Yeah some of us arent as clever as you guys hey
But thankx.. no worries if you're busy...
i got how to do Q16, 25, 30, 31 and 34 .. so no need for those
It's not about being clever dude. You're going to have the same exams in 2 days, so you should have been ready.
Nov 05 p112. You just need to write the balanced equations for the
complete combustion of
one mole of each element.
Mg + 1/2O
2 ----> Mg0
Al + 3/4O
2 ----> 1/2Al
2O
3S + 3/2O
2 ----> SO
3Hence we can note that the number of moles of oxygen required
increases but not constantly from Magnesium to Sulfur.
Answer is
D15. You should first write the balanced equation representing the decomposition of the nitrate. That will be:
Ca(NO
3)
2 -----> CaO + 2NO
2 + 1/2O
2From equation:
1 mole of nitrate gives
1/2 mole of oxygen gas.
164g of nitrate produces (24/2) dm
3 of oxygen gas.
Hence 8.2g of nitrate will emit (12/164 x 8.2) = 0.6dm
3 =
600cm3Answer is
C19. Chiral centres are shown by a carbon atom carrying 4 different group of atoms. No carbon involved in a C=C will be able to show chirality.
Hence this leaves us with only two chiral centres. Starting from the right, they are the
second and the
third carbon atom
Answer is
B24. I'll proceed by elimination.
Propan-2-ol and 2-methyl-propan-2-ol are first to be rejected since they can form only one type of alkene due to free rotation.
2-methylbutan-2-ol can form only 2 types of alkenes due to free rotation. It is connected to three other carbon atoms and should have been able to form 3 alkenes but two of them are identical.
Butan-2-ol should have formed only two types of alkenes but one of them show cis-trans isomerism. Hence it forms 3 types of alkenes.
Answer is
A.
36. A reaction in which the same element is both oxidized and reduced is called a disproportionation reaction.
Is that enough or should I go in more details?
39.
Unburnt hydrocarbons are made to burn completely to form carbon dioxide and water. Hence this is a form of oxidation. Oxygen comes from the oxides of nitrogen which in turn gets reduced.