A reaction is first order. If the initial reactant concentration is 0.0200 M, and 25.0 days later the concentration is 6.25 x 10-4 M, then its half-life is:
Answer:
a. 4900 mL; b. 4900 cm³ c. 4.9x10⁻³ m³
Explanation:
1 L = 1 dm³
4.9 L = 4.9 dm³
1 dm³ = 1000 cm³ → 4900 cm³
1 cm³ = 1 mL (4900 mL)
1 dm³ = 1x10⁻³ m³ → 4.9x10⁻³ m³
Answer:
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Explanation:
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Answer:
A) 6.48 g of OF₂ at the anode.
Explanation:
The gas OF₂ can be obtained through the oxidation of F⁻ (inverse reaction of the reduction presented). The standard potential of the oxidation is the opposite of the standard potential of the reduction.
H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻ E° = -2.15 V
Oxidation takes place in the anode.
We can establish the following relations:
- 1 Faraday is the charge corresponding to 1 mole of e⁻.
- 1 mole of OF₂ is produced when 4 moles of e⁻ circulate.
- The molar mass of OF₂ is 54.0 g/mol.
The mass of OF₂ produced when 0.480 F pass through an aqueous KF solution is:
