A hypothetical covalent molecule, X–Y, has a dipole moment of 1.93 1.93 D and a bond length of 109 pm. 109 pm. Calculate the par
1 answer:
Answer:
q= 110.5 ke
Explanation:
Dipole moment is the product of the separation of the ends of a dipole and the magnitude of the charges.
μ = q * d
μ= Dipole moment (1.93 D)
q= partial charge on each pole
d= separation between the poles(109 pm).
e= electronic charge ( 1.60217662 × 10⁻¹⁹ coulombs)
So,
q= coulombs
q = e
q = 1.105 * 10⁵ e
q= 110.5 ke
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Answer:
0.55 atm
Explanation:
First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by
where
m = 1.00 g is the mass of the gas
Mm = 44.0 g/mol is the molar mass of the gas
Substituting,
Now we can find the pressure of the gas by using the ideal gas law:
where
p is the gas pressure
V = 1.00 L is the volume
n = 0.0227 mol is the number of moles
R = 0.082 L/(atm K mol) is the gas constant
T = 25.0 C + 273 = 298 K is the temperature of the gas
Solving the formula for p, we find
a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
<h3>Further explanation</h3>Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)
R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol
b. the effusion rates of two gases = the square root of the inverse of their molar masses:
M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395
the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide