Weight of the balloon = 2.0 g
Six weights each of mass 30.0 g is added to the balloon.
Total mass of the balloon = 2.0 g + 6*30.0 g = 182 g
Density of salt water = 1.02 g/mL
Calculating the volume from mass and density:

Converting the volume from mL to cubic cm:

Assuming the balloon to be a sphere,
Volume of the sphere =
π

r = 3.49 cm
Radius of the balloon = 3.49 cm
Diameter of the balloon = 2 r = 2*3.49 cm = 6.98cm
<h3>Answer:</h3>
Ionic Bond
<h3>Explanation:</h3>
Types of Bonds can be predicted by calculating the difference in electronegativity.
If, Electronegativity difference is,
Less than 0.4 then it is Non Polar Covalent Bonding
Between 0.4 and 1.7 then it is Polar Covalent Bonding
Greater than 1.7 then it is Ionic Bonding
For Mg and Cl,
E.N of Chlorine = 3.16
E.N of Magnesium = 1.31
________
E.N Difference 1.85 (Ionic Bond)
<h3>Conclusion:</h3>
MgCl₂ being ionic in nature when dissolved in water dissociates into Magnesium and Chloride ions respectively.
Answer:
The correct answer is -1085 KJ/mol
Explanation:
To calculate the formation enthalphy of a compound by knowing its lattice energy, you have to draw the Born-Haber cycle step by step until you obtain each element in its gaseous ions. Find attached the correspondent Born-Haber cycle.
In the cycle, Mg(s) is sublimated (ΔHsub= 150 KJ/mol) to Mg(g) and then atoms are ionizated twice (first ionization: ΔH1PI= 735 KJ/mol, second ionization= 1445 KJ/mol) to give the magnesium ions in gaseous state.
By other hand, the covalent bonds in F₂(g) are broken into 2 F(g) (Edis= 154 KJ/mol) and then they are ionizated to give the fluor ions in gaseous state 2 F⁻(g) (2 x ΔHafinity=-328 KJ/mol). The ions together form the solid by lattice energy (ΔElat=-2913 KJ/mol).
The formation enthalphy of MgF₂ is:
ΔHºf= ΔHsub + Edis + ΔH1PI + ΔH2PI + (2 x ΔHaffinity) + ΔElat
ΔHºf= 150 KJ/mol + 154 KJ/mol + 735 KJ/mol + 1445 KJ/mol + (2 x (-328 KJ/mol) + (-2913 KJ/mol).
ΔHºf= -1085 KJ/mol
Answer:
It is called a salt bridge