Answer:
15.4 g of sucrose
Explanation:
Formula to be applied for solving these question: colligative property of freezing point depression. → ΔT = Kf . m
ΔT = Freezing T° of pure solvent - Freezing T° of solution
Let's replace data given: 0°C - (-0.56°C) = 1.86 C/m . m
0.56°C / 1.86 m/°C = m → 0.301 mol/kg
m → molality (moles of solute in 1kg of solvent)
Our mass of solvent is not 1kg, it is 150 g. Let's convert it from g to kg, to determine the moles of solute: 150 g. 1kg/1000g = 0.150 kg
0.301 mol/kg . 0.150kg = 0.045 moles.
We determine the mass of sucrose, by the molar mass:
0.045 mol . 342 g/1mol = 15.4 g
Answer: The pH of an aqueous solution of .25M acetic acid is 2.7
Explanation:
cM 0 0
So dissociation constant will be:
Give c= 0.25 M and 
= ?
Putting in the values we get:
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=0.25\times 0.0084=0.0021](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.25%5Ctimes%200.0084%3D0.0021)
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log[0.0021]=2.7](https://tex.z-dn.net/?f=pH%3D-log%5B0.0021%5D%3D2.7)
Thus pH is 2.7
Answer:
Kinetic
Explanation:
Kinetic energy kicks in after potienal builds up energy
