H2(g) + F2(g) → 2 HF(g) ΔH=-546 kJ

2 answers:

5 0

Answer: The answer is D, More energy was released in the formation of the 2 HF bonds than in the breaking of the bonds of H2 and F2.

Explanation: Honestly, I guessed on this question and got it right, but I see that right after the arrow on the right side of the equation, there's a 2 in front of the HF(g). On the left side of the equation, it does not match the right side, which led me to believe that there was more energy being released.

Arisa [49]3 years ago

4 0

Answer: help please !!

Explanation:

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The standard emf for the cell using the overall cell reaction below is +2.20 V:

ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: $2Al-6e^{-}\rightarrow 2Al^{3+}$

Reduction: $3I_{2}+6e^{-}\rightarrow 6I^{-}$

---------------------------------------------------------------------------------

Overall: $2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}$

Nernst equation for this cell reaction at $25^{0}\textrm{C}$ -

$E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}$

where n is number of electrons exchanged during cell reaction, $E_{cell}^{0}$ is standard cell emf , $E_{cell}$ is cell emf , $[Al^{3+}]$ is concentration of $Al^{3+}$ and $[Cl^{-}]$ is concentration of $Cl^{-}$