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<u>Answer:</u> The equilibrium concentration of HCl is 
<u>Explanation:</u>
We are given:
Moles of 
= 0.564 moles
Volume of vessel = 1.00 L
Molarity is calculated by using the equation:
Molarity of 
The given chemical equation follows:
<u>Initial:</u> 0.564
<u>At eqllm:</u> 0.564-x x x
The expression of 
for above equation follows:
![K_c=[NH_3][HCl]](https://tex.z-dn.net/?f=K_c%3D%5BNH_3%5D%5BHCl%5D)
The concentration of pure solid and pure liquid is taken as 1.
We are given:
Putting values in above equation, we get:
Negative sign is neglected because concentration cannot be negative.
So, ![[HCl]=2.26\times 10^{-3}M](https://tex.z-dn.net/?f=%5BHCl%5D%3D2.26%5Ctimes%2010%5E%7B-3%7DM)
Hence, the equilibrium concentration of HCl is
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:
<em>2</em><em>.</em><em>6</em><em>0</em><em> </em><em>mol</em><em>2</em>
Explanation:
2.60mol2 is ur answer
Atomic radius aka distance from the nucleus to the outermost energy level. The greater this distance, the less electrostatic attraction between these oppositely charged particles.
