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GT for Chemistry paper 22 AS.

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Q1
i wrote sodium for the colored flame with chlorine.. not sure
2Al(s) + 3Cl2(g)--> AlCl3(s)
when this reaction occurs, the pale yellow color of the gas disappears and a white solid is formed
Argon doesn't react as it is in a stable elemental state because of its full octet.
Sodium dissolves pH 7
Aluminum dissolves pH 6-- ?..
Silicon reacts pH 3
Reaction is redox/disproportionation
Sulfur- 2 lone pairs
Nitrogen- 1 lone pair
Bond A is smaller due to the presence of 2 lone pairs compared to just 1, thus repelling bond pairs more.

Q2
Ch3OH(l)+3/2O2(g) = CO2(g) + 2H2O(l)
enthalpy change is -129 kJ/mol
increase in pressure increases rate of formation due to increased interaction/collisions between reactant molecules
increase in temp. increases rate of formation due to more particles having energy greater than Ea
presence of catalyst increases rate of formation as it lowers Ea
_______The reaction was an irreversible reaction, so the concept of equilibrium shifts cannot be incorporated________

Q3
I don't remember the exact structure of the compound but
to the left, an ester with the tail -CO2H groups were formed
to the right, an ester with the middle -OH group was formed
straight down, an alkene was formed with the elimination of the middle -OH group and a hydrogen from one of the neighboring C's
alkene + cool, dilute KMnO4 gave you a product with the double bond removed and a diol
alkene + hot, conc. KMnO4 gives you something i'm not sure of! :p
NOTE--- I have just talked about the parts of the compounds that have changed, everything else is meant to be the same
I cannot recall the reaction we had to write the types of...
the optical isomer was easy
My cis-trans isomers looked the same as Pal_1010's..-- Am I right?

For the fourth que,
Na--- -CO2H and -OH
K2Cr2O7 & heat---- -OH and -CHO
NaHCO3--- -CO2H-- (I remember cuz its a test :O)
i got the functional group as -oh -- alcohol
moles of hydrogen atoms as 6.67x10^-3
proved the presence of 2 -OH groups by showing the ration of R-OH
and the presence of a ketone group.
The structure of G was CH2(OH)C=OCH2(OH), drawn of course!
the empirical formula was C6H8O7.... But I wrote C7H9O8.. if I have made a mistake in multiplying the correct base ratio of CHO- C1H1.3O1.17 with a factor, do you think I can get at least 2/3 marks?.. :/



Guys, Do correct me where necessary.. :)
 
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I hadn't notice it was SKELETAL formula that was asked for the cis-trans isomer(silly mistake I know) but I saw it like when 5 mins was left, and I hastily drew these. Do you think I'll get my marks?
guys i drew the same but dint show OOH coz i thought it was cis coz of H and trans coz of H so i dint write -OOH.........?
 
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They asked which element burns in colored flame in Chlorine. Only Sulfur burns in with a colored flame in Chlorine.

Sulfur burns with a blue flame.

Cheers :)
 
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They asked which element burns in colored flame in Chlorine. Only Sulfur burns in with a colored flame in Chlorine.

Sulfur burns with a blue flame.

Cheers :)

there was no chlorine mentioned in the question. :p
only which element burns with a colored flame, and the possible answers were. sodium (yellow), phosphorous (blue), sulfur (yellow)
 
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They asked which element burns in colored flame in Chlorine. Only Sulfur burns in with a colored flame in Chlorine.
Sulfur burns with a blue flame.
Cheers :)
in many text books and on wikipedia it says sodium burns in chorine in a yellow flame.....
 
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438
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106
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Q3
I don't remember the exact structure of the compound but
to the left, an ester with the tail -CO2H groups were formed
to the right, an ester with the middle -OH group was formed
straight down, an alkene was formed with the elimination of the middle -OH group and a hydrogen from one of the neighboring C's
alkene + cool, dilute KMnO4 gave you a product with the double bond removed and a diol
alkene + hot, conc. KMnO4 gives you something i'm not sure of! :p

Q3 answers:

HO2CCH2CH(OH)CO2H + CH3OH----------->CH3O2CCH2CH(OH)CO2CH3
HO2CCH2CH(OH)CO2H + CH3CO2H-------->HO2CCH2CH(CO2H)O2CCH3
HO2CCH2CH(OH)CO2H + conc. H2SO4----->HO2CCH=CHCO2H

HO2CCH=CHCO2H + hot conc. KMnO4----->HO2CCHO
HO2CCH=CHCO2H + cold dil. KMnO4------->HO2CCH(OH)CH(OH)CO2H
HO2CCH=CHCO2H + steam/H3PO4--------->I did this one wrong =[.. but it should be HO2CCH2CH(OH)CO2H
 
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