Answer:
During the process of reaching thermal equilibrium heat is transferred between the object. heat is always transferred from the object at the higher temperature to the object with lower temperature. For a gas, the heat transfer is related to a change in temperature.
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
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Answer : The value of 
of the weak acid is, 4.72
Explanation :
First we have to calculate the moles of KOH.
Now we have to calculate the value of of the weak acid.
The equilibrium chemical reaction is:
Initial moles 0.25 0.03 0
At eqm. (0.25-0.03) 0.03 0.03
= 0.22
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[HK]}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BHK%5D%7D%7B%5BHA%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the value of of the weak acid is, 4.72
Hydrogen bonds to either Nitrogen, Oxygen, or Fluorine to experience Hydrogen bonding.
