We know, by law of constant proportion :
Chemical compounds always contains its component in constant fixed ratio or in fixed percentage without depend on its source, method of preparation and mass of compound.
Therefore, percentage of oxygen is same for any mass.
Hence, this is the required solution.
Answer: 5 electrons I think
Explanation:
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:
Answer: 2.25
Explanation: 1:1 mole ratio seen in equation.
A molecule of sucrose (C12H22O11) has 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms. The subscripts also indicate the ratios of the elements.