It would be: 40 + 272 = 313 K
In short, Your Answer would be Option A
Hope this helps!
Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As
For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²
Answer:
1.3 x 10^(-2) atm/s
Explanation:
It follows the stoichiometry. For every mole of O3 that disappears, 1.5 moles (that is, 3/2) of O2 appears:
1.5 * 0.009 atm/s = 0.0135 atm/sec; the answer is 1.3 x 10^(-2) atm/s
I would definitely think its B....