Chemistry: Post your doubts here!

[sidbloom1995]

I also have these questions as doubts-
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Thank you!!!

Q3. we add the first and second ionisation energies of Al: 1820 + 577 = 2397 kJ/mol
now we add the the ionisation energies of each element given and the one that add up to 2397 will be our answer!
so when we add the 1st and 2nd ionisation energy of Co we get: 1640 + 757 = 2397 kJ/mol so our ans is A

Q4. in this one we have to find the one that DOES NOT HAVE 126 NEUTRONS!
they have give us the total nucleon number so when looking at the periodic table, only take the proton number of the element!
Bi = 209 - 83 = 126
Pb = 208 - 82 = 126
Po = 210 - 84 = 126
Tl = 208 - 81 = 127
so D= T l will be our ans as it does not have the magic no. 126 nucleon!

Q7. im not sure about this one but i do know that since Phosphorus is in group 5 it will form bonding like ammonia and will have the bond angle of 107
since they havent given us 107 in our answer we have to chose and answer close to 107! so logically speacking 109 is closer to 107 but wa cant have that as the answer! out angle should be smaller them or equal to 107 so the next closest no. is 104, thus the answer 104!

Q18. again im not sure of this answer so please confirm this one with someone else!
hydrogen bond will be present! so our ans will narrow down to C and D!
since NH3 is basic it ionises the water and thus we will have ions in the solution!
thus our answer is D

 

[mehria]

what's hybridization?? (related with AS-chemistry Syllabus)

 

[daredevil]

Question explaination needed :-/

I didnt get how they calculated it

 

[daredevil]

what's hybridization?? (related with AS-chemistry Syllabus)

splitting up of orbitals ... so that they are at the same energy level.

 

[A star]

what's hybridization?? (related with AS-chemistry Syllabus)

merging of orbitals of different energy level to form orbitals of same energy level inorder to for single covalent bonds

 

[mehria]

splitting up of orbitals ... so that they are at the same energy level.

i knw ... but wat's the difference between sp, sp2, sp3 hybridisations??

 

[Muskan Achhpilia]

Q3. we add the first and second ionisation energies of Al: 1820 + 577 = 2397 kJ/mol
now we add the the ionisation energies of each element given and the one that add up to 2397 will be our answer!
so when we add the 1st and 2nd ionisation energy of Co we get: 1640 + 757 = 2397 kJ/mol so our ans is A

Q4. in this one we have to find the one that DOES NOT HAVE 126 NEUTRONS!
they have give us the total nucleon number so when looking at the periodic table, only take the proton number of the element!
Bi = 209 - 83 = 126
Pb = 208 - 82 = 126
Po = 210 - 84 = 126
Tl = 208 - 81 = 127
so D= T l will be our ans as it does not have the magic no. 126 nucleon!

Q7. im not sure about this one but i do know that since Phosphorus is in group 5 it will form bonding like ammonia and will have the bond angle of 107
since they havent given us 107 in our answer we have to chose and answer close to 107! so logically speacking 109 is closer to 107 but wa cant have that as the answer! out angle should be smaller them or equal to 107 so the next closest no. is 104, thus the answer 104!

Q18. again im not sure of this answer so please confirm this one with someone else!
hydrogen bond will be present! so our ans will narrow down to C and D!
since NH3 is basic it ionises the water and thus we will have ions in the solution!
thus our answer is D

Thank you!!

 

[daredevil]

i knw ... but wat's the difference between sp, sp2, sp3 hybridisations??

sp means s orbital + 1 p orbital

sp2 means s orbital + 2 p orbitals

sp3 means s orbital + 3 p orbitals

the p orbitals that are not hybridized make bonds on their own at their normal energy level...

 

[ahmed abdulla]

Question explaination needed :-/

I didnt get how they calculated it

Quoted from examiner report :

" The data needed a careful analysis,
since the orders were not obvious at first glance. Taking the first and last rows in the table, and
comparing the rates to the concentrations of the complex, it could be seen that as [complex]
doubles, so does the rate. Thus they are proportional to each other, and so the order with respect
to [complex] is one. This could be confirmed by comparing rows one and two. The fact that (CO)
seems to have no effect on the rate indicates that the order with respect to (CO) is zero. "

let me now if you didnt understand

 

[daredevil]

Quoted from examiner report :

" The data needed a careful analysis,
since the orders were not obvious at first glance. Taking the first and last rows in the table, and
comparing the rates to the concentrations of the complex, it could be seen that as [complex]
doubles, so does the rate. Thus they are proportional to each other, and so the order with respect
to [complex] is one. This could be confirmed by comparing rows one and two. The fact that (CO)
seems to have no effect on the rate indicates that the order with respect to (CO) is zero. "

let me now if you didnt understand

how do we know that [CO] has no effect on rate?

 

[ahmed abdulla]

how do we know that [CO] has no effect on rate?

When ratio of Conc : rate are Equal eg 1:1 l 2:2 >> its first order with the taken reagent
When ratio of Conc : rate are 2:1 ie ( conc has higher ratio > its Zero order with the taken reagent
When ratio of Conc : rate are 1:2 ie ( conc has lowerratio > its Second order

 

[mehria]

sp means s orbital + 1 p orbital

sp2 means s orbital + 2 p orbitals

sp3 means s orbital + 3 p orbitals

the p orbitals that are not hybridized make bonds on their own at their normal energy level...

but how r we supposed to knw that wat type of hybridisation is present in a specific molecule ??

 

[ahmed abdulla]

how do we know that [CO] has no effect on rate?

Seems you didnt get it !
take Exp 1&2 ... divide the two concentraions eg exp2 by exp1 .. you get 2.11 .. divide rates.of . exp2 by exp1 > you get 1.45
conc. is not in proportion with rate , and rate DECREASES so its zero order
thats what i remember my teacher told me!

 

[ahmed abdulla]

Can someone have a look at my doubts ??

 

[Abdel Moniem]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
Question 28

 

[sidbloom1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf
Question 28

i will first explain the wrong answers!
D = because it is a tertiary aldehyde so it wont react!
A = the basic CH3CO2CH3 structure is being disturbed, when reacting only the functional groups will react
B = it could have been the answer but in our question, the molecule of CH3CO2CH3 was ionised to -CH2CO2CH3, with H being removed from the 1st carbon but as you can see in this it has reacted with the 3rd carbon of the molecule so this one wont do either!
so our answer is C because it does not disturb the CH3CO2CH3 molecule and the reaction is taking place at 1st carbon!
hope you understood it!
i think it is a little confusing, if you dont get it i will try explaining again!

 

[sidbloom1995]

Question explaination needed :-/

I didnt get how they calculated it

can you post the paper no.?

 

[Abdel Moniem]

thank you for your help but could you explain more iam still confused

i will first explain the wrong answers!
D = because it is a tertiary aldehyde so it wont react!
A = the basic CH3CO2CH3 structure is being disturbed, when reacting only the functional groups will react
B = it could have been the answer but in our question, the molecule of CH3CO2CH3 was ionised to -CH2CO2CH3, with H being removed from the 1st carbon but as you can see in this it has reacted with the 3rd carbon of the molecule so this one wont do either!
so our answer is C because it does not disturb the CH3CO2CH3 molecule and the reaction is taking place at 1st carbon!
hope you understood it!
i think it is a little confusing, if you dont get it i will try explaining again!

 

[sidbloom1995]

Went through my chem paper 4 and had this doubts .. Any help ?
3 e (iv )
5 b (iv )
7 c (iv )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf

Q3. e
first write a balanced equation of reaction between CH3CO2H and NaOH

CH3CO2H + NaOH ----------> CH3CO2-Na+ + H2O

we have 0.25 mol/dm3 of CH3CO2H and 0.1 mol/dm3 of NaOH
since the buffer is of acid and i's salt, all of the alkali will be converted to salt and the acid will be in excess
conc. of salt = 0.1 (same as the alkali)
conc. of acid reacted with alkali: since the mole ratio according to the equation is 1 : 1, so the conc. that reacted will also be same as the alkali, 0.1
conc. of acid remainig: 0.25 - 0.1 = 0.15 mol/dm3
now use the two conc. in the following formula

pH= pKa + log([salt]/[acid])

you have the pKa value and the conc. of salt and acid so just substitute it
pH= 4.76 + log([0.1]/[0.15]) = 4.58
that is your answer

as for Q5 b im unable to answer it

Q7 c
use the formula

no. of carbons = (100/1.1) x ([m+1]/[m])

you have the values of m:m+1 as 5.9:0.2
add them to the formula

no. of C = (100/1.1) x ([0.2]/[5.9]) = 3.08 = 3

so we have 3 carbons
the question says that our structure have an M+2 peak, which is formed due to either Cl37 /Cl35 or Br79/Br81
it aslo states that the peak is about the same height as the M peak
since there height is same we can conclude that it is Br because if it was Cl the difference in the peaks will be of ratio 3:1, meaning one of the two peaks will be more then double the length
as to how many Br are there it's only one because there is an other peak that is M+4 in which there are 2 atoms of higher and lower mass (isotopes)
and since they did not mention this peak here means that there is only one atom of Br
so we have 3 C and 1Br so the rest will be H, which in this case will be 7

the NMR spectrum shows us that there are 6H in one environment and 1 H in a different environment
so we can come up with a structure in which 1 H will be all alone
the structure will be

CH3-CH(Br)-CH3

 

[sidbloom1995]

thank you for your help but could you explain more iam still confused

i hope this helps
if not then you can still ask again (but i hope you get it this time)