Answer : The molal freezing point depression constant of liquid X is, 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of liquid X (solvent) = 450 g = 0.450 kg
Molar mass of urea = 60 g/mole
Formula used :
where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X = 
i = Van't Hoff factor = 1 (for non-electrolyte)
= Molal-freezing-point-depression constant = ?
m = molality
Now put all the given values in this formula, we get
Therefore, the molal freezing point depression constant of liquid X is,
As depth increases, the density of the layers decreases.
Answer:
1120 L.
Explanation:
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In this case, as no conditions of pressure of temperature are given for this problem, we can assume that the scuba diver dives at STP (1 atm and 273.15 K), which means that 1 mole of air would occupy a volume of 22.4 L.
In such a way, since she needs 50.0 moles of air, the following ratio is useful to compute the size (volume) of the tank she needs:
Thereby, we plug in to obtain:
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The particles of light known as photon.
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Answer:
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is 
.
Explanation:
The combined gas equation is,
where,
= initial pressure of gas = 104 kPa
= final pressure of gas = 52 kPa
= initial volume of gas = 
= final volume of gas = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is .
