Answer:
Explanation:
The pressure is constant, so, to calculate the volume, we can use Charles' Law:
\dfrac{V_{1}}{T_{1}} = \dfrac{V_{2}}{T_{2}}
Data:
V₁ = ?; T₁ = 450 K
V₂ = 103.4 L; T₂ = 284.9 K
Calculation:
Answer : The volume in mL of the sodium carbonate stock solution is 364 mL.
Explanation :
According to dilution law:
where,
= molarity of aqueous sodium carbonate
= molarity of aqueous sodium carbonate stock solution
= volume of aqueous sodium carbonate
= volume of aqueous sodium carbonate stock solution
Given:
= 1.00 M
= 1.58 M
= 575 mL
= ?
Now put all the given values in the above formula, we get:
Therefore, the volume in mL of the sodium carbonate stock solution is 364 mL.
The rate of effusion of ammonia (NH₃) in the same apparatus is 63.3 cm/min
<h3>Graham's law of diffusion </h3>
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
<h3>How to determine the rate of ammonia (NH₃) </h3>
- Rate of HCl (R₁) = 43.2 cm/min
- Molar mass of HCl (M₁) = 1 + 35.5 = 36.5 g/mol
- Molar mass of NH₃ (M₂) = 14 + (3×1) = 17 g/mol
R₁/R₂ = √(M₂/M₁)
43.2 / R₂ = √(17 / 36.5)
Cross multiply
43.2 = R₂ × √(17 / 36.5)
Divide both side by √(17 / 36.5)
R₂ = 43.2 / √(17 / 36.5)
R₂ = 63.3 cm/min
Thus, the rate of effusion of ammonia is 63.3 cm/min
Learn more about Graham's law of diffusion:
brainly.com/question/14004529
