Answer:
63.5 w isvthebanswerok is th answer
Radius of Xenon = 1.3Ă—10â’8 cm
Volume = 100 ml = 0.1 L
Pressure P = 1.2 atm = 121.59 Kpa
Temperature = 281 K
R = Gas Constant = 8.31 J mol^-1 K^-1
Now find the number of atoms
PV = nRT => n = PV / RT
n = (121.59 x 0.1) / (8.31 x 281) = / 2335.11 = 0.0052
Number of atoms in a mole is same as Avogadro constant A, which is 6.02 x
10^23 particles.
n = number of atoms= 0.0052
N = number of particles
Avogadro constant A = 6.02 x 10^23
n = N/A => N = n x A = 0.0052 x 6.02 x 106^23 = 3.13 x 10^20
Volume of Xe atom which would be a sphere = (4/3) x pi x r^3
Volume = = (4/3) x 3.14 x (1.3Ă—10â’8)^3 = 9.2 x 10^-24
Volume occupied by these particles = n x Volume = 3.13 x 10^20 x 9.2 x
10^-24 = 0.00288
Fraction of volume will be = 0.00288 / 0.1 = 0.0288
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Answer:
45.4 L
Explanation:
Using Ideal gas equation for same mole of gas as
Given ,
V₁ = 27.9 L
V₂ = ?
P₁ = 732 mmHg
P₂ = 385 mmHg
T₁ = 30.1 ºC
T₂ = -13.6 ºC
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (30.1 + 273.15) K = 303.25 K
T₂ = (-13.6 + 273.15) K = 259.55 K
Using above equation as:


Solving for V₂ , we get:
<u>V₂ = 45.4 L</u>
Answer:
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