2 LiOH(s)→ Li 2 O(s)+ H 2 O(l)Δ r H° = 379.1 KJ/mol

The Answer of 2 LiOH(s)→ Li 2 O(s)+ H 2 O(l)Δ r H° = 379.1 KJ/mol

2 LiOH(s)→ Li 2 O(s)+ H 2 O(l)Δ r H° = 379.1 KJ/mol

The Answer of 2 LiOH(s)→ Li 2 O(s)+ H 2 O(l)Δ r H° = 379.1 KJ/mol

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2 LiOH(s)→ Li₂O(s)+ H₂O(l)Δ_rH° = 379.1 KJ/mol

Question 56

Multiple Choice

2 LiOH(s) → Li₂O(s) + H₂O(l) Δ_rH° = 379.1 kJ/mol
LiH(s) + H₂O(l) → LiOH(s) + H₂(g) Δ_rH° = -111.0 kJ/mol
2 H₂(g) + O₂(g) → 2 H₂O(l) Δ_rH° = -285.9 kJmol
Compute Δ_rH° in kJ/mol for 2 LiH(s) + O₂(g) → Li₂O(s) + H₂O(l)

A) +125.2 kJ/mol
B) -17.7 kJ/mol
C) -128.8 kJ/mol
D) -303.6 kJ/mol
E) + 128.8 kJ/mol

2 LiOH(s)→ Li2O(s)+ H2O(l)ΔrH° = 379.1 KJ/mol

Multiple Choice

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2 LiOH(s)→ Li₂O(s)+ H₂O(l)Δ_rH° = 379.1 KJ/mol

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