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
I think this question is talking about electrons (since they flow freely in the electron cloud of an atom).
Answer:
Kc = 4.774 * 10¹³
Explanation:
the desired reaction is
2 NO₂(g) ⇋ N₂(g) + 2 O₂(g)
Kc =[N₂]*[O₂]² /[NO₂]²
Since
1/2 N₂(g) + 1/2 O₂(g) ⇋ NO(g)
Kc₁= [NO]/(√[N₂]√[O₂]) → Kc₁²= [NO]²/([N₂][O₂])
and
2 NO₂(g) ⇋ 2 NO(g) + O₂(g)
Kc₂= [NO]²*[O₂]/[NO₂]² → 1/Kc₂= [NO₂]²/([NO]²[O₂])
then
Kc₁²* (1/Kc₂) = [NO]²/([N₂]*[O₂]) *[NO₂]²/([NO]²[O₂]) = [NO₂]²/([N₂]*[O₂]²) = 1/Kc
Kc₁² /Kc₂ = 1/Kc
Kc= Kc₂/Kc₁² =1.1*10⁻⁵/(4.8*10⁻¹⁰)² = 4.774 * 10¹³
Answer:
2C₈H₁₈(g) + 25O₂(g)→16CO₂(g) + 18H₂O
Explanation:
To balance an equation, the moles of one element on one side of the equation should be the same as those on the other side of the equation. This is because (as a law of thermodynamics), in a chemical reaction, the matter is not destroyed nor created - atoms are only rearranged.
D. divide 4.56 by molar mass of mg(OH)2
