The atomic radii of mg2+ and fions are 0.072 and 0.133 nm, respectively. (a) calculate the force of attraction between these two ions at their equilibrium interionic separation (i. e., when the ions just touch one another) b) what is the force of repulsion at this same separation distance

Explanation:

(a) Attractive force

To calculate the attractive force (F), we can use Coulomb's Law:

where  

Q₁ and Q₂ are the charges on the ions,

r is the distance between them, and

k = the Coulomb constant

Data:

Q₁ = 2+

Q₂ = 1-

r₊ = 0.072 nm

r₋ = 0.133 nm

k = 8.988 × 10⁹ N· m²C⁻²

Calculations:

Q₁ = (+2) × 1.602 × 10⁻¹⁹ C = +3.204 × 10⁻¹⁹ C

Q₂ = (-1) × 1.602 × 10⁻¹⁹ C = -1.602 × 10⁻¹⁹ C

r = r₊ + r₋ = 0.072 nm + 0.133 nm = 0.205 nm = 0.205 × 10⁻⁹ m

(b) Repulsive force

The equilibrium position is reached when the ions just touch each other. If they come any closer, the nuclear repulsions will outweigh the Coulomb attraction.

Thus, the repulsive force is equal and opposite to the attractive force.

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol