I would expect silane because all the rest have an overall dipole movement
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
![\frac{\Delta[NO_{2}]}{\Delta t} = rate](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20rate)
![\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%2A%5Cfrac%7B3600%20s%7D%7B1%20h%7D%20%3D%202.56%20mol.L%5E%7B-1%7D.h%7B-1%7D)
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!
Answer:
The answer is 0.188L
Explanation:
I did the problem on paper and put the answer on the pictures. I'm sorry if I didn't explain it well but I hope I helped.
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
Explanation:
Pressure of a gas is the combined force with which the molecules bombard a unit area of the wall of the container.
1 atm = 760mmHg
= 760torr
= 101325Pa
= 101325Nm⁻²
1mmHg = 1torr
1Pa = 1Nm⁻²
101.325 kPa and 101,325 Pa are the same
1000Pa = 1kPa
101,325 Pa and 1 atm
1atm and 101.325 kPa