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first and second ionization of energies of Be are 900 kjmol and 1750 kjmol ,calculate approxmate percentage of Be+2 IONS,IF 1GRAM OF Be absorb 150 kj of energy

first and second ionization of energies of Be are 900 kjmol and 1750 kjmol ,calculate approxmate percentage of Be+2 IONS,IF 1GRAM OF Be absorb 150 kj of energy
 

Grade:10

1 Answers

Arun
25758 Points
2 years ago
Amount of Mg atoms= 
24
1
​ 
 =4.167×10 
−2
 mol
Energy absorbed in ionizing Mg to Mg 
+
 =4.167×10 
−2
 ×740
=30.84KJ
Energy absorbed in ionizing Mg 
+
  to Mg 
2+
 =(50−30.84)KJ
=19.16KJ
Amount of Mg 
+
  converted to Mg 
2+
 = 
1450
19.16
​ 
 
=1.321×10 
−2
 mol
Amount of Mg 
+
  remaining as such=4.167×10 
−2
 −1.321×10 
−2
 
=2.846×10 
−2
 mol
Composition of final mixture would be as follows:
% of Mg 
+
 = 
4.167×10 
−2
 
2.846×10 
−2
 
​ 
 ×100=68.3%
% of Mg 
2+
 =100−68.3=31.7 %

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