How much heat will be released if 9.75 g of aluminum reacts with excess ammonium nitrate (NH₄NO₃) according to the following equ
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Answer:
The energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J
Explanation:
Let the charge on the cation be q₁
Also let the charge on the anion be q₂
A cation q₁ with a valence of 1, has a charge of 1 X 1.602×10⁻¹⁹C = 1.602×10⁻¹⁹C
An anion q₂ with a valence of 3, has a charge of 3 X 1.602×10⁻¹⁹C = 4.806 ×10⁻¹⁹C
The distance between the two charges is 7.5nm = 7.5 X10⁻⁹m
Energy of attraction =
Where k is coulomb's constant = 8.99 X 10⁹ Nm₂/C₂
Energy of attraction =
Energy of attraction = 1.231 X 10⁻¹¹ J
Therefore, the energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J