Answer:
How does the equilibrium change with the removal of hydrogen (H2) gas from this equation? 2H2S ⇌ 2H2(g) + S2(g) A. ... Equilibrium shifts left to produce less reactant.
Explanation:
option A is the correct answer
Equilibrium shifts right to produce more product.
I hope it will help you.
They are all transioning in states of matter
Answer:
Explanation:
Hello,
In this case, since the rate equation turns out as shown below due to the first-order kinetics:
Its integration results:
However, the rate constant is computed by considering the given half-life time as follows:
In such a way, the required time in minutes to diminish the concentration by 66.8% of the initial turns out:
Best regards.
The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
mass of CO₂ formed is 591.8 g
Answer:
He counted the first hour (The one where he was sitting at home) in his speed calculations.