Answer:
250J
Explanation:
Given parameters:
Spring constant = 500N/m
Extension = 1m
Unknown:
Elastic potential energy = ?
Solution:
The elastic potential energy of a body is the energy stored within an elastic string.
EPE = 
k e²
k is the spring constant
e is the extension
Now;
EPE = x 500 x 1² = 250J
Answer:
0.204.
Explanation:
- For the following reaction:
<em>CO(g) + Cl₂(g) ⇌ COCl₂(g), </em>
Kp = (P of COCl₂)/(P of CO)(P of Cl₂)
P of COCl₂ = 0.22 atm, P of CO = 0.83 atm, P of Cl₂ = 1.3 atm.
∴ Kp = (P of COCl₂)/(P of CO)(P of Cl₂) = (0.22 atm)/(0.83 atm)(1.3 atm) = 0.204.
(K+ + MnO4- )+(K+ +Cl- )
This would be a one to one ratio as the charges of each component are equal
Answer:
Explanation:
Given:
V1 = 200 ml
T1 = 20 °C
= 20 + 273
= 293 K
P1 = 3 atm
P2 = 2 atm
V2 = 400 ml
Using ideal gas equation,
P1 × V1/T1 = P2 × V2/T2
T2 = (2 × 400 × 293)/200 × 3
= 234400/600
= 390.67 K
= 390.67 - 273
= 117.67 °C
