Calculations

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[Thung Su Ling]

I have two questions that i desperately need help here.
They are from Paper 1 of year 2006. Questions 17 and 47
Question 17...how do we manage to calculate the oxidation number for oxygen to get an answer of -2?
Question 47... i know that this question we have three forms of half equation added up but what is it??And what ion is IO3-??what ion is this??what reaction is this??is it the reaction between iodine and water....to form hypoiodic acid and hypoiodous acid??if so what are all the half equaations involved>??

[Sim Thiam Huat]

Q.17. I consider this as a trick. there is no calculation involved, but just to follow the rule: Rule no. 2, Text book page 107 which says " The oxidation number of oxygen in a compound is always -2 except ......."

Q47. Students do not have to know or worry about the equation. I think this equation is from Form 6 chemistry. Students just need to apply the concepts and the calculation strategy and they can get the answer easiliy like this...
1st,  put 3 waters on the right which will mean 6 hydrogen ion on the left.
2nd, count the charges on the left : 6+ from hydrogen ion and 1- fron IO3 ion , the net is 5+ so just put in 5 electrons to balance the whole thing .
Answer : q-6; r-5 ; s-3
Hope it helps u.

[Thung Su Ling]

Somehow after analysing it for a while, i found that the only answer to Q17 is students must know that O is always O2- and for 47 is about balancing of the whole equations. thanks for helping.

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