Answer: 18.65L
Explanation:
Given that,
Original volume of oxygen (V1) = 30.0L
Original temperature of oxygen (T1) = 200°C
[Convert temperature in Celsius to Kelvin by adding 273.
So, (200°C + 273 = 473K)]
New volume of oxygen V2 = ?
New temperature of oxygen T2 = 1°C
(1°C + 273 = 274K)
Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law
V1/T1 = V2/T2
30.0L/473K = V2/294K
To get the value of V2, cross multiply
30.0L x 294K = 473K x V2
8820L•K = 473K•V2
Divide both sides by 473K
8820L•K / 473K = 473K•V2/473K
18.65L = V2
Thus, the new volume of oxygen is 18.65 liters.
Answer:
Boiling point for the solution is 100.237°C
Explanation:
We must apply colligative property of boiling point elevation
T° boiling solution - T° boiling pure solvent = Kb . m
m = molalilty (a given data)
Kb = Ebulloscopic constant (a given data)
We know that water boils at 100°C so let's replace the information in the formula.
T° boiling solution - 100°C = 0.512 °C/m . 0.464 m
T° boiliing solution = 0.512 °C/m . 0.464 m + 100°C → 100.237 °C
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