An antifreeze solution is made by mixing ethylene glycol (ρ = 1116 kg/m3) with water. suppose the specific gravity of such a solution is 1.05. assuming that the total volume of the solution is the sum of its parts, determine the volume percentage of ethylene glycol in the solution.

Answer is: the volume percentage of ethylene glycol in the solution is 43.1%.
V(solution) = 1 m³.
d(solution) = 1050 kg/m³.
m(solution) = 1050 kg.
d(ethylene glycol) = 1116 kg/m³.
d(water) = 1000 kg/m³.
m(solution) = d(ethylene glycol) · V(ethylene glycol) + d(water) · V(water). 
V(ethylene glycol) = 1m³ - V(water).
1050 kg = 1116 kg/m³ · (1m³ - V(water)) + 1000 kg/m³ · V(water).
1050 kg = 1116 kg - 1116·V(water) + 1000·V(water).
116 kg/m³ ·V(water) = 66kg.
V(water) = 0.569 m³.
V(ethylene glycol) = 1 m³ - 0.569 m³ = 0.431 m³.
percentage = 0.431 m³ ÷ 1 m³ · 100% = 43.1%.

Forgot the question hahaha

Exothermic,  physical change after it is already baked