Answer is: <span>the volume percentage of ethylene glycol in the solution is 43.1%.
V(solution) = 1 m</span>³.
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%.
The answer to your question is frequency.
If you would like further information about the subject, private message me and id be happy to explain in further detail for you!. :0
Answer:
Explanation:
C is oxidised because C up to C+4
PbO is reduced because PbO from Pb2+ down to Pb0
