Question

One mole of magnesium in the vapour state absorbed

Answer 1

Answer:

$Energy absorbed in the ionisation of 1 mole of MgMg to Mg^+Mg+(g) =750=750 kJ</p><p></p><p></p><p>Energy left unused =1200-750\ = 450=1200−750 =450 kJ</p><p></p><p></p><p>% of Mg^+Mg+(g) converted into Mg^{2+}Mg2+(g)</p><p></p><p></p><p>=\dfrac {450}{1450} \times 100 =31=1450450×100=31%</p><p></p><p></p><p>Hence, the % of Mg^+ = 100-31=69Mg+=100−31=69%</p><p></p><p>$

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Answer 2

Answer:

One mole of magnesium in the vapour state absorbed 1200 kJ mol^{-1}

−1

energy . If the first and second ionisation energies of Mg are 750 kJ mol^{-1}

−1

and 1450 kJ mol^{-1}

−1

respectively

Explanation:

Energy absorbed in the ionisation of 1 mole of MgMg to Mg^+Mg

+

(g) =750=750 kJ

Energy left unused =1200-750\ = 450=1200−750 =450 kJ

% of Mg^+Mg

+

(g) converted into Mg^{2+}Mg

2+

(g)

=\dfrac {450}{1450} \times 100 =31=

1450

450

×100=31%

Hence, the % of Mg^+ = 100-31=69Mg

+

=100−31=69%