Answer:
1. 2 M
2. 2 M
Explanation:
1. Determination of the final concentration.
Initial Volume (V₁) = 2 L
Initial concentration (C₁) = 6 M
Final volume (V₂) = 6 L
Final concentration (C₂) =?
The final concentration can be obtained as follow:
C₁V₁ = C₂V₂
6 × 2 = C₂ × 6
12 = C₂ × 6
Divide both side by 6
C₂ = 12 / 6
C₂ = 2 M
Therefore, the final concentration of the solution is 2 M
2. Determination of the final concentration.
Initial Volume (V₁) = 0.5 L
Initial concentration (C₁) = 12 M
Final volume (V₂) = 3 L
Final concentration (C₂) =?
The final concentration can be obtained as follow:
C₁V₁ = C₂V₂
12 × 0.5 = C₂ × 3
6 = C₂ × 3
Divide both side by 3
C₂ = 6 / 3
C₂ = 2 M
Therefore, the final concentration of the solution is 2 M
This problem could be solved through the Graham’s law of
effusion (also known as law of diffusion). This law states that the ratio of
the effusion rate of the first gas and effusion rate of the second gas is
equivalent to the square root of the ratio of its molar mass. Thus the answer
would be 0.1098.
Answer:
278 balloons can be aired.
Explanation:
We apply the Ideal Gases Law to determine the total available volume for each baloon of 0.75L. So we need to divide the volume we obtain by the volume of each baloon.
We determine the moles of CO₂
375 g. 1 mol / 44g = 8.52 moles
V = (n . R . T) / P → (8.52 mol . 0.082 . 298K) / 1 atm = 208.3 L
That's the total volume from the tank, so we can inflate (208.3 / 0.75) = at least 278 balloons
The answer is 2.18 moles MgSi03.