Question

If the rate of disappearance of dinitrogen pentoxide is 0.0300 M/s, what is the rate of appearance of nitrogen dioxide? Use your equations from part a

Answer 1

equations from part a?

Answer 2

C3H8 + 10 S Æ 3 CS2 + 4 H2S

Answer 3

[AI-generated answer — Explain Bot]

In order to determine the rate of appearance of nitrogen dioxide (NO₂), we need to consider the balanced chemical equation and the stoichiometry between dinitrogen pentoxide (N₂O₅) and nitrogen dioxide.

The balanced chemical equation is:
N₂O₅ → 2NO₂

From this equation, we can see that for every 1 mole of N₂O₅ that disappears, 2 moles of NO₂ are formed.

If the rate of disappearance of N₂O₅ is given as 0.0300 M/s (Molarity per second), we can use this information to find the rate of appearance of NO₂.

To do this, we need to express the rate of appearance of NO₂ in the same units (M/s). We can use the stoichiometry of the reaction to convert between the two rates:

Rate of disappearance of N₂O₅ = 0.0300 M/s
Rate of appearance of NO₂ = 2 * (Rate of disappearance of N₂O₅)

Substituting the given value into the equation, we have:
Rate of appearance of NO₂ = 2 * 0.0300 M/s
Rate of appearance of NO₂ = 0.0600 M/s

Therefore, the rate of appearance of nitrogen dioxide is 0.0600 M/s.