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QUESTIONS ANSWERS AND COMMENTS

The substances are grouped by oxidation number of sulfur in the following table  [Pg.222]

Using the rules in Box 2.1, rule 1 tells us that in Br (aq), the oxidation number of bromine is -1 rule 2 tells us that in Br2(g) it is zero. Rules 4 and 7 imply that in SO2 and S04 , the oxidation numbers of sulfur are +4 and +6, respectively. Thus the unbalanced equation takes the form  [Pg.222]

sulfur is oxidized (+4 to +6) and bromine is reduced (zero to -1). The sulfur and bromine are not present in the right proportions because the total change in oxidation number is +1. It can be made zero by doubling the bromine on each side so that the total change in bromine oxidation number becomes -2 to balance the sulfur change of +2  [Pg.222]

As there are four oxygens on the right and only two on the left, we add 2H20(1) to the left-hand side  [Pg.223]

As there are no hydrogens on the right and four on the left, we now add 4H (aq) to the right-hand side  [Pg.223]

SUMMARY OF PART 1 LEARNING OUTCOMES FOR PART 1 QUESTIONS ANSWERS AND COMMENTS EXERCISES ANSWERS AND COMMENTS FURTHER READING ACKNOWLEDGEMENTS... [Pg.4]

LEARNING OUTCOMES FOR PART 3 QUESTIONS ANSWERS AND COMMENTS ACKNOWLEDGEMENTS... [Pg.6]

See other pages where QUESTIONS ANSWERS AND COMMENTS is mentioned: [Pg.29]    [Pg.7]    [Pg.113]    [Pg.113]    [Pg.115]    [Pg.117]    [Pg.119]    [Pg.121]    [Pg.5]    [Pg.108]    [Pg.109]    [Pg.111]    [Pg.113]    [Pg.115]    [Pg.117]    [Pg.119]    [Pg.182]    [Pg.183]    [Pg.216]    [Pg.217]    [Pg.219]    [Pg.221]    [Pg.223]    [Pg.78]    [Pg.79]    [Pg.81]    [Pg.83]    [Pg.85]    [Pg.222]    [Pg.223]    [Pg.225]    [Pg.227]    [Pg.229]    [Pg.231]    [Pg.233]    [Pg.235]    [Pg.237]    [Pg.241]   


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