Answer:
1171.12 mL
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (milliliters)
V2 = final volume (milliliters)
T1 = initial temperature (Kelvin)
T2 = final temperature (Kelvin)
According to the information provided in this question:
P1 = 300 mmHg
P2 = 140 mmHg
V1 = 400 mL
V2 = ?
T1 = 0°C = 273K
T2 = 100°C = 100 + 273 = 373K
Using P1V1/T1 = P2V2/T2
300 × 400/273 = 140 × V2/373
120000/273 = 140V2/373
120000 × 373 = 273 × 140V2
44760000 = 38220V2
V2 = 44760000 ÷ 38220
V2 = 1171.115
The new volume is 1171.12 mL
the answer is B. ribosomes make protien
Mass / density = volume
321g / 0.84 = 382.142857
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,
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