A solution composed of 30.6g NH3 in 81.3g of H20. Calculate the mole fraction for NH3 and H20
2 answers:
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Answer:
The answer to your question is 32.44 moles
Explanation:
Data
moles of Na₂CO₃ = ?
volume = 9.54 l
concentration = 3.4 M
Formula
Molarity =
Solve for number of moles
number of moles = Molarity x volume
Substitution
Number of moles = (3.4)( 9.54)
Simplification
Number of moles = 32.44
Answer: 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal
Explanation:
To calculate the moles :
The balanced chemical equuation is:
According to stoichiometry :
4 moles of produce == 2 moles of
Thus 0.556 moles of will produce=
of
Mass of
Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.
Answer:
El potencial celular estándar, is +0.46 V
Explanation:
Las reacciones de media célula son;
Media reacción del ánodo Cu²⁺ + 2e⁻ ↔ Cu, E ° = 0.34 V
Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V
Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V
y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V
que es más alta que la del cobre presente, por lo tanto, el cobre se oxidará en el ánodo
Por lo tanto, en el ánodo, tendremos
Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)
En el cátodo
2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)
El potencial celular estándar, = +0.46 V