If the rate of appearance of o2 in the reaction: 2o3((g) is 0.250 m/s over the first 5.50 s, how much oxygen will form during this time?

Given:

Rate of appearance of O2 = 0.250 M/s

Time period = 5.50 s

To determine:

The concentration of O2 formed

Explanation:

2O3 (g) ↔ 3O2 (g)

Rate of appearance of O2 = 1/3 * Δ[O2]/Δt

Based on the given data:

0.250 M/s = 1/3 * [O2]/5.50 s

[O2 ] = 0.250 Ms⁻¹ * 3 * 5.50 s = 4.125 M

Ans: Amount of oxygen formed is 4.13 M

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6