Answer:
–2.23 L
Explanation:
We'll begin by calculating the final volume. This can be obtained as follow:
Initial pressure (P₁) = 1.03 atm
Initial volume (V₁) = 3.62 L
Final pressure (P₂) = 2.68 atm
Final volume (V₂) =?
P₁V₁ = P₂V₂
1.03 × 3.62 = 2.68 × V₂
3.7286 = 2.68 × V₂
Divide both side by 2.68
V₂ = 3.7286 / 2.68
V₂ = 1.39 L
Finally, we shall determine the change in volume. This can be obtained as follow:
Initial volume (V₁) = 3.62 L
Final volume (V₂) = 1.39 L
Change in volume (ΔV) =?
ΔV = V₂ – V₁
ΔV = 1.39 – 3.62
ΔV = –2.23 L
Thus, the change in the volume of her lung is –2.23 L.
NOTE: The negative sign indicate that the volume of her lung reduced as she goes below the surface!
Answer:In determining the energy of activation, why was it prudent to run the slowest trial done at room temperature in the hot water bath and the fastest trial done at room temperature in the cold water bath?
Explanation:
Answer:
C.0.28 V
Explanation:
Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:
The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:
Cu²⁺ + 2e⁻ → Cu E° = 0.52V
Ag⁺ + 1e⁻ → Ag E° = 0.80V
As the Cu will be oxidized:
Cu → Cu²⁺ + 2e⁻
The cell potential is:
E°Cell = E°cathode(reduced) - E°cathode(oxidized)
E°cell = 0.80V - (0.52V)
E°cell = 1.32V
Right answer is:
<h3>C.0.28 V
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