Answer:
108 kPa
Step-by-step explanation:
To solve this problem, we can use the <em>Combined Gas Laws</em>:
p₁V₁/T₁ = p₂V₂/T₂ Multiply each side by T₁
p₁V₁ = p₂V₂ × T₁/T₂ Divide each side by V₁
p₁ = p₂ × V₂/V₁ × T₁/T₂
Data:
p₁ = ?; V₁ = 34.3 L; T₁ = 31.5 °C
p₂ = 122.2 kPa; V₂ = 29.2 L; T₂ = 21.0 °C
Calculations:
(a) Convert temperatures to <em>kelvins
</em>
T₁ = (31.5 + 273.15) K = 304.65 K
T₂ = (21.0 + 273.15) K = 294.15 K
(b) Calculate the <em>pressure
</em>
p₁ = 122.2 kPa × (29.2/34.3) × (304.65/294.15)
= 122.2 kPa × 0.8542 × 1.0357
= 108 kPa
Answer:
2nd one down
Explanation: distance divided by time interval
I would say 2 because co2 goes out and o goes in
Answer:
The rate of reaction of a zero-order reaction is 0.0020 mol/L.
Explanation:
The rate expression of the zero order kinetic are :
![R=k[A]^o](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5Eo)
[A]= initial concentration of reactant
k = rate constant
R = rate of reaction
We have :
Rate constant of the reaction , k = 0.0020 mol/L s

R = 0.0020 mol/L s
The rate of reaction of a zero-order reaction is 0.0020 mol/L.