<u>Given:</u>
Rate of appearance of O2 = 0.250 M/s
Time period = 5.50 s
<u>To determine:</u>
The concentration of O2 formed
<u>Explanation:</u>
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
Answer:
0.9 moles of Hg contain 5.418×10²³ atoms
Explanation:
As we know, Avogadro's number (NA) is the amount of particles that are contained in 1 mol → 6.02×10²³
Let's make a rule of three for this:
6.02×10²³ atoms are contained in 1 mol of Hg
5.418×10²³ atoms would be contained in (5.418×10²³ . 1) / NA = 0.9 moles
True, if you have rough surface the more friction you have. The More slipery, less friction.
Answer:
I think the answer of the question is C?
