Answer: Freezing point of a solution will be 
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
= freezing point constant = 
m= molality
Weight of solvent (benzene)= 1480 g =1.48 kg
Molar mass of solute (octane) = 114.0 g/mol
Mass of solute (octane) = 220 g
Thus the freezing point of a solution will be
Answer:
Acceleration:
Speed/Time
Change in speed or velocity over a specific amount of time
Speed:
Distance/Time
Change in distance over a specific amount of time
Velocity:
Distance/Time
Speed in a given direction
Answer:
57.6g
Explanation:
So, if in one mole of water, 16 g of oxygen atom is present. Then, in 3.6 moles of water, the mass of oxygen present will be 3.6×16=57.6g. Therefore, the amount of oxygen present in 3.6 g water is option (B)- 57.6 g.
The physical properties of alkenes and alkynes are generally similar to those of alkanes or cycloalkanes with equal numbers of carbon atoms. Alkynes have higher boiling points than alkanes or alkenes, because the electric field of an alkyne, with its increased number of weakly held π electrons, is more easily distorted, producing stronger attractive forces between molecules.
Answer:
2.19 x 10^-12.
Explanation:-
The relation between Ka and Kb for an acid and it's conjugate base is
Ka x Kb = Kw where Kw = ionic product of water.
So Kb = 10^-14 / (4.57 x 10 ^ -3)
= 2.19 x 10^-12
