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Chemistry: Post your doubts here!

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upload_2015-6-8_20-33-7-png.54840

Help! it's 1 and 2 only
How would each react and what will it give in each case? And what would 3 give since it's wrong??
 
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upload_2015-6-8_22-18-10.png
Why D?
Isn't it
Mg = 0.5mol O2
Al = 0.75mol O2
S = 1mol O2 (since SO2 is formed right as SO3 required a catalyst)??
 
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IE naturally increases across the period because the nuclear charge is higher, so Mg wil lhavve more IE than Na. It will also have more IE than Al, because Al has an electron in the outer p shell which has more energy and is also farther than the nucleus. (Mg outer shell => 3s2, Al outer shell => 3s2 3p1)
Thanks alot i got it :D
 
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View attachment 54847
Why D?
Isn't it
Mg = 0.5mol O2
Al = 0.75mol O2
S = 1mol O2 (since SO2 is formed right as SO3 required a catalyst)??
6A*student credits.
This is oxidation reactions forming oxides
eg. 1 Mg + 0.5O2 -> 1MgO
1 Al + 0.75O2 -> 0.5Al2O3
1 S + 1O2 -> 1 SO2 because its in excess oxygen 1SO2 + 0.5 O2 -> 1SO3 so overall equation is 1S + 1.5O2 -> 1SO3

the oxidation ratio is 0.5:0.75: 1.5 = 2:3:6 = 1:1.5:3 so answer is D
 
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Actually, the most straightforward way to tackle this question is to refer to the list of IE and ionic radii in the data booklet as suggested by the question, and see which option fits the description.
11421453_941066172580240_1385222293_n.jpg

Can some1 please explain this
 
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6A*student credits.
This is oxidation reactions forming oxides
eg. 1 Mg + 0.5O2 -> 1MgO
1 Al + 0.75O2 -> 0.5Al2O3
1 S + 1O2 -> 1 SO2 because its in excess oxygen 1SO2 + 0.5 O2 -> 1SO3 so overall equation is 1S + 1.5O2 -> 1SO3

the oxidation ratio is 0.5:0.75: 1.5 = 2:3:6 = 1:1.5:3 so answer is D
Dayum xD Some questions they want you to take in account the catalyst... others they dont :S
 
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