92). So definitely this.
D Ruled out. Sum of ionic charges is 0 in all cases.
Q31: D in a heartbeat.
2 is ruled out the second you see it. P4O10 will react with the ammonia produced. This alone eliminates A B C. 3 is wrong anyway and even an O Level kid will tell you that!
36: D again.
1- Increased surface area = more catalyst present = reduced NOx conc.
2- More exhaust gases same surface area. Catalyst being used completely. Some gas escapes without making contact with the catalyst. NOx goes up
3- Higher temperature = much more NOx produced.
Abby can you answer the question of paper5 posted ab ahmed abdullah
Jazak Allah khair for yr help,May Allah grant you with good grades !pH = pKa + lg([salt]/[acid]) (or pH = pKa - lg([acid]/[salt]) )
5.5 = -lg(1.79 × 10^-5) + lg([Sa]/[A])
5.5 - 4.75 = lg([Sa]/[A])
lg([Sa]/[A]) = 0.75
[Sa]/[A] = 10^0.75
[Sa]/[A]= 5.62
S = 5.62A
A+S = 100
100-A=5.62A
A = 15.1 = 15
So S = 85.
Acid = 15, Salt = 85. Is this the answer?
Edit: Somehow [ S ] wasn't being posted without the gaps so I just changed it to Sa.
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