Answer:
C
Explanation:
It looks pretty reasonable to me
Answer:

Explanation:
Given , molarity of glycerol= 
Volume= 1 L.
Therefore, No of moles of glycerol= 
Now, volume of water needed, V=998.8 mL.
Density is given as= 0.9982 g/mL.
Therefore, mass of water = 
Now, molality=
Hence, this is the required solution.
0.17 M is the is the molal concentration of this solution
Explanation:
Data given:
freezing point of glucose solution = -0.325 degree celsius
molal concentration of the solution =?
solution is of glucose=?
atomic mass of glucose = 180.01 grams/mole
freezing point of glucose = 146 degrees
freezing point of water = 0 degrees
Kf of glucose = 1.86 °C
ΔT = (freezing point of solvent) - (freezing point of solution)
ΔT = 0.325 degree celsius
molality =?
ΔT = Kfm
rearranging the equation:
m = 
m= 0.17 M
molal concentration of the glucose solution is 0.17 M
Displacement = √(3² + 4²)
Displacement = 5 meters north east
Velocity = displacement / time
Velocity = 5 / 35
Velocity = 0.14 m/s northeast
Answer:
P1 = 2.5ATM
Explanation:
V1 = 28L
T1 = 45°C = (45 + 273.15)K = 318.15K
V2 = 34L
T2 = 35°C = (35 + 273.15)K = 308.15K
P1 = ?
P2 = 2ATM
applying combined gas equation,
P1V1 / T1 = P2V2 / T2
P1*V1*T2 = P2*V2*T1
Solving for P1
P1 = P2*V2*T1 / V1*T2
P1 = (2.0 * 34 * 318.15) / (28 * 308.15)
P1 = 21634.2 / 8628.2
P1 = 2.5ATM
The initial pressure was 2.5ATM