Answer:
D) Oxygen is oxidized and hydrogen is reduced.
Explanation:
In the electrolysis of water, an electric current passes through an electrolytic solution (e.g. aqueous NaCl), leading to the following redox reaction.
H₂O(l) → H₂(g) + 1/2 O₂(g)
The corresponding half-reactions are:
Reduction: 2 H₂O(l) + 2 e⁻ → H₂(g) + 2 OH⁻
Oxidation: 2 H₂O(l) → O₂(g) + 4 H⁺(aq) + 4 e⁻
As we can see, H in water is reduced (its oxidation number decreases from 1 to 0), while O in water is oxidized (its oxidation number increases from -2 to 0).
Answer:
Molarity = 0.08 M
Explanation:
Given data:
Mass of copper sulfate = 3.8 g
Volume of water = 250 cm³ (250/1000 = 0.25 L)
Concentration of solution = ?
Solution:
Number of moles of copper sulfate:
Number of moles = mass/molar mass
Number of moles = 3.8 g/ 159.6 g/mol
Number of moles = 0.02 mol
Concentration:
Molarity = Number of moles / volume in L
By putting values,
Molarity = 0.02 mol / 0.25 L
Molarity = 0.08 mol/L
Molarity = 0.08 M
Answer:
KCl ⇒ 1.205x10²⁴ molecules
O₂ ⇒ 1.807x10²⁴ molecules
KClO₃ ⇒ 1.205x10²⁴ molecules
Explanation:
In order to calculate the number of particles from the number of moles, we have to use Avogadro's number, which states <em>the number of particles in one mol</em>:
- In 1 mol there are 6.023x10²³ particles (ions, molecules or atoms).
So now we <u>multiply the number of moles of each substance by Avogadro's number</u>:
- KCl ⇒ 2 mol * 6.023x10²³ molecules/mol = 1.205x10²⁴ molecules
- O₂ ⇒ 3 mol * 6.023x10²³ molecules/mol = 1.807x10²⁴ molecules
- KClO₃ ⇒ 2 mol * 6.023x10²³ molecules/mol = 1.205x10²⁴ molecules
Answer:
The particles in the substance become less active
