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for Q40 primary or secondary alcohol oxidize to form aldehyde and ketone respectively. u reverse the process that is called reduction
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
Do you know how to do Qs 33 , 32, 20, 16, 11 in 01/M/J/2003?
I am a bit confused about the answers to those Qs.
for Q11http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
Do you know how to do Qs 33 , 32, 20, 16, 11 in 01/M/J/2003?
I am a bit confused about the answers to those Qs.
Ahaa Thank you so much! That helped alot bro!
sorry if i am annoying you !it forms an aldehyde 1st and then gets converted into an acid!
for Q11
lone pairs make compounds polar more lone pairs means more polar so ammonia can not be more polar than water.
stronger base will pull those H+ protons with greater ease.
for Q16
for option A u get a complex ion where Cl has oxidation number of -1
for option B u get HCl again -1
for option C u get NaClO here u get +1
for option D u get NaClO3 here u get +5 so this is the correct answer.
for q32
right hand side of the molecule is polar as the oxygen atom has lone pairs any compound that has lone pair is considered polar.
like gets like which means polar molecule dissolves in polar and non polar dissolves in nonpolar. so polar cant dissolve in non polar oil u know oil does not dissolve in water.
for q33
refer to this diagram
when u break the bonds in graphite and diamond u get carbon gas carbon gas would be at the same energy that is the hess law.
View attachment 28218
You have been helped already..maybe next time
because when the double bond broke, the C was bonded to only one other C and one H and double bonded to an O which is a CHO group of aldehyde. Always just join an O atom whenever u break a double bond using KMnO4. then check if the C is bonded to 1 or 2 other C. If its bonded to 1 C then make an acid. If its with 2 other C then its a ketonesorry if i am annoying you !
y not a ketone .... how u knew it gonna be an acid
For Q11 - for A XYn-->X + nYFor question 11, I also had a doubt...
Q21--> A compound will not have cis trans isomers in one case, which is when 2 identical groups are present on the same carbon atom, so instead of R here u shd have CH2CH3 which is the same as C2H5 in D
For question 29, I guess this comes from ur knowledge (I mean just memorisation of the reactions and their conditions)
h4rriet Can u please check out question number 11 ?
For Q11 - for A XYn-->X + nY
there are n X-Y bonds in the molecule. Enthalpy change is the energy needed to break these n bonds. They all break and form atoms of X and Y. So dividing the enthalpy by n we get the energy for one bond.
no problem
there is no affect of the Cl-Cl bond or Br-Br bond here. The main bond here is H-Cl and H-Br. U should refer to there bonds to answer the question. Bond energies are diff according to the elements they are bonded to so we cant relate Cl-Cl and Br-Br to H-Cl and H-BrThe answer is C. Why cant the it be D?
caan you explain again :/For Q11 - for A XYn-->X + nY
there are n X-Y bonds in the molecule. Enthalpy change is the energy needed to break these n bonds. They all break and form atoms of X and Y. So dividing the enthalpy by n we get the energy for one bond.
For question 11, I also had a doubt...
Q21--> A compound will not have cis trans isomers in one case, which is when 2 identical groups are present on the same carbon atom, so instead of R here u shd have CH2CH3 which is the same as C2H5 in D
For question 29, I guess this comes from ur knowledge (I mean just memorisation of the reactions and their conditions)
h4rriet Can u please check out question number 11 ?
11 is A, if I'm not mistaken. If you take n to be 4, for example, you'll see there are 4 X-Y bonds, so the total energy can be divided by 4 to get the energy for one bond. Not so for the B, C and D.
In A) it says XYn ..... but in D) it says nXY , so doesnt D) tell us that there are 'n' X-Y bonds.....
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