<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
Answer:
D. All of the above
Explanation:
Because all of these have something to do with how glaciers are affecting the planet.
Vegetable soup is a mixture.
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
1 gramo de metano aporta 50.125 kilojoules.
Octano
1 gramo de metano aporta 48.246 kilojoules.
Answer:
D) 5.15
Explanation:
Step 1: Write the equation for the dissociation of HCN
HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)
Step 2: Calculate [H⁺] at equilibrium
The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.
α% = [H⁺]eq / [HCN]₀ × 100%
[H⁺]eq = α%/100% × [HCN]₀
[H⁺]eq = 0.0070%/100% × 0.10 M
[H⁺]eq = 7.0 × 10⁻⁶ M
Step 3: Calculate the pH
pH = -log [H⁺] = -log 7.0 × 10⁻⁶ = 5.15
