u want an explanation for them or the answers?
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étudiante
étudiante
for finding oxidation numbers, u need to add up the oxidation no of all the elements and = charge on ion... the element for which u dunno will be x n u need to solve for x
So for MnO4... it will be x+ 4(-2) = -1 solve for x, x= +7
for SO2... x+ 2(-2) = 0 solve for x again, x= +4
for simple ions, its the charge itself.... 2+ for Mn2+
for SO4... x+ 2(-4) = -2 x= +6
So for MnO4... it will be x+ 4(-2) = -1 solve for x, x= +7
for SO2... x+ 2(-2) = 0 solve for x again, x= +4
for simple ions, its the charge itself.... 2+ for Mn2+
for SO4... x+ 2(-4) = -2 x= +6
étudiante
second 1.... m not sure how to explain here... but its kinda gonna be like a probability tree thingy in maths... Cl-37 having a probabilty of 1/4
leadingguy
sorrY fr late replY, ....xD
sOrrY to say dat bt yes , I am aware of how 2 calclculate oxidation no, I am cnfusd at balancing them
(((
I am aware of the fact tdat chlorine 37 is having a raio of 1:3 with cl 35 ... xD bT hw is 1/9 fr m+4 ) thnx alot
hope m nt distrUbng??
sOrrY to say dat bt yes , I am aware of how 2 calclculate oxidation no, I am cnfusd at balancing them
(((I am aware of the fact tdat chlorine 37 is having a raio of 1:3 with cl 35 ... xD bT hw is 1/9 fr m+4 ) thnx alot
hope m nt distrUbng??