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.
No, the properties of a substance are not affected by the amount of a substance.
The second stage of photosynthesis also called Calvin stage produces glucose
Unlike solid matter, where particles are tightly packed and slightly vibrating, or gas, where particles go around everywhere and are extremely loose, a liquid has particles that are loosely packed but are still in slight contact with each other. Hope that's good enough
Answer:
The value is 
Explanation:
From the question we are told that

The initial volume of the fluorocarbon gas is 
The final volume of the fluorocarbon gas is
The initial temperature of the fluorocarbon gas is 
The final temperature of the fluorocarbon gas is 
The initial pressure is 
The final pressure is 
Generally the equation for adiabatically reversible expansion is mathematically represented as
![T_2 = T_1 * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }](https://tex.z-dn.net/?f=T_2%20%3D%20%20T_1%20%20%2A%20%5B%20%5Cfrac%7BV_1%7D%7BV_2%7D%20%5D%5E%7B%5Cfrac%7BR%7D%7BC_v%7D%20%7D)
Here R is the ideal gas constant with the value

So
=> 
Generally adiabatic reversible expansion can also be mathematically expressed as

=>
=> 
=>
So

=> 