<span>11.2G is the answer to this problem.
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The neutral grain spirit is essentially known as rectified spirits or ethyl alcohol.
<h3>Meaning of
neutral grain spirit</h3>
neutral grain spirit can be defined as raw material that is very concentrated with alcohol used to produce strong alcoholic products.
neutral grain spirit as the name implies is neutral in nature that is it can be blended with any other alcoholic product.
In conclusion, The neutral grain spirit is essentially known as rectified spirits or ethyl alcohol.
Learn more about ethyl alcohol: brainly.com/question/1049383
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An electron because that is the only part able to be lost or gained without nuclear action needed
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
PH = −log [H+] = − log [5.4 × 10−3] ≈ 2.27 or 2.3.
or basically 2