Answer:
The temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C
Explanation:
Here we make use of the Clausius-Clapeyron equation;

Where:
P₁ = 1 atm =The substance vapor pressure at temperature T₁ = 282°C = 555.15 K
P₂ = 0.2 atm = The substance vapor pressure at temperature T₂
= The heat of vaporization = 28.5 kJ/mol
R = The universal gas constant = 8.314 J/K·mol
Plugging in the above values in the Clausius-Clapeyron equation, we have;


T₂ = 440.37 K
To convert to Celsius degree temperature, we subtract 273.15 as follows
T₂ in °C = 440.37 - 273.15 = 167.22 °C
Therefore, the temperature at which the liquid vapor pressure will be 0.2 atm = 167.22 °C.
The answer is; D
When both kids splash at both ends of the pool, the waves formed from the different direction travel and meet in the middle of the pool (since they are traveling with the same speed). When the waves meet, they interact and form constructive waves. This means that when a crest meets a crest, the height doubles. This also occurs for the troughs. Therefore the sizes of the waves at the middle are twice the individual waves.
It is C). <span>It decreases as the concentration of products increases.</span>