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jenyasd209 [6]
3 years ago
13

An antifreeze solution is made by mixing ethylene glycol (ρ = 1116 kg/m3) with water. suppose the specific gravity of such a sol

ution 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.
Chemistry
1 answer:
Harman [31]3 years ago
3 0
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%.

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At standard temperature and pressure, the volume of a tire is 3.5L. What is the new pressure if the temperature outside is 296k and its weight causes the volume of the gas is 2.0 L?

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The new pressure is: 1.896 atm

Explanation:

At standard temperature and pressure, we have:

P_1 = 1atm

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Determine the new pressure

Using combined gas law, we have:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

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\frac{1 * 3.5}{273.15} =\frac{P_2*2.0}{296}

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