Create a path of mechanisms for the reaction.
2 answers:
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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
The number of atoms present in 0.231 g of sodium is 6.02 x 10²¹ atoms.
The given parameters;
- <em>mass of the reacting sodium = 0.231 g</em>
- <em>1 mole of an atom = 6.02 x 10²³ atoms</em>
- <em>the atomic mass of sodium is 23 g/mol</em>
The number of moles of 0.231 g of sodium available is calculated as;
The number of atoms of 0.01 mole of sodium available is calculated as;
1 mole ---------- 6.02 x 10²³ atoms
0.01 ----------------- ?
= 0.01 x 6.02 x 10²³ atoms
= 6.02 x 10²¹ atoms.
Thus, the number of atoms present in 0.231 g of sodium is 6.02 x 10²¹ atoms.
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