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Reply 1
Jan 05 fudge up:

2a)i) Write the equation for each of the following:
Sodium oxide with water
aluminium oxide with dil. HCl
aluminuim hydroxide with dil NaOH
Sulphur trioxide with water
(i dids all that :smile: )

ii) State how the acid-base character of the oxide of an element can be used to identify the metallic or non-metallic nature of the element. (2)

iii) State how the reactions in (i) show the way in which the metallic character of the element changes across the period Na to Ar in the P. Table. (1)


The bit in bold is where me fudged up...can sum1 help???
Reply 2
Another fudge up:
Complete the equation:
PbO2 + HCl -> PbCl2 + _______ + _______
2 marks
Reply 3
mackin boi
Jan 05 fudge up:

2a)i) Write the equation for each of the following:
Sodium oxide with water
aluminium oxide with dil. HCl
aluminuim hydroxide with dil NaOH
Sulphur trioxide with water
(i dids all that :smile: )

ii) State how the acid-base character of the oxide of an element can be used to identify the metallic or non-metallic nature of the element. (2)

iii) State how the reactions in (i) show the way in which the metallic character of the element changes across the period Na to Ar in the P. Table. (1)


The bit in bold is where me fudged up...can sum1 help???


iirc, ii) metals will form basic or amphoteric oxides and non-metals will form acidic oxides. In group 4, Carbon and Silicon both form acidic oxides (Carbon dioxide and silicon dioxide). Going down the group, lead and tin form amphoteric oxides.

iii) Metalic to non-metallic across the period because oxides initally show basic character but change to acidic.

Finally,

PbO2 + 4HCl -> PbCl2 + 2H20 + Cl2
Reply 4
And the SURVER SAYS:
kpg
iirc, ii) metals will form basic or amphoteric oxides and non-metals will form acidic oxides. In group 4, Carbon and Silicon both form acidic oxides (Carbon dioxide and silicon dioxide). Going down the group, lead and tin form amphoteric oxides.

iii) Metalic to non-metallic across the period because oxides initally show basic character but change to acidic.

Finally,

PbO2 + 4HCl -> PbCl2 + 2H20 + Cl2


The Markscheme Says:
ii) acidic oxide -non metal (1)
basic oxide - metal (1)
if oxide ommited twice (max 1)

iii)Metallic to non-metallic/decrease in metallic character because oxides change from basic to acidic/ decrease in basic character

OR

Metallic to non-metallic/decrease in metallic character with reference to at least 2 reactions

and the same for the equations
Reply 5
Well I did get 87/90 on this paper in January so I should be getting them right :wink:
Reply 6
kpg
Well I did get 87/90 on this paper in January so I should be getting them right :wink:


whoa....u crazyyy lol
Reply 7
wots ur secret of gettin 87/90! :biggrin:
does anybody know the points that you get marked on for the solubilities of sulphates and hydorxides
Reply 9
the hell?! 87/90?!

I revised real hard for this paper and i got 33 which is U i think. I cant get more than C in unit4 past papers
Reply 10
Asian cutie
does anybody know the points that you get marked on for the solubilities of sulphates and hydorxides


group two sulphates down the group get a slightly lower lattice energy, and the hydroation enthalpy falls, so overall the enthalpy of solution becomes more endothermic. hydroxides, the lattice energy falls quite a bit and the hydration enthalpy falls less, so overall the enthalpy of solution becomes more exothermic.
Reply 11
elniro
the hell?! 87/90?!

I revised real hard for this paper and i got 33 which is U i think. I cant get more than C in unit4 past papers


All I did was learn off by heart..

-All the period 3 oxides properties + few reactions
-All the period 3 chloride properties + few reactions
-Born haber definitions and the chemistry behind the solubility
-How to do the acid base calculations + graphs
-Organic chem reactions

Simply a case of write down, look at it, write it down again, look at it, write it down again etc. etc. till I could do it off by heart. Not much understanding required really. Other little bits like "Why is the prediction for Lattice enthalpy better with compound X than Y" or "Why does reduction with LiAlH4 work with carbonyl compounds and not alkenes" etc. that are only worth 1/2 marks - I just picked up from doing about 4 or 5 past papers.
Reply 12
Ahh cool, cheers for the tip. I revised really hard in January but only got 60 so decided to retake it cos I know I'm gonna lose so many marks in the synoptic. Ye unit 4 seems just a case of remembering everything which is why I did really bad in January.
Has anyone got a good clear list of the organic reactions because I'm finding them really hard to get straight in my head? There are too many of them!
Reply 14
Grignard:

A grignard is formed by adding a halogenoalkane to magnesium in dry ether and heating under reflux. The grignard reagent formed can then react with a C=O bond, by adding the compound and then acidifying with dilute HCL, in four different cases:
1. Grignard + carbon dioxide (solid) makes a carboxilic acid
2. Grignard + methanal makes a primary alcohol
3. Grignard + any other aldehyde makes a secondary alcohol
4. Grignard + a ketone makes a tertiary alcohol
Grignard reagents must be kept dry because they react with water to form alkanes, which is a waste.

thats a start!

Can someone answer why LiAlH4 must be used in dry conditions?
Reply 15
because LiAlH4 reacts very vigorously with water
Reply 16
BCHL85
because LiAlH4 reacts very vigorously with water

More detail
LiAlH4 + 4H2O -> LiOH + Al(OH)3 + 4H2
LiAlH4, will reduce water and it tends to explode
Reply 18
Can I just check, LiAlH4 will not reduce C=C bonds, because it is not strong enough?
And so thats why you must use a mickle catalyst to add hydrigen to it?
Reply 19
Nope. Basically, LiAlH4 an NaBH4 act as a source of hydrogen ions. The C=O is is highly polar due to the large difference in electronegativity so the Carbon is slightly positive and the oxygen is slightly negative. The hydride ion (hydrogen with a negative charge and a lone pair of electrons) acts as a nucleophile and is attracted towards the slightly positive carbon atom. This isn't possible with alkenes because the C=C bond isn't polar.

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