The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity of the urea solution ;

Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL


(1 g = 0.001 kg)
Molality of the urea solution ;


The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Answer:
Imo : Gas particles are in constant, random motion. The volume of gas particles is negligible in comparison to the volume of the container. There are no attractive forces between gas particles.
To answer this question a balanced chemical equation is necessary. The correct equation is: N2 + 3H2 = 2NH3
From this equation, one mole of nitrogen react with 3 moles of hydrogen to give 2 moles of ammonia.
Therefore, the mole ratio of NH3 to N2 is 2:1
Answer:
Total partial pressure, Pt = 821 mm Hg
Partial pressure of Helium, P1 = 105 mm Hg
Partial pressure of Nitrogen, P2 = 312 mm Hg
Partial pressure of Oxygen, P3 = ? mm Hg
According to Dalton's law of Partial pressures,
Pt = P1 + P2 + P3
So, <u>P3 = 404 mm Hg</u>