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Explanation:
2 Li(s) +Cl₂→ 2 Li⁺ (aq) + 2Cl⁻ (aq)
The cell potential of the reaction above is +4.40V
<em><u>calculation</u></em>
Cell potential =∈° red - ∈° oxidation
in reaction above Li is oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04
Cl is reduce from oxidation state 0 to -1 therefore the ∈°red = +1.36 V
cell potential is therefore = +1.36 v -- 3.04 = + 4.40 V
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Answer:
384.2 K
Explanation:
First we convert 27 °C to K:
- 27 °C + 273.16 = 300.16 K
With the absolute temperature we can use <em>Charles' law </em>to solve this problem. This law states that at constant pressure:
Where in this case:
We input the data:
300.16 K * 1600 m³ = T₂ * 1250 m³
And solve for T₂:
T₂ = 384.2 K
