V
1
/T
1
=V
2
/T
2
(900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL.
Change the 900 to 800, and the 300 to 27, then change the 405 to 132. And solve
A redox reaction --> a reaction whereby oxidation & reduction occurs
Reduction:
Charge of Cl2 = 0
Charge of Cl- in NaCl = -1
Hence, since charge of Cl2 decreased from 0 in Cl2 to -1 in NaCl, reduction occured.
Oxidation:
Charge of Na = 0
Charge of Na+ in NaCl = +1
Hence, since charge of Na increased from 0 in Na to +1 in NaCl, oxidation occured.
Since both oxidation & reduction occured in the reaction, it is a redox reaction.
It is definitely not A. B is an effect. I would say C because D is more of a conservative answer , C is more of a liberal answer, and we currently live in a liberally swayed world. They are probably looking for C. It is not in your nature to be bad.
Chlorine (Cl) forms a salt when it is combined with a metal. This element belongs in <span>halogens.
</span>
Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed
