<h3><u>Answer</u>;</h3>
1.0875 x 10-2 atm
<h3><u>Explanation;</u></h3>
2O3(g) → 3O2(g)
rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t
The average rate of disappearance of ozone ... is found to
be 7.25 × 10–3 atm over a certain interval of time.
This means (ignoring time)
∆[O3]/∆t = -7.25 × 10^–3 atm
(it is disappearing, thus the negative sign)
rate = -(1/2)∆[O3]/∆t
rate = -(1/2)*(-7.25 × 10^–3 atm)
= 3.625 × 10^–3 atm
Now use the other part of the expression:
rate = +(1/3)∆[O2)∆t
3.625 × 10–3 atm = +(1/3)∆[O2)/t
∆[O2)/∆t = (3)*(3.625× 10^–3 atm)
= 1.0875 x 10-2 atm over the same time interval
What are you asking on this question?
Answer: The reactivity of acetic acid with various chemicals (A)
Explanation: engenuity 2021
Answer:
81.5 L
Explanation:
We can use the combined gas law equation that gives the relationship among pressure, temperature and volume of gases for a fixed amount of gas.
P1V1 / T1 = P2V2 / T2
where P1 - pressure, V1 - volume and T1 - temperature at the first instance
P2 - pressure, V2 - volume and T2 - temperature at the second instance
substituting the values in the equation
1240 Torr x 47.2 L / 298 K = 730 Torr x V2 / 303 K
V2 = 81.5 L
the new volume the gas would occupy when the conditions have changed is 81.5 L
