Idk tbh srry man am in quarantine so I have a big packet to do
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
<em />
Answer:
The concentration of sucrose will decrease.
Explanation:
The balloon contains 50% sucrose solution.
Diffusion is the net movement of the substance from the region of the higher concentration to the region of the lower concentration.
Given that the membrane is only permeable to water not sucrose. So, movement of sucrose will not take place. <u>When the balloon is placed in a beaker of water, by the process of diffusion, the water moves from beaker to balloon and thus, the balloon will expand and the concentration of sucrose solution will decrease.</u>
53 if your talking about electrons. 74 if your talking about neutrons. energy levels=5
Hope this helps☺️