Answer:
41.3 s
Explanation:
Let t₁ represent the time taken for SO₂ to effuse.
Let t₂ represent the time taken for Ar to effuse.
Let M₁ represent the molar mass of SO₂
Let M₂ represent the molar mass of Ar
From the question given above,
Time taken (t₁) for SO₂ = 52.3 s
Time taken (t₂) for Ar =?
Molar mass (M₁) of SO₂ = 32 + (16×2) = 32 + 32 = 64 g/mol
Molar mass (M₂) of Ar = 40 g/mol
Finally, we shall determine the time taken for Ar to effuse by using the Graham's law equation as shown below:
t₂ / t₁ = √(M₂ / M₁)
t₂ / 52.3 = √(40 / 64)
t₂ / 52.3 = √0.625
t₂ / 52.3 = 0.79
Cross multiply
t₂ = 52.3 × 0.79
t₂ = 41.3 s
Thus, the time taken for the amount of Ar to effuse is 41.3 s
Explanation:
the coefficient of hydrogen is 3
<span>
some elements have their outer electrons more tightly bound than
others. Those who have less tightly bound electrons are more reactive.
After this it gets more complex in explaining why they are bound with
different strengths. </span>
The chemical equation given is:
<span>2x(g) ⇄ y(g)+z(s)</span>
Answer: the higher the amount of x(g) the more the forward reacton will occur and the higher the amounts of products y(g) and z(s) will be obtained at equilibrium.
Justification:
As Le Chatellier's priciple states, any change in a system in equilibrium will be compensated to restablish the equilibrium.
The higher the amount, and so the concentration, of X(g), the more the forward reaction will proceed to deal witht he high concentration of X(g), leading to an increase on the concentration of the products y(g) and z (s).
